{"id":3408,"date":"2017-11-05T23:50:06","date_gmt":"2017-11-05T23:50:06","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3408"},"modified":"2024-09-04T14:23:28","modified_gmt":"2024-09-04T14:23:28","slug":"resolucao-comentada-fisica-termica-2017-2016-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/fisica-termica\/termologia-2017-2016-2015\/resolucao-comentada-fisica-termica-2017-2016-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada \u2013 F\u00edsica T\u00e9rmica \u2013 2017 \u2013 2016 \u2013 2015"},"content":{"rendered":"

Resolu\u00e7\u00e3o Comentada \u2013 Vestibulares Recentes <\/span><\/span><\/span><\/p>\n

F\u00edsica T\u00e9rmica \u2013 2017 \u2013 2016 \u2013 2015<\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- D<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

03-<\/b><\/span><\/span><\/span> <\/b>(V) Temperatura\u00a0\"\"\u00a0grandeza f\u00edsica\u00a0<\/b><\/span><\/span>que representa a\u00a0<\/b><\/span><\/span>medida do estado de agita\u00e7\u00e3o m\u00e9dio das mol\u00e9culas de um corpo.<\/b><\/span><\/span><\/span><\/span><\/p>\n

(F) Calor\u00a0\"\"\u00a0energia t\u00e9rmica\u00a0<\/b><\/span><\/span>que passa, de\u00a0<\/b><\/span><\/span>forma espont\u00e2nea<\/b><\/span><\/span>, do corpo de<\/b><\/span><\/span>maior temperatura para o de menor temperatura.<\/b><\/span><\/span><\/span><\/span><\/p>\n

(F) Fus\u00e3o<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>mudan\u00e7a de estado f\u00edsico\u00a0<\/b><\/span><\/span>sofrida por um\u00a0<\/b><\/span><\/span>s\u00f3lido ao se transformar em l\u00edquido\u00a0<\/b><\/span><\/span>quando\u00a0<\/b><\/span><\/span>recebe uma certa quantidade de calor.<\/b><\/span><\/span><\/span><\/span><\/p>\n

(V ) Evapora\u00e7\u00e3o\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>passagem do estado l\u00edquido<\/b><\/span><\/span>\u00a0para o\u00a0<\/b><\/span><\/span>estado gasoso\u00a0<\/b><\/span><\/span>que <\/b><\/span><\/span>ocorre de forma lenta.<\/b><\/span><\/span><\/span><\/span><\/p>\n

(F) Equil\u00edbrio t\u00e9rmico\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>condi\u00e7\u00e3o f\u00edsica<\/b><\/span><\/span>\u00a0na qual\u00a0<\/b><\/span><\/span>dois ou mais corpos\u00a0<\/b><\/span><\/span>ficam com a\u00a0<\/b><\/span><\/span>mesma temperatura (<\/b><\/span><\/span>temperatura de equil\u00edbrio t\u00e9rmico).\u00a0As\u00a0<\/b><\/span><\/span>trocas de calor podem n\u00e3o cessar<\/b><\/span><\/span>, e os corpos\u00a0<\/b><\/span><\/span>podem<\/b><\/span><\/span>\u00a0continuar recebendo e cedendo\u00a0<\/b><\/span><\/span>calor na mesma propor\u00e7\u00e3o<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

(V ) Convec\u00e7\u00e3o\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>processo de transmiss\u00e3o de calor\u00a0<\/b><\/span><\/span>que ocorre devido \u00e0 <\/b><\/span><\/span>movimenta\u00e7\u00e3o de massas,\u00a0<\/b><\/span><\/span>em especial, nos\u00a0<\/b><\/span><\/span>l\u00edquidos e nos gases\u00a0<\/b><\/span><\/span>e essa <\/b><\/span><\/span>movimenta\u00e7\u00e3o\u00a0<\/b><\/span><\/span>( massas mais quentes, menos densas sobem e as frias mais densas descem).\u00a0<\/b><\/span><\/span>Essa movimenta\u00e7\u00e3o ocorre at\u00e9 que a massa total fique com a mesma temperatura.<\/b><\/span><\/span><\/span><\/span><\/p>\n

(F) Caloria\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>quantidade de calor necess\u00e1ria\u00a0<\/b><\/span><\/span>para que\u00a0<\/b><\/span><\/span>1g<\/b><\/span><\/span>\u00a0de\u00a0<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span>\u00a0tenha sua\u00a0<\/b><\/span><\/span>temperatura alterada em 1\u00b0C .<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>I.<\/b><\/span><\/span><\/span>\u00a0As\u00a0<\/b><\/span><\/span><\/span>prateleiras<\/b><\/span><\/span><\/span>\u00a0de uma\u00a0<\/b><\/span><\/span><\/span>geladeira dom\u00e9stica<\/b><\/span><\/span><\/span>\u00a0s\u00e3o\u00a0<\/b><\/span><\/span><\/span>grades vazadas<\/b><\/span><\/span><\/span>, para\u00a0<\/b><\/span><\/span><\/span>facilitar fluxo de energia t\u00e9rmica at\u00e9 o congelador por [\u2026]<\/b><\/span><\/span><\/span><\/p>\n

Situa\u00e7\u00e3o I\u00a0\"\"\u00a0Convec\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

II.<\/b><\/span><\/span><\/span>\u00a0O\u00a0<\/b><\/span><\/span><\/span>\u00fanico processo de troca de calor que pode ocorrer no v\u00e1cuo \u00e9 por [\u2026].<\/b><\/span><\/span><\/span><\/p>\n

Situa\u00e7\u00e3o II\u00a0\"\"\u00a0Radia\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

III.\u00a0<\/b><\/span><\/span><\/span>Em uma\u00a0<\/b><\/span><\/span><\/span>garrafa t\u00e9rmica<\/b><\/span><\/span><\/span>, \u00e9 mantido\u00a0<\/b><\/span><\/span><\/span>v\u00e1cuo entre as paredes duplas de vidro para evitar que o calor saia ou entre por [\u2026.].<\/b><\/span><\/span><\/span><\/p>\n

Situa\u00e7\u00e3o III\u00a0\"\"\u00a0Condu\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

O calor \u00e9 conduzido de um ponto a outro do corpo sem que haja deslocamento das part\u00edculas.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Explicando microscopicamente o fen\u00f4meno:<\/b><\/span><\/span><\/span>\u00a0a regi\u00e3o pr\u00f3xima da chama tem o movimento vibrat\u00f3rio de suas mol\u00e9culas aumentado, adquirindo assim maior energia cin\u00e9tica, que \u00e9 transferida atrav\u00e9s de choques \u00e0s part\u00edculas vizinhas, que tamb\u00e9m aumentam seu movimento vibrat\u00f3rio.<\/b><\/span><\/span><\/span><\/p>\n

Atrav\u00e9s desse transporte de energia<\/b><\/span><\/span><\/span>, toda a barra \u00e9 aquecida.<\/b><\/span><\/span><\/span><\/p>\n

A condu\u00e7\u00e3o\u00a0n\u00e3o ocorre no v\u00e1cuo,\u00a0<\/b><\/span><\/span><\/span>pois<\/b><\/span><\/span><\/span>\u00a0ela\u00a0precisa de um meio material para se propagar.<\/b><\/span><\/span><\/span><\/p>\n

Na\u00a0garrafa t\u00e9rmica,<\/b><\/span><\/span><\/span>\u00a0\u00e9\u00a0mantido o v\u00e1cuo entre as paredes de vidro para evitar trocas de calor por\u00a0<\/b><\/span><\/span><\/span>condu\u00e7\u00e3o e\u00a0convec\u00e7\u00e3o<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

Rela\u00e7\u00e3o\u00a0<\/b><\/span><\/span>entre as\u00a0<\/b><\/span><\/span>escalas Celsius e Fahrenheit\u00a0\"\"\u00a0\"\"\u00a0=\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0\"\"\u00a0=\"\"\u00a0\"\"\u00a0\"\"\u00a0=\u00a0\"\"\u00a0\"\"\u00a09C = 540\u00a0\"\"\u00a0<\/b><\/span><\/span>C = 60oC.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- C<\/b><\/span><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

07-<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como\u00a0<\/b><\/span><\/span>toda energia emitida pelo r\u00e1dio<\/b><\/span><\/span>\u00a0foi\u00a0<\/b><\/span><\/span>absorvida pela \u00e1gua<\/b><\/span><\/span>\u00a0e empregada exclusivamente para\u00a0<\/b><\/span><\/span>elevar sua temperatura sempre no estado l\u00edquido indo do ponto de congelamento (0\u00a0oC) at\u00e9 o ponto<\/b><\/span><\/span>\u00a0de<\/b><\/span><\/span>ebuli\u00e7\u00e3o (100\u00a0oC),<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>energia liberada<\/b><\/span><\/span>\u00a0pelo r\u00e1dio foi de\u00a0<\/b><\/span><\/span>Q = m.c.(\u03b8 \u2013 \u03b8<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>)<\/b><\/span><\/span>\u00a0= 1,3.1.(100 \u2013 0)\u00a0\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

Q = 130 cal.<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como o\u00a0<\/b><\/span><\/span>experimento durou \u2206t = 1 h = 3600 s,<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>pot\u00eancia m\u00e9dia liberada pelo r\u00e1dio<\/b><\/span><\/span>\u00a0\u00e9 fornecida por<\/b><\/span><\/span><\/span><\/span><\/p>\n

P =\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0=\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>= 0,14 W<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

09-<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

Calor Latente<\/b><\/span><\/span><\/span><\/span><\/p>\n

O\u00a0<\/b><\/span><\/span><\/span>calor recebido por um corpo pode fazer com que sua temperatura aumente\u00a0<\/b><\/span><\/span><\/span>ou que ele\u00a0<\/b><\/span><\/span><\/span>sofra uma<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>mudan\u00e7a de estado.\u00a0<\/b><\/span><\/span><\/span>No segundo caso voc\u00ea tem uma\u00a0<\/b><\/span><\/span><\/span>quantidade de calor latente<\/b><\/span><\/span><\/span>,\u00a0ou seja, o\u00a0<\/b><\/span><\/span><\/span>calor<\/b><\/span><\/span><\/span> <\/b>latente se refere \u00e0 quantidade de calor que um corpo recebe\u00a0apenas para mudar de estado, o que ocorre \u00e0 temperatura constante.<\/b><\/span><\/span><\/span><\/p>\n

Constata-se que a\u00a0<\/b><\/span><\/span><\/span>quantidade de calor latente (Q) cedida ou retirada de uma subst\u00e2ncia para que ela\u00a0<\/b><\/span><\/span><\/span>mude de estado o que sempre ocorre \u00e0 <\/b><\/span><\/span><\/span>temperatura constante\u00a0<\/b><\/span><\/span><\/span>\u00e9\u00a0<\/b><\/span><\/span><\/span>diretamente proporcional \u00e0 sua massa (m), e depende das caracter\u00edsticas da subst\u00e2ncia (L).\u00a0<\/b><\/span><\/span><\/span>Matematicamente:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

O\u00a0<\/b><\/span><\/span><\/span>calor de fus\u00e3o ou de vaporiza\u00e7\u00e3o L\u00a0\u00e9 caracter\u00edstica do material de que o objeto \u00e9 feito\u00a0independendo da massa ou do volume do objeto.\u00a0<\/b><\/span><\/span><\/span>Assim, se o <\/b><\/span><\/span><\/span>objeto possui calor de fus\u00e3o LF\u00a0= 50 J\/kg\u00a0<\/b><\/span><\/span><\/span>ele ter\u00e1\u00a0<\/b><\/span><\/span><\/span>sempre esse valor <\/b><\/span><\/span><\/span>independente de sua massa ou volume, pois o\u00a0<\/b><\/span><\/span><\/span>material \u00e9 o mesmo.<\/b><\/span><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

11-<\/b><\/span><\/span><\/span> <\/b>a)<\/b><\/span><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span><\/span>quantidade de calor,<\/b><\/span><\/span><\/span>\u00a0em\u00a0<\/b><\/span><\/span><\/span>cal,<\/b><\/span><\/span><\/span>\u00a0necess\u00e1ria para\u00a0<\/b><\/span><\/span><\/span>elevar a temperatura do copo com \u00e1gua l\u00edquida de<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>20 \u00baC para 50 \u00baC.<\/b><\/span><\/span><\/span><\/p>\n

Copo\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0para a\u00a0<\/b><\/span><\/span><\/span>temperatura<\/b><\/span><\/span><\/span>\u00a0do copo subir de\u00a0<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 20<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> C para t = 50\u00ba C, sendo o calor espec\u00edfico do copo c<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0,2 cal\/g<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C e m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0= 200 g\u00a0\"\"\u00a0Qc\u00a0= mc.c.(t \u2013 to) =<\/b><\/span><\/span><\/span>100.0,2.(50 \u2013 20<\/b><\/span><\/span><\/span>)\u00a0\"\"\u00a0Qc\u00a0= 600 cal.<\/b><\/span><\/span><\/span><\/p>\n

\u00c1gua\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>\u00a0para a\u00a0<\/b><\/span><\/span><\/span>temperatura<\/b><\/span><\/span><\/span>\u00a0da \u00e1gua subir de\u00a0<\/b><\/span><\/span><\/span>to\u00a0= 20<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C para t = 50\u00ba C, sendo o calor espec\u00edfico da c<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>\u00a0= 1 cal\/g<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C e m<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>\u00a0= 100 g\u00a0\"\"\u00a0Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>\u00a0= ma.c<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>.(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) =\u00a0<\/b><\/span><\/span><\/span>200.1.(50 \u2013 20<\/b><\/span><\/span><\/span>)\u00a0\"\"\u00a0Q<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0= 6000 cal.<\/b><\/span><\/span><\/span><\/p>\n

Quantidade de calor pedida<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0Qt\u00a0= 600 + 6000\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>Qt\u00a0= 6600 cal.<\/b><\/span><\/span><\/span><\/p>\n

b)<\/b><\/span><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span><\/span>massa de vapor de \u00e1gua<\/b><\/span><\/span><\/span>, em\u00a0<\/b><\/span><\/span><\/span>gramas,<\/b><\/span><\/span><\/span>\u00a0necess\u00e1ria para\u00a0<\/b><\/span><\/span><\/span>elevar a temperatura do copo com \u00e1gua<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>l\u00edquida at\u00e9 atingir o equil\u00edbrio t\u00e9rmico a 50 \u00baC.<\/b><\/span><\/span><\/span><\/p>\n

Pelo enunciado a\u00a0<\/b><\/span><\/span><\/span>temperatura inicial do vapor de \u00e1gua era t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 120\u00a0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C,<\/b><\/span><\/span><\/span>que deve se\u00a0<\/b><\/span><\/span><\/span>transformar<\/b><\/span><\/span><\/span>\u00a0em\u00a0<\/b><\/span><\/span><\/span>\u00e1gua l\u00edquida a 50\u00a0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C. Etapas:<\/b><\/span><\/span><\/span><\/p>\n

1<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>\u00a0\"\"\u00a0Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>quantidade de calor cedida<\/b><\/span><\/span><\/span>\u00a0pelo\u00a0<\/b><\/span><\/span><\/span>vapor de \u00e1gua<\/b><\/span><\/span><\/span>\u00a0para passar de\u00a0<\/b><\/span><\/span><\/span>to\u00a0= 120\u00ba C para t =<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>100\u00ba C, sem mudan\u00e7a de estado<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= mv.cv.(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>\u00a0= mv.0,5.(100 \u2013 120)\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= \u2013 10mv.<\/b><\/span><\/span><\/span><\/p>\n

2\u00aa\u00a0\"\"\u00a0Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>quantidade de calor cedida pelo vapor de \u00e1gua<\/b><\/span><\/span><\/span>\u00a0para passar de<\/b><\/span><\/span><\/span>vapor de \u00e1gua a 100\u00ba C<\/b><\/span><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span><\/span>\u00e1gua l\u00edquida a 100\u00ba C (condensa\u00e7\u00e3o, liquefa\u00e7\u00e3o)<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0= mv.L<\/b><\/span><\/span><\/span>= mV.(-540)\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0= \u2013 540mv.<\/b><\/span><\/span><\/span><\/p>\n

3<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>\u00a0\"\"\u00a0Q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>quantidade de calor cedida pela \u00e1gua l\u00edquida<\/b><\/span><\/span><\/span>\u00a0para passar de <\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 100\u00ba C para t = 50\u00ba C, sem mudan\u00e7a de estado<\/b><\/span><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0= mv.c<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>.(t \u2013 to)<\/b><\/span><\/span><\/span>\u00a0= mv.1.(50 \u2013 100)\"\"\u00a0Q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0= \u2013 50mv.<\/b><\/span><\/span><\/span><\/p>\n

Quantidade de calor\u00a0<\/b><\/span><\/span><\/span>cedida pelo vapor de \u00e1gua<\/b><\/span><\/span><\/span>\u00a0para se transformar\u00a0<\/b><\/span><\/span><\/span>em \u00e1gua l\u00edquida a 50oC<\/b><\/span><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/span><\/p>\n

Q\u2019 = Q1\u00a0+ Q2\u00a0+ Q3\u00a0=<\/b><\/span><\/span><\/span>\u00a0\u2013 10 mv\u00a0\u2013 540mv\u00a0\u2013 50mv\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>Q\u2019<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= \u2013 600mv.<\/b><\/span><\/span><\/span><\/p>\n

Essa quantidade de calor\u00a0<\/b><\/span><\/span><\/span>Q\u2019 cedida pelo vapor de \u00e1gua para ir de 120\u00ba C a \u00e1gua a 50\u00ba C<\/b><\/span><\/span><\/span>\u00a0\u00e9 a\u00a0<\/b><\/span><\/span><\/span>mesma<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>recebida pelo sistema (copo + \u00e1gua) para ir de 20\u00ba C a 50\u00ba C (achada no item Qt\u00a0= 6600 cal)<\/b><\/span><\/span><\/span>, ou seja, no\u00a0<\/b><\/span><\/span><\/span>equil\u00edbrio t\u00e9rmico <\/b><\/span><\/span><\/span>a soma dessas quantidades de calor\u00a0<\/b><\/span><\/span><\/span>trocadas entre o vapor e o copo com \u00e1gua deve ser nula.<\/b><\/span><\/span><\/span><\/p>\n

– 600mv\u00a0+ 6600 = 0\u00a0\"\"\u00a0mv\u00a0= 6600\/600\u00a0\"\"\u00a0mv\u00a0= 11 g.<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span> As\u00a0<\/b><\/span><\/span>alternativas (A) e (E) est\u00e3o erradas,<\/b><\/span><\/span>\u00a0pois a\u00a0<\/b><\/span><\/span>temperatura de fus\u00e3o ou de vaporiza\u00e7\u00e3o n\u00e3o\u00a0<\/b><\/span><\/span>depende da\u00a0<\/b><\/span><\/span>massa do corpo, apenas do material que o constitui.<\/b><\/span><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>(C) tamb\u00e9m est\u00e1 errada,\u00a0<\/b><\/span><\/span>pois a\u00a0<\/b><\/span><\/span>subst\u00e2ncia n\u00e3o \u00e9 a \u00e1gua que sofre fus\u00e3o a 0<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>\u00a0C.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Na\u00a0<\/b><\/span><\/span>fus\u00e3o desse corpo de massa m = 20 g que ocorre a 50\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>C\u00a0<\/b><\/span><\/span>voc\u00ea deve usar a express\u00e3o o\u00a0<\/b><\/span><\/span>calor latente fornecida por Q = m.LF\u00a0\"\"<\/b><\/span><\/span>\u00a0(200 \u2013 100) = 20LF\u00a0\"\"\u00a0LF\u00a0=\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>LF\u00a0= 5 cal\/g.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span> C\u00e1lculo da\u00a0<\/b><\/span><\/span>quantidade de calor (energia t\u00e9rmica) necess\u00e1ria para aquecer\u00a0<\/b><\/span><\/span>o l\u00edquido de\u00a0<\/b><\/span><\/span>calor espec\u00edfico c = 0,58 cal\/g<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>C com sua temperatura variando de 0o\u00a0at\u00e9 78<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>C,\u00a0<\/b><\/span><\/span>sem mudan\u00e7a de estado\u00a0<\/b><\/span><\/span>Q<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>\u00a0=m.c.(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>)\u00a0<\/b><\/span><\/span>= m.0,58.(78 \u2013 0)\u00a0\"\"<\/b><\/span><\/span>Q<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>\u00a0= 45,24.m.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

14-<\/b><\/span><\/span><\/span> <\/b>Para resolver esse exerc\u00edcio,\u00a0<\/b><\/span><\/span>precisamos encontrar a massa de \u00e1gua<\/b><\/span><\/span>, ent\u00e3o vamos utilizar as outras informa\u00e7\u00f5es que j\u00e1 temos:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Analisando pela\u00a0<\/b><\/span><\/span>quantidade de calor<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Onde:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Q \u00e9 a quantidade de calor<\/b><\/span><\/span><\/span><\/span><\/p>\n

m \u00e9 a massa<\/b><\/span><\/span><\/span><\/span><\/p>\n

c \u00e9 o calor espec\u00edfico<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u0394T \u00e9 a varia\u00e7\u00e3o de temperatura<\/b><\/span><\/span><\/span><\/span><\/p>\n

Isolando o m:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"Substituindo os valores:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"Como todos os valores est\u00e3o em gramas<\/b><\/span><\/span>, ent\u00e3o nosso\u00a0<\/b><\/span><\/span>resultado vai sair em gramas<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"Agora que temos a massa, podemos usar\u00a0<\/b><\/span><\/span>a outra f\u00f3rmula de quantidade de calor<\/b><\/span><\/span>, para\u00a0<\/b><\/span><\/span>calcularmos a vaporiza\u00e7\u00e3o:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"Onde:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Q \u00e9 a quantidade de calor<\/b><\/span><\/span><\/span><\/span><\/p>\n

m \u00e9 a massa<\/b><\/span><\/span><\/span><\/span><\/p>\n

L \u00e9 o calor latente de vaporiza\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R \u2013 C<\/b><\/span><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

 <\/p>\n

Calculando a massa de le<\/b><\/span><\/span>ite\u00a0\"\"\u00a0<\/b><\/span><\/span>densidade do leite =\u00a0\"\"\u00a0\"\"mL\u00a0= dL.VL\u00a0\"\"<\/b><\/span><\/span>\u00a0mL\u00a0= 1 (g\/mL).VL(g)\u00a0\"\"\u00a0<\/b><\/span><\/span>mL\u00a0= VL\u00a0(g)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Calculando a massa de caf\u00e9<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>densidade do caf\u00e9 =\u00a0\"\"\u00a0\"\"\u00a0mc\u00a0= dc.Vc\u00a0\"\"<\/b><\/span><\/span>\u00a0mc\u00a0= 1g\/mL.30 mL\u00a0\"\"\u00a0<\/b><\/span><\/span>mc\u00a0= 30 g<\/b><\/span><\/span><\/span><\/span><\/p>\n

Ap\u00f3s a mistura ambos<\/b><\/span><\/span>\u00a0ter\u00e3o a\u00a0<\/b><\/span><\/span>mesma temperatura\u00a0<\/b><\/span><\/span>(de equil\u00edbrio t\u00e9rmico)\u00a0\"\"<\/b><\/span><\/span>\u00a0= 40\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>C.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- B<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>
\nPrinc\u00edpio das trocas de calor<\/b><\/span><\/span><\/p>\n

Colocando\u00a0<\/b><\/span><\/span>v\u00e1rios corpos a diferentes temperaturas\u00a0<\/b><\/span><\/span>no\u00a0<\/b><\/span><\/span>interior de um recipiente adiab\u00e1tico,<\/b><\/span><\/span>\u00a0haver\u00e1\u00a0<\/b><\/span><\/span>trocas de calor entre eles<\/b><\/span><\/span>, at\u00e9\u00a0<\/b><\/span><\/span>atingirem o equil\u00edbrio t\u00e9rmico.<\/b><\/span><\/span><\/p>\n

Assim, como o\u00a0<\/b><\/span><\/span>recipiente \u00e9 adiab\u00e1tico,\u00a0a\u00a0quantidade de calor cedida pelos corpos mais quentes\u00a0\u00e9igual \u00e0 quantidade de calor recebida pelos mais frios.<\/b><\/span><\/span><\/p>\n

Princ\u00edpio da Conserva\u00e7\u00e3o da energia<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A teoria acima \u00e9\u00a0<\/b><\/span><\/span>v\u00e1lida para mais de um corpo<\/b><\/span><\/span>\u00a0e, como a quantidade de calor\u00a0<\/b><\/span><\/span>recebida \u00e9 positiva e<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>cedida \u00e9 negativa, tem-se que<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0 Q1<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>+ Q2\u00a0+ Q3\u00a0+ \u2026. + QN\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 0 \u00a0ou<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>m1<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>.c1<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>.(<\/b><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u2013\u00a0<\/b><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>) + m2.c2.(\u00a0<\/b><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u2013\u00a0<\/b><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span>2) + m3.c3.(\u00a0<\/b><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u2013<\/b><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span>3) + \u2026\u00a0+ mN.cN.(<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u2013\u00a0<\/b><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span>N) = 0.<\/b><\/span><\/span><\/p>\n

Enunciado desse princ\u00edpio das trocas de calor:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Seis horas depois<\/b><\/span><\/span>,\u00a0<\/b><\/span><\/span>abre-se<\/b><\/span><\/span>\u00a0a garrafa e\u00a0<\/b><\/span><\/span>mede-se a temperatura da \u00e1gua<\/b><\/span><\/span>, obtendo-se\u00a0<\/b><\/span><\/span>\u03b8 =<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>16\u00b0C.
\n\"\"<\/b><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span> <\/b>Trata-se de dilata\u00e7\u00e3o superficial (uma das \u00e1reas) fornecida pela express\u00e3o a seguir:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Na\u00a0<\/b><\/span><\/span>express\u00e3o acima, a letra grega \u03b2 (beta) \u00e9 uma grandeza constante<\/b><\/span><\/span>,\u00a0<\/b><\/span><\/span>caracter\u00edstica do material,<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>denominada\u00a0<\/b><\/span><\/span>coeficiente de dilata\u00e7\u00e3o superficial m\u00e9dio.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Valor de<\/b><\/span><\/span>\u00a0uma das \u00e1reas\u00a0\"\"\u00a0S<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>\u00a0= L<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>.L<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span>= 20\u00d720\u00a0<\/b><\/span><\/span>= 400 cm<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>\u00a0\"\"\u00a0\u03b2 = 2.\"\"\u00a0<\/b><\/span><\/span>= 2.2.10-5<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 4.10<\/b><\/span><\/span>-5<\/b><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>C<\/b><\/span><\/span>-1<\/b><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u2206S = S<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>.\u03b2.(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>)\u00a0<\/b><\/span><\/span>= 400.4.10<\/b><\/span><\/span>-5<\/b><\/span><\/sup><\/span>.(120 \u2013 20) = 160000.10<\/b><\/span><\/span>-5<\/b><\/span><\/sup><\/span>\u00a0\"\"\u00a0\u2206S =\u00a016.10<\/b><\/span><\/span>-1<\/b><\/span><\/sup><\/span> cm<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span> J\u00e1 que o\u00a0<\/b><\/span><\/span>enunciado\u00a0<\/b><\/span><\/span>afirma que o\u00a0<\/b><\/span><\/span>coeficiente de dilata\u00e7\u00e3o do vidro utilizado \u00e9 desprez\u00edvel comparado ao do \u00e1lcool,\u00a0<\/b><\/span><\/span>voc\u00ea deve levar em conta apenas a dilata\u00e7\u00e3o (\u2206V) do \u00e1lcool de coeficiente de dilata\u00e7\u00e3o volum\u00e9trica\u00a0\"\"\u00a0= 11.10<\/b><\/span><\/span>-4<\/b><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>C<\/b><\/span><\/span>-1<\/b><\/span><\/sup><\/span>, cujos volume \u00e9 o do bulbo V<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>\u00a0= 2 cm<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span>, pois ele\u00a0<\/b><\/span><\/span>\u00e9 totalmente preenchido com \u00e1lcool at\u00e9 a base do tubo.\u00a0<\/b><\/span><\/span>A\u00a0<\/b><\/span><\/span>varia\u00e7\u00e3o de temperatura\u00a0<\/b><\/span><\/span>\u00e9 de\u00a0\u2206\"\" <\/b><\/span><\/span>=\u00a0\"\"<\/b><\/span><\/span> <\/b>– \"\"<\/b><\/span><\/span> o<\/b><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span>= 80 \u2013 30\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a0= 50\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>C.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Substituindo esses valores na\u00a0<\/b><\/span><\/span>express\u00e3o da dilata\u00e7\u00e3o volum\u00e9trica\u00a0\"\"\u00a0\u2206V = Vo.\"\"\u00a0= 2.11.10 <\/b><\/span><\/span>-4<\/b><\/span><\/sup><\/span>.50\u00a0\"\"\u00a0\u2206V =\u00a0<\/b><\/span><\/span>1100.10<\/b><\/span><\/span>–<\/b><\/span><\/sup><\/span>4<\/b><\/span><\/sup><\/span>\u00a0= 0,11 cm<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Mas, o\u00a0<\/b><\/span><\/span>volume do cilindro\u00a0<\/b><\/span><\/span>que\u00a0<\/b><\/span><\/span>constitui o tubo\u00a0<\/b><\/span><\/span>\u00e9 fornecido por\u00a0<\/b><\/span><\/span>\u2206V = \u00e1rea da basexaltura = 1.10<\/b><\/span><\/span>-2<\/b><\/span><\/sup><\/span>.h.<\/b><\/span><\/span><\/span><\/span><\/p>\n

O\u00a0<\/b><\/span><\/span>volume de l\u00edquido dilatado\u00a0<\/b><\/span><\/span>preenche esse cilindro\u00a0<\/b><\/span><\/span>at\u00e9 a altura h\u00a0<\/b><\/span><\/span>de\u00a0\"\"<\/b><\/span><\/span>\u2206V\u00a0<\/b><\/span><\/span>= 1.10<\/b><\/span><\/span>-2<\/b><\/span><\/sup><\/span>\u00a0.h\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>0,11 = 0,01h\u00a0\"\"<\/b><\/span><\/span>\u00a0h =\u00a0\"\"\u00a0\"\"\u00a0h = 11,0 cm.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R \u2013 B<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

19-<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

20-<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

Capacidade t\u00e9rmica (C) \u2013 Calor espec\u00edfico (c)<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0Define-se\u00a0<\/b><\/span><\/span>capacidade t\u00e9rmica (C) ou capacidade calor\u00edfica\u00a0<\/b><\/span><\/span>de um corpo como sendo o\u00a0<\/b><\/span><\/span>produto da massa desse corpo pelo calor espec\u00edfico da subst\u00e2ncia de que ele \u00e9 constitu\u00eddo,\u00a0<\/b><\/span><\/span>ou seja\u00a0\u00a0\"\"\u00a0\u00a0\u00a0<\/b><\/span><\/span>C = m.c\u00a0\u00a0\"\"\u00a0 como Q = m.c.\u0394\u03b8\u00a0\u00a0\"\"\u00a0\u00a0Q = C.\u0394\u03b8\u00a0\u00a0\"\"\u00a0 ou\u00a0Q = C.(\u03b8 \u2013 \u03b8<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>).<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

No caso do exerc\u00edcio\u00a0\"\"<\/b><\/span><\/span>\u00a0Q = C.\"\"\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a0=\u00a0\"\"\u00a0(I).<\/b><\/span><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>dilata\u00e7\u00e3o t\u00e9rmica (varia\u00e7\u00e3o de volume \u2206V)\u00a0<\/b><\/span><\/span>sofrida pelo bloco \u00e9\u00a0<\/b><\/span><\/span>fornecida por \u2206V = V<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>.\u00a0\"\".\"\",\u00a0<\/b><\/span><\/span>onde o\u00a0<\/b><\/span><\/span>coeficiente de dilata\u00e7\u00e3o volum\u00e9trica\u00a0\"\"vale 3.\"\"\u00a0<\/b><\/span><\/span>sendo\u00a0\"\"<\/b><\/span><\/span>\u00a0o coeficiente de dilata\u00e7\u00e3o linear do bloco\u00a0\"\"\u00a0\u2206V = Vo.\u00a0\"\".\"\"\u00a0(II).<\/b><\/span><\/span><\/span><\/span><\/p>\n

Substituindo\u00a0<\/b><\/span><\/span>(I) em (II)\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u2206V = V<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>.\u00a0\"\".\"\"\u00a0.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Portanto, a\u00a0<\/b><\/span><\/span>varia\u00e7\u00e3o de volume\u00a0\u2206V\u00a0do bloco \u00e9 diretamente proporcional a V<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>,\u00a0\"\"\u00a0e inversamente proporcional a C.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span><\/p>\n

Dilata\u00e7\u00e3o linear dos s\u00f3lidos<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Trata-se da dilata\u00e7\u00e3o de uma das dimens\u00f5es de um corpo, como por exemplo, seu comprimento.\u00a0<\/b><\/span><\/span><\/span>Considere uma\u00a0<\/b><\/span><\/span><\/span>haste met\u00e1lica de comprimento L<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0e \u00e0\u00a0<\/b><\/span><\/span><\/span>temperatura t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span>\u00a0Quando\u00a0<\/b><\/span><\/span><\/span>aquecida<\/b><\/span><\/span><\/span>\u00a0ter\u00e1\u00a0<\/b><\/span><\/span><\/span>comprimento L a uma temperatura t.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0<\/span>Caracter\u00edsticas:<\/b><\/span><\/span><\/span><\/p>\n

A dilata\u00e7\u00e3o linear \u0394L \u00e9 diretamente proporcional \u00e0 varia\u00e7\u00e3o de temperatura \u0394t<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

A dilata\u00e7\u00e3o linear \u0394L \u00e9 diretamente proporcional ao comprimento inicial Lo.<\/b><\/span><\/span><\/span><\/p>\n

A dilata\u00e7\u00e3o linear \u0394L depende do material de que \u00e9 constitu\u00eddo o corpo.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u0394L\u00a0\"\"\u00a0 dilata\u00e7\u00e3o linear<\/b><\/span><\/span><\/span><\/p>\n

\u03b1\u00a0\u00a0\"\"\u00a0coeficiente de dilata\u00e7\u00e3o linear m\u00e9dio, caracter\u00edstica do material que constitui a barra.<\/b><\/span><\/span><\/span><\/p>\n

Lo\u00a0\u00a0\"\"\u00a0comprimento inicial<\/b><\/span><\/span><\/span><\/p>\n

L\u00a0\u00a0\"\"\u00a0comprimento final<\/b><\/span><\/span><\/span><\/p>\n

\u0394t\u00a0\u00a0\"\"\u00a0 intervalo de temperatura<\/b><\/span><\/span><\/span><\/p>\n

Analisando as alternativas<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

a)<\/b><\/span><\/span><\/span>\u00a0num dia de ver\u00e3o em que a\u00a0<\/b><\/span><\/span><\/span>temperatura variar 20 \u00b0C<\/b><\/span><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span><\/span>comprimento de uma barra de ferro de 10,0 m sofrer\u00e1 uma varia\u00e7\u00e3o de 2,6 cm<\/b><\/span><\/span><\/span><\/p>\n

a \u2013 Falsa<\/b><\/span><\/span><\/span><\/p>\n

\u0394L = Lo.\u03b1. \u0394t\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u0394L = 10.13.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>.20<\/b><\/span><\/span><\/span><\/p>\n

\u0394L = 26. 10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\u00a0m = 0,26 cm<\/b><\/span><\/span><\/span><\/p>\n

b)<\/b><\/span><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span><\/span>coeficiente de dilata\u00e7\u00e3o superficial do ferro \u00e9 169.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

b \u2013 Falsa<\/b><\/span><\/span><\/span><\/p>\n

\uf062<\/span>\u00a0= 2.\u03b1\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>\uf062<\/b><\/span><\/span><\/span>\u00a0= 2. 13.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00baC<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>\uf062<\/b><\/span><\/span><\/span>\u00a0= 26.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00baC<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span><\/p>\n

c)<\/b><\/span><\/span><\/span>\u00a0para\u00a0<\/b><\/span><\/span><\/span>cada 1 \u00b0C de varia\u00e7\u00e3o de temperatura<\/b><\/span><\/span><\/span>, o\u00a0<\/b><\/span><\/span><\/span>comprimento de uma barra de 1,0 m desse material varia 13.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>m<\/b><\/span><\/span><\/span>\u22c5<\/b><\/span><\/span><\/span><\/p>\n

c \u2013 Verdadeira<\/b><\/span><\/span><\/span><\/p>\n

\u0394L = Lo.\u03b1. \u0394t\u00a0\"\"\u00a0\u0394L = 1.13.10-6.1\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>\u0394L = 13.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0m<\/b><\/span><\/span><\/span><\/p>\n

d)<\/b><\/span><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span><\/span>coeficiente de dilata\u00e7\u00e3o volum\u00e9trica do ferro \u00e9 39.10<\/b><\/span><\/span><\/span>-18<\/b><\/span><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

d \u2013 Falsa<\/b><\/span><\/span><\/span><\/p>\n

\uf067<\/span>\u00a0= 3.\u03b1\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>\uf067<\/b><\/span><\/span><\/span>\u00a0= 3.13.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>\uf067<\/b><\/span><\/span><\/span>\u00a0= 39.10-6<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00baC<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

R- C<\/b><\/span><\/span><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

Transforma\u00e7\u00e3o adiab\u00e1tica<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0Nela, o\u00a0<\/b><\/span><\/span>sistema n\u00e3o troca calor com o meio externo (Q = 0) e isso ocorre porque o g\u00e1s est\u00e1 termicamente isolado ou porque a transforma\u00e7\u00e3o \u00e9 muito r\u00e1pida de modo que qualquer troca de calor com possa ser considerada.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0Se Q = 0\u00a0\"\"<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span>\u0394U = Q \u2013 W\u00a0\u00a0\"\"<\/b><\/span><\/span>\u00a0 \u0394U = 0 \u2013 W\u00a0\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span>\u0394U = \u2013 W.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\u00a0<\/span>R- D<\/b><\/span><\/span><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

R- A<\/b><\/span><\/span><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0Para calcularmos a temperatura final\u00a0<\/b><\/span><\/span>precisamos primeiro do volume final. Podemos calcular visto que sabemos que o \u00eambolo se deslocou 10 cm:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Volume inicial = 6.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span>\u00a0m\u00b3<\/b><\/span><\/span><\/span><\/span><\/p>\n

Volume final = Volume inical +\u00a0\"\"(volume com 10 cm de deslocamento)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Passando 10 cm para metros (dividir por 100):<\/b><\/span><\/span><\/span><\/span><\/p>\n

10 cm = 0,1 m<\/b><\/span><\/span><\/span><\/span><\/p>\n

Calculando \"\":<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0= \u00e1rea interna X deslocamento<\/b><\/span><\/span><\/span><\/span><\/p>\n

Substituindo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0= 2.10<\/b><\/span><\/span>-2<\/b><\/span><\/sup><\/span>.0,1<\/b><\/span><\/span><\/span><\/span><\/p>\n

Calculando:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0= 2.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span>\u00a0m\u00b3<\/b><\/span><\/span><\/span><\/span><\/p>\n

Agora calculando o volume final:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Volume final = 6.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span>\u00a0+ 2.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span><\/span><\/span><\/p>\n

Volume final = 8.10-3\u00a0m\u00b3<\/b><\/span><\/span><\/span><\/span><\/p>\n

Sabendo o volume final podemos utilizar a f\u00f3rmula a seguir, j\u00e1 que se trata de um g\u00e1s ideal:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Onde:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Pi\u00a0e Pf\u00a0\u00e9 a press\u00e3o inicial e final<\/b><\/span><\/span><\/span><\/span><\/p>\n

Vi\u00a0e Vf\u00a0\u00e9 o volume inicial e final<\/b><\/span><\/span><\/span><\/span><\/p>\n

Ti\u00a0e Tf\u00a0\u00e9 a temperatura inicial e final<\/b><\/span><\/span><\/span><\/span><\/p>\n

Como \u00e9\u00a0<\/b><\/span><\/span>uma expans\u00e3o isob\u00e1rica (press\u00e3o constante) podemos cortar a press\u00e3o em ambos os lados<\/b><\/span><\/span>, restando apenas:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0=\u00a0\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

Substituindo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0=\u00a0\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

Isolando a temperatura final:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Tf\u00a0=\u00a0\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

Calculando:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Tf\u00a0= 400 K (temperatura final)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Achamos a temperatura, agora\u00a0<\/b><\/span><\/span>vamos para a segunda parte do exerc\u00edcio. <\/b><\/span><\/span>Para\u00a0<\/b><\/span><\/span>calcularmos a quantidade de calor vamos utilizar a primeira lei da termodin\u00e2mica. Como se trata de uma transforma\u00e7\u00e3o isob\u00e1rica:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como a\u00a0<\/b><\/span><\/span>press\u00e3o \u00e9 constante<\/b><\/span><\/span>\u00a0o nosso\u00a0<\/b><\/span><\/span>trabalho s\u00f3 depende da varia\u00e7\u00e3o do volume:<\/b><\/span><\/span><\/span><\/span><\/p>\n

W = P.\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

Substituindo.\u00a0<\/b><\/span><\/span>N\u00f3s calculamos a varia\u00e7\u00e3o anteriormente (2.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span>):<\/b><\/span><\/span><\/span><\/span><\/p>\n

W = 2.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>. 2.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span><\/span><\/span><\/p>\n

Calculando:<\/b><\/span><\/span><\/span><\/span><\/p>\n

W = 4.10<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>\u00a0J<\/b><\/span><\/span><\/span><\/span><\/p>\n

O exerc\u00edcio j\u00e1 nos forneceu\u00a0<\/b><\/span><\/span>a varia\u00e7\u00e3o de energia interna (600 J)<\/b><\/span><\/span>. Vamos ent\u00e3o\u00a0<\/b><\/span><\/span>substituir na equa\u00e7\u00e3o da primeira lei:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0= Q \u2013 W<\/b><\/span><\/span><\/span><\/span><\/p>\n

Isolando a quantidade de calor:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Q =\u00a0\"\"\u00a0+ W<\/b><\/span><\/span><\/span><\/span><\/p>\n

Substituindo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Q = 600 + 400<\/b><\/span><\/span><\/span><\/span><\/p>\n

Calculando:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Q = 1000 J (quantidade de calor)<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

26-<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>A)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>A\u00a0<\/b><\/span><\/span>energia interna de certa massa de um g\u00e1s<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>perfeito<\/b><\/span><\/span>\u00a0\u00e9\u00a0<\/b><\/span><\/span>fun\u00e7\u00e3o exclusiva da temperatura\u00a0<\/b><\/span><\/span>desse g\u00e1s<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0\u00a0<\/span>Aumento de temperatura\u00a0\u00a0\"\"<\/b><\/span><\/span>\u00a0\u00a0(\u0394T > 0) \u00a0\"\"\u00a0\u00a0aumento de energia interna\u00a0\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span>(\u0394U > 0).<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Diminui\u00e7\u00e3o de temperatura\u00a0\u00a0\"\"<\/b><\/span><\/span>\u00a0\u00a0(\u0394T < 0)\u00a0\u00a0\"\"\u00a0\u00a0diminui\u00e7\u00e3o de energia interna\u00a0\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span>(\u0394U<0).<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Temperatura constante<\/b><\/span><\/span>\u00a0\"\"\u00a0\u00a0(\u0394T = 0)\u00a0\u00a0\"\"\u00a0\u00a0energia interna constante\u00a0\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span>(\u0394U = 0).<\/b><\/span><\/span><\/span><\/span><\/p>\n

Vamos ent\u00e3o\u00a0<\/b><\/span><\/span>calcular a varia\u00e7\u00e3o da energia interna em cada transforma\u00e7\u00e3o:<\/b><\/span><\/span><\/span><\/span><\/p>\n

I: E\u00a0\"\"\"\"\"\"\u00a0=\u00a0\"\"\u00a0\"\"\u00a01.105.0,6\/TE\u00a0= 3.105.0,2\/TF\u00a0\"\"\u00a0<\/b><\/span><\/span>6.104\u00a0\/ TE\u00a0= 6.104\/TF\u00a0\"\"<\/b><\/span><\/span>\u00a0TE\u00a0= TF\u00a0\"\"\u00a0energia interna constante.<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

II:<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>F\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>3.105.0,2\/TF\u00a0= 6.105.0,6\/TH\u00a0\"\"\u00a0<\/b><\/span><\/span>6.104\/TE\u00a0= 6.104\/TF\u00a0\"\"<\/b><\/span><\/span>\u00a06.104TG\u00a0= 12.104TF\u00a0\"\"\u00a0TF\u00a0> TG\u00a0\"\"\u00a0TG\u00a0= 2TF\u00a0\"\"\u00a0energia interna n\u00e3o constante.<\/b><\/span><\/span><\/span><\/span><\/p>\n

III:<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>G\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>6.105.0,2\/TG\u00a0= 4.105.0,6\/TH\u00a0\"\"\u00a0<\/b><\/span><\/span>12.104\/TG\u00a0= 24.104\/TH\u00a0\"\"<\/b><\/span><\/span>\u00a024.104TG= 12.104TH\u00a0\"\"\u00a0TH\u00a0= 2TH\u00a0\"\"\u00a0TH\u00a0\"\"\u00a0TG\u00a0\"\"\u00a0energia interna n\u00e3o constante.<\/b><\/span><\/span><\/span><\/span><\/p>\n

IV:\u00a0<\/b><\/span><\/span>H\u00a0\"\"\"\"\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>4.105.0,2\/TH\u00a0= 1.105.0,6\/TE\u00a0\"\"\u00a0<\/b><\/span><\/span>8.104\/TH\u00a0= 6.104\/TE\u00a0\"\"<\/b><\/span><\/span>\u00a08.104TE= 6.104TH\u00a0\"\"\u00a0TE\u00a0\"\"\u00a0TH\u00a0\"\"\u00a0energia interna n\u00e3o constante.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- A energia interna \u2206U \u00e9 constante apenas em I.<\/b><\/span><\/span><\/span><\/span><\/p>\n

B)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>Em\u00a0<\/b><\/span><\/span>I n\u00e3o ocorre varia\u00e7\u00e3o\u00a0<\/b><\/span><\/span>de\u00a0<\/b><\/span><\/span>energia interna.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Em\u00a0<\/b><\/span><\/span>II e III houve aumento de temperatura\u00a0<\/b><\/span><\/span>e, consequentemente\u00a0<\/b><\/span><\/span>aumento de energia interna.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Em\u00a0<\/b><\/span><\/span>IV\u00a0<\/b><\/span><\/span>houve\u00a0<\/b><\/span><\/span>diminui\u00e7\u00e3o de temperatura\u00a0<\/b><\/span><\/span>e, consequentemente\u00a0<\/b><\/span><\/span>diminui\u00e7\u00e3o de energia interna.<\/b><\/span><\/span><\/span><\/span><\/p>\n

O aumento de enet=rgia interna ocorreu apenas em II e III e, como a energia interna \u00e9 diretamente proporcional \u00e0 temperatura absoluta, de A:<\/b><\/span><\/span><\/span><\/span><\/p>\n

II. TG\u00a0= 2TF\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 2<\/b><\/span><\/span><\/span><\/span><\/p>\n

III. TH\u00a0= 2TG\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 2<\/b><\/span><\/span><\/span><\/span><\/p>\n

C)<\/b><\/span><\/span> Veja informa\u00e7\u00e3o abaixo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

No caso doe exerc\u00edcio vamos calcular a \u00e1rea do ciclo lembrando que nos trechos FG e HE o trabalho \u00e9 nulo (transforma\u00e7\u00e3o isovlum\u00e9trica).<\/b><\/span><\/span><\/span><\/span><\/p>\n

C\u00e1lculo do trabalho W no trecho GH onde ele fornecido pela \u00e1rea do trap\u00e9zio e \u00e9 positivo (sentido hor\u00e1rio):<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

C\u00e1lculo do trabalho W no trecho EF onde ele \u00e9fornecido pela \u00e1rea do trap\u00e9zio e \u00e9 negativo (sentido anti-hor\u00e1rio):<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

W =<\/b><\/span><\/span>\u00a02.105\u00a0\u2013 0,8.105<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 1,2.105\u00a0J<\/b><\/span><\/span><\/span><\/span><\/p>\n

Po\u00a0=\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a0Po\u00a0=\u00a0\"\"\u00a0= 0,02.105\u00a0\"\"\u00a0Po\u00a0= 2.103\u00a0W\u00a0\"\"\u00a0como\u00a0<\/b><\/span><\/span>s\u00e3o 50 ciclos\u00a0\"\"<\/b><\/span><\/span>\u00a0Po\u00a0= 50.2.103\u00a0\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

Po\u00a0= 100 kW.<\/b><\/span><\/span><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

A seguir, breve resumo te\u00f3rico:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Transforma\u00e7\u00e3o isob\u00e1rica<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>Ocorre \u00e0 press\u00e3o constante.<\/b><\/span><\/span>\u00a0A\u00a0<\/b><\/span><\/span>varia\u00e7\u00e3o de temperatura (\u0394T)<\/b><\/span><\/span>\u00a0provoca uma<\/b><\/span><\/span>varia\u00e7\u00e3o de energia interna (\u0394U) do sistema\u00a0<\/b><\/span><\/span>e a\u00a0<\/b><\/span><\/span>varia\u00e7\u00e3o de volume (\u0394V) produz trabalho.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Parte\u00a0do calor (Q) recebido pelo sistema\u00a0<\/b><\/span><\/span>\u00e9\u00a0<\/b><\/span><\/span>armazenada sob forma de energia interna\u00a0<\/b><\/span><\/span>e\u00a0<\/b><\/span><\/span>parte<\/b><\/span><\/span>\u00a0\u00e9\u00a0<\/b><\/span><\/span>transformada em trabalho,\u00a0<\/b><\/span><\/span>de modo que\u00a0<\/b><\/span><\/span>\u0394U = Q \u2013 W.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

No caso do exerc\u00edcio, s\u00e3o fornecidos:<\/b><\/span><\/span><\/span><\/span><\/p>\n

P =\u00a0<\/b><\/span><\/span>constante = 1 atm\u00a0<\/b><\/span><\/span>= 1.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>\u00a0N\/m<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>\u00a0(Pa)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Volume inicial\u00a0\"\"\u00a0V<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span>= 1000 cm<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span>\u00a0= 10<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span>\u00a0cm<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span>\u00a0= 10<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span>.(10<\/b><\/span><\/span>-6<\/b><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span>m<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span>)\u00a0\"\"\u00a0<\/b><\/span><\/span>Vo\u00a0= 10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span>m<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span><\/span><\/span><\/p>\n

Volume final que foi aumentado 300%\u00a0\"\"<\/b><\/span><\/span>\u00a0V =(\u00a0\"\"\u00a0).10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>V = 3.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span>\u00a0m<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span><\/span><\/span><\/p>\n

C\u00e1lculo do<\/b><\/span><\/span>\u00a0trabalho mec\u00e2nico,\u00a0<\/b><\/span><\/span>em joules,\u00a0<\/b><\/span><\/span>realizado pelo g\u00e1s durante\u00a0<\/b><\/span><\/span>essa transforma\u00e7\u00e3o\u00a0\"\"\u00a0<\/b><\/span><\/span>W = P.\u2206V =\u00a010<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>.(3.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span>\u00a0\u2013 1.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span>)\u00a0\"\"\u00a0W = 10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>.2.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span>\u00a0\"\"\u00a0W = 2.10<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>\u00a0J<\/b><\/span><\/span><\/span><\/span><\/p>\n

Energia interna de um g\u00e1s perfeito<\/b><\/span><\/span><\/span><\/span><\/p>\n

A energia interna (U) de um\u00a0<\/b><\/span><\/span>g\u00e1s perfeito monoat\u00f4mico\u00a0corresponde<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00e0 soma das energias cin\u00e9ticas m\u00e9dias (Ec) de todas as suas mol\u00e9culas e, pela lei de Joule \u00e9 fornecida por:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A energia interna de certa massa de um g\u00e1s perfeito \u00e9 fun\u00e7\u00e3o exclusiva da temperatura desse g\u00e1s<\/b><\/span><\/span><\/span><\/span><\/p>\n

Se U = (3\/2).n.R.T e PV = n.R.T\u00a0\"\"\u00a0U = (3\/2).PV\u00a0\"\"\u00a0\u2206U = (3\/2).P.\u2206V =\u00a0<\/b><\/span><\/span>(\"\"\u00a0. 10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>.2.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0\u2206U = 3.10<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>\u00a0J.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

28-<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>a)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o de Clapeyron ou Equa\u00e7\u00e3o de um g\u00e1s ideal<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Pelo gr\u00e1fico:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Ponto A\u00a0\"\"\u00a0PA\u00a0= 2.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>\u00a0N\/m<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>\u00a0\"\"\u00a0VA\u00a0= 2.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span>\u00a0m<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span>\u00a0\"\"\u00a0TA<\/b><\/span><\/span><\/span><\/span><\/p>\n

Ponto B\u00a0\"\"\u00a0PB\u00a0= 5.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>\u00a0N\/m<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>\u00a0\"\"\u00a0VB\u00a0= 6.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span>\u00a0m<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span>\u00a0\"\"\u00a0TB<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Primeiro Princ\u00edpio da Termodin\u00e2mica ou Princ\u00edpio da Conserva\u00e7\u00e3o da energia<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

O\u00a0<\/b><\/span><\/span>Primeiro Princ\u00edpio da Termodin\u00e2mica (Princ\u00edpio da Conserva\u00e7\u00e3o da Energia)<\/b><\/span><\/span>\u00a0afirma que:\u00a0<\/b><\/span><\/span>\u201cA energia n\u00e3o pode ser criada nem destru\u00edda, mas apenas transformada\u201d<\/b><\/span><\/span><\/span><\/span><\/p>\n

Exemplo num\u00e9rico:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Suponha que um<\/b><\/span><\/span>\u00a0sistema isolado receba Q = 300 J de calor.\u00a0<\/b><\/span><\/span>Se, por exemplo,<\/b><\/span><\/span>\u00a0\u2206U = 50 J dessa energia forem absorvidos pelo sistema aumentando sua energia interna de 50 J,\u00a0<\/b><\/span><\/span>ent\u00e3o a parte restante,<\/b><\/span><\/span>\u00a0W = 250 J, ser\u00e1 fornecida ao ambiente sob forma de trabalho.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Ent\u00e3o, Q = W + \u2206U<\/b><\/span><\/span>\u00a0ou\u00a0\u2206U = Q \u2013 W.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Como nas transforma\u00e7\u00f5es\u00a0<\/b><\/span><\/span>ACDEB e AFB os estados inicial (A, de temperatura TA)) e final (B, de<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>temperatura TB) coincidem,<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>varia\u00e7\u00e3o de energia interna \u2206U ser\u00e1 a mesma nas duas transforma\u00e7\u00f5es<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>\u2206UACDEB\u00a0= \u2206UAFB.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u2206UACDEB\u00a0= Q1\u00a0\u2013 W1\u00a0e \u2206UAFB\u00a0= Q2\u00a0\u2013 W2\u00a0\"\"\u00a0Q1\u00a0\u2013 Q2\u00a0= W1\u00a0\u2013 W2.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Mas,\u00a0<\/b><\/span><\/span>em\u00a0<\/b><\/span><\/span>toda transforma\u00e7\u00e3o\u00a0<\/b><\/span><\/span>representada no\u00a0<\/b><\/span><\/span>diagrama PxV, o trabalho realizado\u00a0\u00e9 numericamente igual \u00e0 \u00e1rea.<\/b><\/span><\/span><\/span><\/span><\/p>\n

C\u00e1lculo do W1\u00a0no trecho ACDEB:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

C\u00e1lculo do W2\u00a0no trecho AFB:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Q1\u00a0\u2013 Q2\u00a0= W1\u00a0\u2013 W2\u00a0= 17.10<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>\u00a0\u2013 8.10<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>\u00a0\"\"\u00a0Q1\u00a0\u2013 Q2\u00a0= 9.10<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>\u00a0J<\/b><\/span><\/span><\/span><\/span><\/p>\n

29<\/span><\/span><\/span>–<\/b><\/span><\/span><\/span><\/p>\n

\u00a0I.<\/b><\/span><\/span><\/span>\u00a0O\u00a0<\/b><\/span><\/span><\/span>trabalho total<\/b><\/span><\/span><\/span>\u00a0realizado no\u00a0<\/b><\/span><\/span><\/span>ciclo ABCA \u00e9 nulo<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

I.<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Falsa<\/b><\/span><\/span><\/span>\u00a0\u2014 Em toda\u00a0<\/b><\/span><\/span><\/span>transforma\u00e7\u00e3o c\u00edclica<\/b><\/span><\/span><\/span>\u00a0representada no\u00a0<\/b><\/span><\/span><\/span>diagrama PxV<\/b><\/span><\/span><\/span>, o\u00a0<\/b><\/span><\/span><\/span>trabalho realizado<\/b><\/span><\/span><\/span>\u00a0\u00e9<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

fornecido pela \u00e1rea do ciclo<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Se o\u00a0<\/b><\/span><\/span><\/span>ciclo \u00e9 realizado no\u00a0<\/b><\/span><\/span><\/span>sentido hor\u00e1rio<\/b><\/span><\/span><\/span>, o trabalho \u00e9\u00a0<\/b><\/span><\/span><\/span>positivo.<\/b><\/span><\/span><\/span><\/p>\n

Se o\u00a0<\/b><\/span><\/span><\/span>ciclo \u00e9 realizado<\/b><\/span><\/span><\/span>\u00a0no\u00a0<\/b><\/span><\/span><\/span>sentido anti-hor\u00e1rio<\/b><\/span><\/span><\/span>, o trabalho \u00e9\u00a0<\/b><\/span><\/span><\/span>negativo.<\/b><\/span><\/span><\/span><\/p>\n

II.<\/b><\/span><\/span><\/span>\u00a0A\u00a0<\/b><\/span><\/span><\/span>energia interna\u00a0<\/b><\/span><\/span><\/span>do g\u00e1s no\u00a0<\/b><\/span><\/span><\/span>estado C \u00e9 maior que no estado A.<\/b><\/span><\/span><\/span><\/p>\n

II.<\/b><\/span><\/span><\/span>\u00a0Correto\u00a0<\/b><\/span><\/span><\/span>\u2014\u00a0<\/b><\/span><\/span><\/span>no\u00a0<\/b><\/span><\/span><\/span>trecho\u00a0<\/b><\/span><\/span><\/span>CA<\/b><\/span><\/span><\/span>\u00a0a transforma\u00e7\u00e3o \u00e9\u00a0<\/b><\/span><\/span><\/span>isovolum\u00e9trica onde o volume V1\u00a0\u00e9 constante<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Observe nas figuras abaixo que quanto\u00a0<\/b><\/span><\/span><\/span>maior a temperatura<\/b><\/span><\/span><\/span>\u00a0maior \u00e9 o movimento vibrat\u00f3rio das<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

mol\u00e9culas do g\u00e1s e\u00a0consequentemente maior a energia cin\u00e9tica (Ec) de suas mol\u00e9culas, o que implica\u00a0<\/b><\/span><\/span><\/span>num aumento da energia interna (U) do g\u00e1s.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

III.<\/b><\/span><\/span><\/span>\u00a0Durante a\u00a0<\/b><\/span><\/span><\/span>transforma\u00e7\u00e3o\u00a0\"\"<\/b><\/span><\/span><\/span>, o g\u00e1s\u00a0<\/b><\/span><\/span><\/span>recebe\u00a0<\/b><\/span><\/span><\/span>calor<\/b><\/span><\/span><\/span>\u00a0e realiza\u00a0<\/b><\/span><\/span><\/span>trabalho.<\/b><\/span><\/span><\/span><\/p>\n

Observe no gr\u00e1fico que no\u00a0<\/b><\/span><\/span><\/span>trecho AB<\/b><\/span><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span><\/span>volume<\/b><\/span><\/span><\/span>\u00a0est\u00e1\u00a0<\/b><\/span><\/span><\/span>aumentando<\/b><\/span><\/span><\/span>\u00a0(VB\u00a0> VA) e o\u00a0<\/b><\/span><\/span><\/span>trabalho realizado \u00e9<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>positivo<\/b><\/span><\/span><\/span>, pois, numa\u00a0<\/b><\/span><\/span><\/span>isob\u00e1rica W = P.(VB\u00a0\u2013 VA)<\/b><\/span><\/span><\/span>\u00a0e, sendo\u00a0<\/b><\/span><\/span><\/span>P positiva e constante e (VB\u00a0\u2013 VA) > 0<\/b><\/span><\/span><\/span>, o\u00a0<\/b><\/span><\/span><\/span>trabalho ser\u00e1 WAB\u00a0> 0.<\/b><\/span><\/span><\/span><\/p>\n

A transforma\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>BC \u00e9 isot\u00e9rmica<\/b><\/span><\/span><\/span>\u00a0(mesma temperatura,\u00a0<\/b><\/span><\/span><\/span>TB\u00a0= TC<\/b><\/span><\/span><\/span>) e, como vimos em\u00a0<\/b><\/span><\/span><\/span>I, TB=TC\u00a0> TA<\/b><\/span><\/span><\/span>\u00a0o que implica que\u00a0<\/b><\/span><\/span><\/span>UB\u00a0> UA<\/b><\/span><\/span><\/span>\u00a0fazendo com que a<\/b><\/span><\/span><\/span>varia\u00e7\u00e3o de energia interna\u00a0<\/b><\/span><\/span><\/span>\u2206U = UB\u00ad\u00a0\u2013 UA\u00a0seja positiva (\u2206UAB\u00a0> 0.<\/b><\/span><\/span><\/span><\/p>\n

Pela\u00a0<\/b><\/span><\/span><\/span>segunda lei da termodin\u00e2mica\u00a0<\/b><\/span><\/span><\/span>\u2206U = Q \u2013 W<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>QAB\u00a0=\u2206UAB\u00a0+ WAB<\/b><\/span><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/span>como \u2206UAB\u00a0e WAB\u00a0s\u00e3o positivos, QAB\u00a0> 0 e o g\u00e1s recebe calor e realiza trabalho. \u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- E\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n


\nRendimento real \u03b7 do motor<\/b><\/span><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>pot\u00eancia total\u00a0<\/b><\/span><\/span>absorvida por esse motor em\u00a0<\/b><\/span><\/span>\u2206t = 1 s\u00a0<\/b><\/span><\/span>vale\u00a0<\/b><\/span><\/span>Pt\u00a0=\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>=\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>=\u00a0\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

Pt\u00a0= 3700 J\/s\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>Pt\u00a0= 3700 W.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Pot\u00eancia \u00fatil\u00a0<\/b><\/span><\/span>do motor\u00a0\"\"\u00a0Pu\u00a0= 2,5\u00d7740\u00a0\"\"\u00a0<\/b><\/span><\/span>Pu\u00a0= 1850 W.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Rendimento do motor\u00a0\"\"\u00a0\u03b7m<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>=Pu\/Pt<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>=\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0\u03b7m\u00a0=\u00a00,5 (50%)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Rendimento de um motor de Carnot<\/b><\/span><\/span><\/span><\/span><\/p>\n

Temperatura Tf\u00a0da fonte fria\u00a0\"\"\u00a0Tf<\/b><\/span><\/span>\u00a0=927 +273\u00a0<\/b><\/span><\/span>= 1200 K.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Temperatura TQ\u00a0da fonte quente\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>=\u00a0\"\"\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>=\u00a0\"\"\u00a0\"\"\u00a0\"\"\u00a0=\u00a0\"\"\u00a0\"\"\u00a0403 \u2013 160 = 9C\u00a0\"\"\u00a0<\/b><\/span><\/span>C = 27\u00a0oC\u00a0\"\"\u00a0TQ\u00a0=\u00a0<\/b><\/span><\/span>27+273\u00a0<\/b><\/span><\/span>= 300 K.<\/b><\/span><\/span><\/span><\/span><\/p>\n

O rendimento\u00a0\u03b7C\u00a0de um motor de Carnot\u00a0<\/b><\/span><\/span>operando entre as temperaturas acima\u00a0<\/b><\/span><\/span>\u00e9 fornecido pela express\u00e3o\u00a0\"\"\u00a0\u03b7C\u00a0= 1 \u2013 TQ\/Tf\u00a0=\u00a0<\/b><\/span><\/span>1 \u2013\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 1 \u2013 0,25\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u03b7C\u00a0= 0,75 (75%)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Raz\u00e3o pedida<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>\u03b7C\/\u03b7m\u00a0=\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>\u03b7C\/\u03b7m\u00a0= 1,5<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

T\u00a0\"\"\u00a0temperatura da fonte quente<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>T = 1 160,6\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>F\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>=\u00a0\"\"\u00a0\"\"\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>=\u00a0\"\"\u00a0\"\"\u00a0\"\"\u00a0=\u00a0\"\"\u00a0\"\"\u00a09K \u2013 2457 = 5643\u00a0\"\"\u00a0K =\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>T = 900 K<\/b><\/span><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>temperatura da fonte fria<\/b><\/span><\/span>\u00a0\u00e9 a da\u00a0<\/b><\/span><\/span>fonte quente menos 270oC\u00a0<\/b><\/span><\/span>= 270 + 273<\/b><\/span><\/span>= 543 K\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>T1<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 900 \u2013 543\u00a0\"\"\u00a0<\/b><\/span><\/span>T1\u00a0= 357 K.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Rendimento da m\u00e1quina t\u00e9rmica\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0= 1 \u2013 T1\/T\u00a0<\/b><\/span><\/span>= 1 –\u00a0\"\"\u00a0= 1 \u2013 0,4\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a0= 0,6 = 60%<\/b><\/span><\/span><\/span><\/span><\/p>\n

R \u2013 ?<\/b><\/span><\/span><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

a)<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

S\u00e3o dados:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Q1<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>quantidade de calor\u00a0<\/b><\/span><\/span>recebido da\u00a0<\/b><\/span><\/span>fonte quente num ciclo\u00a0\"\"\u00a0Q1\u00a0= 4,0.103\u00a0J<\/b><\/span><\/span><\/span><\/span><\/p>\n

Q2<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>quantidade de calor\u00a0<\/b><\/span><\/span>rejeitado para a\u00a0<\/b><\/span><\/span>fonte fria num ciclo\u00a0\"\"\u00a0Q2\u00a0= 2,8.103\u00a0J<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b)<\/b><\/span><\/span> C\u00e1lculo do trabalho\"\"\u00a0realizado em cada ciclo\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= Q1\u00a0\u2013 Q2<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 4,0.103\u00a0\u2013 2,8.103\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 1,2.103\u00a0J<\/b><\/span><\/span><\/span><\/span><\/p>\n

Como s\u00e3o 3000 ciclos o trabalho total ser\u00e1\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>total<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>=<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>1,2.103×3000\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>total<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 3,6.106\u00a0J.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u03b7 = 1 \u2013\u00a0\"\"<\/b><\/span><\/span>\u00a0= 1 –\u00a0\"\"\u00a0= 0,5\u00a0\"\"\u00a0<\/b><\/span><\/span>\u03b7 = 50 %.<\/b><\/span><\/span><\/span><\/span><\/p>\n

T1\u00a0= 27 + 273 = 300 K\u00a0\"\"\u00a0\u03b7 = 1 \u2013\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a00,5 = 1 \u2013\u00a0\"\"\u00a0\"\"\u00a0\u2013 0,5 = \u2013\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>T = 600 K.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/span><\/span><\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a>
\n<\/span><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o Comentada \u2013 Vestibulares Recentes F\u00edsica T\u00e9rmica \u2013 2017 \u2013 2016 \u2013 2015 01- R- D 03- (V) Temperatura\u00a0\u00a0grandeza f\u00edsica\u00a0que representa a\u00a0medida do estado de agita\u00e7\u00e3o m\u00e9dio das mol\u00e9culas de um corpo. (F) Calor\u00a0\u00a0energia t\u00e9rmica\u00a0que passa, de\u00a0forma espont\u00e2nea, do corpo demaior temperatura para o de menor temperatura. (F) Fus\u00e3o\u00a0\u00a0mudan\u00e7a de estado f\u00edsico\u00a0sofrida por um\u00a0s\u00f3lido ao se transformar em l\u00edquido\u00a0quando\u00a0recebe uma<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3406,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-3408","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3408","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=3408"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3408\/revisions"}],"predecessor-version":[{"id":3410,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3408\/revisions\/3410"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3406"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3408"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}