{"id":2036,"date":"2016-03-08T00:37:19","date_gmt":"2016-03-08T00:37:19","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2036"},"modified":"2024-08-21T14:37:36","modified_gmt":"2024-08-21T14:37:36","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-calor-especifico-sensivel-e-trocas-de-calor-sem-mudanca-de-estado","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/fisica-termica\/calorimetria\/calor-sensivel-especifico-e-trocas-de-calor-sem-mudanca-de-estado\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-calor-especifico-sensivel-e-trocas-de-calor-sem-mudanca-de-estado\/","title":{"rendered":"Calor espec\u00edfico (sens\u00edvel) e trocas de calor sem mudan\u00e7a de estado"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre<\/b><\/span><\/span><\/span><\/p>\n

Calor espec\u00edfico (sens\u00edvel) e trocas de calor sem mudan\u00e7a de estado<\/b><\/span><\/span><\/span><\/p>\n

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

01<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe os dados do gr\u00e1fico\u00a0 —<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span> <\/b>R- A\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que eles possuem a mesma massa (m) e recebem a mesma quantidade de calor (Q)\u00a0\u00a0 —\u00a0 Q=m.c.\u0394\u03b8\u00a0 —\u00a0 \u0394\u03b8=Q\/m.c\u00a0 —\u00a0 como Q e m s\u00e3o constantes, observe que a varia\u00e7\u00e3o de temperatura (\u0394\u03b8) \u00e9 inversamente proporcional ao calor espec\u00edfico (c)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

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

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>1 minuto \u2013 2 kcal\u00a0 —\u00a0 60min \u2013 Q\u00a0 —\u00a0 Q=60.2\u00a0 —\u00a0 Q=120 kcal\u00a0 —\u00a0 Q=m.c.\u0394t\u00a0 —\u00a0 120=m.1.80\u00a0 —\u00a0 m=1,5 kg\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>P\u00e3o + hamb\u00farguer\u00a0 —\u00a0 82,5 + 292,5\u00a0 —\u00a0 375,0 kcal<\/b><\/span><\/span><\/p>\n

Batata\u00a0 —\u00a0 50.6=300\u00a0 ———————\u00a0 300,0 kcal<\/b><\/span><\/span><\/p>\n

\u00c1gua de c\u00f4co\u00a0 ——————————\u00a0\u00a0\u00a0 42,0 kcal<\/b><\/span><\/span><\/p>\n

Sorvete\u00a0 —\u00a0 50.3=150\u00a0 ——————–\u00a0\u00a0 150,0 kcal<\/b><\/span><\/span><\/p>\n

Soma\u00a0 —\u00a0 867,0\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>capacidade t\u00e9rmica\u00a0 —\u00a0 C=Q\/\u0394\u03b8=10.55\/(45 \u2013 20)\u00a0 —\u00a0 C=22 cal\/<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 calor espec\u00edfico\u00a0 —\u00a0 C=m.c\u00a0 —\u00a0 22=100.c\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

c=0,22 cal,g.<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=Q\/\u0394\u03b8<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q=C<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>.\u0394\u03b8<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q=30C<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 C<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>=Q\/\u0394\u03b8<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q=C<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>.\u0394\u03b8<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q=60C<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 30C<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=60C<\/b><\/span><\/span>Y\u00a0\u00a0<\/b><\/span><\/span><\/sub>—\u00a0 C<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=2C<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Q=m.c.\u0394\u03b8\u00a0 —\u00a0 3.000=1.000.c.5\u00a0 —\u00a0 c=3\/5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>c=0,6cal\/g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) O asfalto, por ser mais escuro recebe mais calor\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

b) Observe no gr\u00e1fico\u00a0 — \u0394\u03b8<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>=56 \u2013 46\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u0394\u03b8<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>=10<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

c) c\u00e1lculo da massa de asfalto\u00a0 —\u00a0 d=m\/v\u00a0 —\u00a0 2,3.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=m\/(0,1.10.000)\u00a0 —\u00a0 m=2,3.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 m=2,3.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0kg\u00a0 —\u00a0 Q=m.c.\u0394\u03b8\u00a0 —\u00a0 \u00a0Q=2,3.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>.0,75.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.(58 \u2013 30)\u00a0 —\u00a0<\/b><\/span><\/span>Q=48,3.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>J=4,83.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>kJ<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Entre -10<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C e 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0\u00a0\u00a0 Q=m.c.\u0394\u03b8\u00a0 —\u00a0 100=m.c.(0 \u2013 (-10))\u00a0 —\u00a0 m.c=10\u00a0 —\u00a0 entre 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C e 10<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 (Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span><\/sup>\u2013 100)=m.c.\u0394\u03b8\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

(Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 100)=10.10\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=200<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a)\u00a0<\/b><\/span><\/span>\u00e1gua=40<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 \u00f3leo=60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

b)\u00a0<\/b><\/span><\/span>\u00e1gua\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 no mesmo tempo de aquecimento, mesmo Q, (t=1,5min) sofre menor varia\u00e7\u00e3o de temperatura (40<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C) contra o \u00f3leo (60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C).<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) c<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>=Q\/m(40 \u2013 0)\u00a0 —\u00a0 c<\/b><\/span><\/span>\u00f3leo<\/b><\/span><\/span><\/sub>=Q\/m(60 \u2013 0)\u00a0 —\u00a0 c<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\/c<\/b><\/span><\/span>\u00f3leo<\/b><\/span><\/span><\/sub>=Q\/40m x 60m\/Q=3\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>c<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\/c<\/b><\/span><\/span>\u00f3leo<\/b><\/span><\/span><\/sub>=3\/2<\/b><\/span><\/span><\/p>\n

b) c<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\/c<\/b><\/span><\/span>\u00f3leo<\/b><\/span><\/span><\/sub>=3\/2\u00a0 —\u00a0 1\/c<\/b><\/span><\/span>\u00f3leo<\/b><\/span><\/span><\/sub>=3\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>c<\/b><\/span><\/span>\u00f3leo<\/b><\/span><\/span><\/sub>=2\/3 cal\/g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) A potencia P dissipada pelo resistor e dada por P = U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R = 144\/100 = 1,44 W\u00a0 —\u00a0 portanto, ao final de uma hora, o calor recebido pelo volume de \u00e1gua ser\u00e1\u00a0 —\u00a0 Q = 1,44 \u00d7 3600 = 5,2 \u00d7 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0J\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0Q=5,2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>J<\/b><\/span><\/span><\/p>\n

b) De acordo com o gr\u00e1fico, vemos que a capacidade t\u00e9rmica C correspondente aquela massa de \u00e1gua \u00e9\u00a0 —\u00a0 C = Q \/\u0394T = 1,44 \u00d7 3600\/12 = 4,32 \u00d7 10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0J\/\u00b0C\u00a0 —\u00a0 portanto, a massa ser\u00e1\u00a0 —\u00a0 m = C\/c = 4,32 \u00d710<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/ 4,2 = 102,8 g\u00a0 —\u00a0 assim sendo, o volume V ser\u00e1 V V= m\/\u03bc = 102,8\/1\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0V=102,8 cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.\u0394T<\/b><\/span><\/span>A\u00ad<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 20=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.\u0394.40\u00a0 —\u00a0 m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=1\/2\u00a0 —\u00a0 c<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=1\/2m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.\u0394T<\/b><\/span><\/span>B\u00ad<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 40=30m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.\u0394.40\u00a0 —\u00a0 m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=1\/30\u00a0 —\u00a0 c<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=1\/30m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 c<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/c<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=1\/2m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0x 30m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/1\u00a0 —<\/b><\/span><\/span>\u00a0c<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/c<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=15<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Em 20s a fonte t\u00e9rmica fornece\u00a0 —\u00a0 Q=60.20=1.200 cal\u00a0 —\u00a0 Q=m.c.\u0394T\u00a0 —\u00a0 1.200=mc.(60 \u2013 20)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0mc=C=30 cal\/<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) C<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/\u0394\u03b8<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=300\/40\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>C<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=7,5 J\/<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 C<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/\u0394\u03b8<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=300\/60\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>C<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=5,0 J\/<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0<\/b><\/span><\/span><\/p>\n

b) c<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=2c<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 C<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 7,5=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 C<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 5=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u00a0—\u00a0 7,5\/5=m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>c<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.2c<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=3<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Q=m.c.\u0394t=250.1.80\u00a0 —\u00a0 Q=20.000 cal\u00a0\u00a0 —\u00a0 como o microondas fornece 10.000 calorias a cada 1 minuto, ele fornecer\u00e1<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>20.000 calorias em 2 minutos\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- 2 minutos<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>d=m\/v\u00a0 —\u00a0 1,2=m\/8\u00a0 —\u00a0 m=9,5g\u00a0 —\u00a0 Q=m.c.\u0394t=9,6.0,24.(37-20)\u00a0 —\u00a0 Q=39,68 cal em 1 minuto\u00a0 —\u00a0 em 24h\u00a0 —\u00a0 Q=24x60x39,68\u00a0 —\u00a0 Q=56.401,92 cal\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>01. Correta\u00a0 —\u00a0 quanto menor o calor espec\u00edfico , mais o corpo absorve calor<\/b><\/span><\/span><\/p>\n

02. Errada\u00a0 —\u00a0 \u00e9 maior, pois ela perde menos calor que a terra, por ter maior calor espec\u00edfico<\/b><\/span><\/span><\/p>\n

04. Errada\u00a0 —\u00a0 a massa de ar junto \u00e0 terra, mais aquecida sobe, cedendo lugar a massa de ar mais fria que vem do mar\u00a0 —\u00a0 durante o dia a brisa sopra do mar para a terra\u00a0 —\u00a0 \u00e0 noite ocorre o contr\u00e1rio.<\/b><\/span><\/span><\/p>\n

08- Errada\u00a0 —\u00a0 veja 04<\/b><\/span><\/span><\/p>\n

16- Errada\u00a0 —\u00a0 a \u00e1gua resfriou mais lentamente.<\/b><\/span><\/span><\/p>\n

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

64- Errada\u00a0 —\u00a0 possuem calores espec\u00edficos diferentes.<\/b><\/span><\/span><\/p>\n

R- (01 + 32)=33<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quantidade de calor absorvida pelo pur\u00ea\u00a0 —\u00a0 Q=m.c.\u0394\u03b8=1.000×1,8x(50 \u2013 20)\u00a0 —\u00a0 Q=54.000calx4,18\u00a0 —\u00a0 Q=225.720J\u00a0 —\u00a0 80% da pot\u00eancia de 1.200W \u00e9 transferida para o prato\u00a0 —\u00a0 P=Q\/\u0394t\u00a0 —\u00a0 0,8×1.200=225.720\/\u0394t\u00a0 —\u00a0 \u0394t=235,125s\/60=3,918 minutos\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Se a densidade da \u00e1gua \u00e9 1kg\/L, a massa de 1L de \u00e1gua \u00e9 m=1kg\u00a0 —\u00a0 Q=m.c.\u0394\u03b8=1.000×4,18x(100 \u2013 20)\u00a0 —\u00a0 Q=334.400J\u00a0 —\u00a0 pot\u00eancia do ebulidor\u00a0 —\u00a0 P=Q\/\u0394t=334.400\/2×60\u00a0 —\u00a0 P=2.786,7W\u00a0 —\u00a0 P=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 2.786,7=120<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 R=14.400\/2.786,7\u00a0 —\u00a0 R=5,17 \u038f\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>40% de 13kg=0,4×13=5,2kg\u00a0 —\u00a0 6.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>cal \u2013 1kg\u00a0 —\u00a0 Q \u2013 5,2kg\u00a0 —\u00a0 Q=6.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>x5,2\u00a0 —\u00a0 Q=31,2.10<\/b><\/span><\/span>6\u00a0<\/b><\/span><\/span><\/sup>cal\u00a0 —\u00a0 Q=m.c.<\/b><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>t\u00a0 —\u00a0 31,2.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>=m.1.(100 \u2013 20)\u00a0 —\u00a0 m=0,39.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0g=39.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0g\u00a0 —\u00a0 1L \u2013 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>g\u00a0 —\u00a0 xL \u2013 39.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>g\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>x=390 L<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>P=3 kW\u00a0 —\u00a0 Q=m.c.<\/b><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>t=30×4.200x(60 \u2013 25)\u00a0 —\u00a0 Q=4.410.000 J\u00a0 —\u00a0 P=Q\/\u0394t\u00a0 —\u00a0 3.000=4.410.000\/\u0394t\u00a0 —\u00a0 \u0394t=1.470s\/60=24,5s\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

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

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.c.(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) + m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.c.(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) = 0\u00a0 —\u00a0 m.c.(t<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 70) = 2m.c.(t<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 10) = 0\u00a0 —\u00a0 3t<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=90\u00a0 —\u00a0 t<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/span><\/sub>.c<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/span><\/sub>.(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>) + m<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span><\/sub>.c<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span><\/sub>.(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>) = 0\u00a0 —\u00a0 250.c<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/span><\/sub>.(20 \u2013 40) + 200.1.(20\u2013 0) = 0\u00a0<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>-5.000c<\/b><\/span><\/span><\/span>l\u00a0<\/b><\/span><\/span><\/span><\/sub>+ 4.000 = 0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/span><\/sub>=0,8 cal\/g<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span>cal<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>cal<\/b><\/span><\/span><\/sub>.(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) + m<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) + m<\/b><\/span><\/span>al<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>al<\/b><\/span><\/span><\/sub>.(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=0\u00a0 —\u00a0 40.(80 \u2013 90) + 110.1.(80 \u2013 90) + m<\/b><\/span><\/span>al<\/b><\/span><\/span><\/sub>.0,2.(80 \u2013 20)=0\u00a0 — \u00a0-400 \u2013 1.100 + 12m<\/b><\/span><\/span>al<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0m<\/b><\/span><\/span>al<\/b><\/span><\/span><\/sub>=125g<\/b><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>.(t<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 t<\/b><\/span><\/span>ox<\/b><\/span><\/span><\/sub>) + m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>.(t<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 t<\/b><\/span><\/span>oa<\/b><\/span><\/span><\/sub>) = 0\u00a0 —\u00a0 1.000.c<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>.(40 \u2013 100) + 300.1.(40 \u2013 20) = 0\u00a0 —\u00a0 1.000.c<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>.(-60) + 300.(20) = 0\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

c<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>=0,1 cal\/g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C.<\/b><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Primeira experi\u00eancia\u00a0 —\u00a0 \u00e1gua \u2013 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>.(31 \u2013 20)=1.100c<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 objeto \u2013 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=m.c.(31 \u2013 t)\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=-Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1.100c<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=mc(t \u2013 31) (I)<\/b><\/span><\/span><\/p>\n

Segunda experi\u00eancia\u00a0 —\u00a0 \u00e1gua \u2013 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>(40 \u2013 20)=2.000c<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 objetos \u2013 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2mc(40 \u2013 t)\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=-Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 — 1.000c<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=mc(t \u2013 40) (II)<\/b><\/span><\/span><\/p>\n

(I)\/(II)\u00a0 —\u00a0 1.100c<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span><\/sub>\/1.100c<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span><\/sub>=mc(t \u2013 31)\/mc(t \u2013 40)\u00a0 —\u00a0 11\/10=(t \u2013 31)\/(t \u2013 40)\u00a0 —\u00a0 11t \u2013 10t=440 \u2013 310\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>t=130<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Princ\u00edpio da Conserva\u00e7\u00e3o da Energia. Nos sistemas fechados, as trocas internas ocorrem com conserva\u00e7\u00e3o da quantidade de energia.<\/b><\/span><\/span><\/p>\n

b) 10.c.(38 \u2013 32) + m.c.(38 \u2013 100)=0\u00a0 —\u00a0 60 \u2013 62m=0\u00a0 —\u00a0 m=0,97\u00a0 —\u00a0 V\u22481L<\/b><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>mar.car.\u0394t + m\u00e1gua.c\u00e1gua.\u0394t = 0\u00a0 —\u00a0 mar.0,25.\u0394t + m\u00e1gua.1.\u0394t = 0\u00a0 —\u00a0 mar\/m\u00e1gua=1\/0,25\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

35-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) m<\/b><\/span><\/span>af<\/b><\/span><\/span><\/sub>.c.(t<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) + m<\/b><\/span><\/span>aq<\/b><\/span><\/span><\/sub>.c.(t<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=0\u00a0 —\u00a0 100.c.(30 – 20) + m<\/b><\/span><\/span>aq<\/b><\/span><\/span><\/sub>.c.(30 – 50)=0\u00a0 —\u00a0 100.c.(10) + m<\/b><\/span><\/span>aq<\/b><\/span><\/span><\/sub>.c.(-20)=0\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0m<\/b><\/span><\/span>aq<\/b><\/span><\/span><\/sub>=50L\u00a0<\/b><\/span><\/span><\/p>\n

b) 0,20L \u2013 1s\u00a0 —\u00a0 50L \u2013 t s\u00a0 —\u00a0 0,2t=50\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0t=250s=4min e 10s<\/b><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1=m<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\/8×250\u00a0 —\u00a0 m<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>=2.000g (massa de \u00e1gua ingerida)\u00a0 —\u00a0 quantidade de calor fornecida pelo organismo da menina para elevar a temperatura dessa massa de \u00e1gua de 6,5<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C at\u00e9 36,5<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q=m.c.(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=2.000.1.30\u00a0 —\u00a0 Q=60.000cal=60kcal\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>.c.(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>) + m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>.c.(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>) = 0\u00a0 —\u00a0 \u00a0m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>.c.(26 – 20) + m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>.c.(26 \u2013 80) = 0\u00a0 —\u00a0 6m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>\u00a0\u2013 54m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>=0\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>=9m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>=10\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

9m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=10\u00a0 —\u00a0 m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=1L\u00a0 —\u00a0 m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=9L\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>um com 9L e outro com 1L.<\/b><\/span><\/span><\/p>\n

38-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Energia cin\u00e9tica dissipada\u00a0 —\u00a0 E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=30.000.(20)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=6,0.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>J\u00a0 —\u00a0 dessa energia, cada freio absorve sob forma de calor\u00a0 — Q<\/b><\/span><\/span>freio<\/b><\/span><\/span><\/sub>=6.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\/6\u00a0 —\u00a0 Q<\/b><\/span><\/span>freio<\/b><\/span><\/span><\/sub>=10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>J\u00a0 —\u00a0 Q<\/b><\/span><\/span>freio<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>freio<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>freio<\/b><\/span><\/span><\/sub>.\u0394t<\/b><\/span><\/span>freio<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>=8.1.500. \u0394t<\/b><\/span><\/span>freio<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u0394t<\/b><\/span><\/span>freio<\/b><\/span><\/span><\/sub>=83<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

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

\"\"<\/p>\n

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

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

\"\"<\/p>\n

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

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

\"\"<\/p>\n

43-<\/b><\/span><\/span><\/span> <\/b>a) A com C\u00a0 —\u00a0 m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) + m<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>.(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) = 0\u00a0 —\u00a0 m.c<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.(30 \u2013 60) + 3m.c<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>.(30 \u2013 20) = 0\u00a0 —\u00a0 c<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=c<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 B com C\u00a0 —\u00a0 2m.c<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.(25 \u2013 40) + 3m.c<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>.(25 \u2013 20) = 0\u00a0 —\u00a0 -30c<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0+ 15c<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 c<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=c<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\/2 ou c<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=c<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/2\u00a0 —\u00a0 A com B\u00a0 —\u00a0 m.c<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.(t \u2013 60) + 2mc<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.(t \u2013 40)=0\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

m.c<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/span><\/sub>.(t \u2013 60) + 2mc<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/span><\/sub>\/2.(t \u2013 40)=0\u00a0 —\u00a0 t \u2013 60 = 40 \u2013 t\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0t=50<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/span><\/p>\n

b) c<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/span><\/sub>=0,5 cal\/g<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sup>C\u00a0 — c<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/span><\/sub>=c<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/span><\/sub>\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0c<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/span><\/sub>=0,25 cal\/g<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/span><\/p>\n

44-<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span>m.c.(T<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span><\/sub>\u00a0\u2013 T) + m\/2.c.(T<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span><\/sub>\u00a0\u2013 T\/2) + m\/4.c.(T<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span><\/sub>\u00a0\u2013 T\/4) = 0\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span><\/sub>\u00a0\u2013 T + T<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span><\/sub>\/2 \u2013 T\/4 + T<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span><\/sub>\/4 \u2013 T\/16=0\u00a0 —\u00a0 28T<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span><\/sub>=21T\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>=3T\/4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

45-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) A soma dos calores trocados deve ser igual a zero.\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ … + Q<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0= 0\u00a0 —\u00a0 (m.c.DT)<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ (m.c.DT)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ … + (m.c.DT)<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0= 0\u00a0 —\u00a0 Como o l\u00edquido \u00e9 o mesmo em todos os recipientes\u00a0 —\u00a0 (m.DT)<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ (m.DT)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ … + (m.DT)<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0= 0\u00a0 —\u00a0 m.(T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0\u2013 T) + (m\/2).(T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0\u2013 T\/2) + …+ (m\/2<\/b><\/span><\/span>N-1<\/b><\/span><\/span><\/sup>).(T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0\u2013 T\/2<\/b><\/span><\/span>N-1<\/b><\/span><\/span><\/sup>) = 0\u00a0 —\u00a0 Ssimplificadas as massas m\u00a0 —\u00a0 (T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0\u2013 T) + (1\/2).(T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0\u2013 T\/2) + …+ (1\/2<\/b><\/span><\/span>N-1<\/b><\/span><\/span><\/sup>).(T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0\u2013 T\/2<\/b><\/span><\/span>N-1<\/b><\/span><\/span><\/sup>) = 0\u00a0 —\u00a0 T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0\u2013 T + (1\/2).T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0\u2013 T\/2<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ …+ (1\/2<\/b><\/span><\/span>N-1<\/b><\/span><\/span><\/sup>)T<\/b><\/span><\/span>f\u00a0<\/b><\/span><\/span><\/sub>\u00a0\u2013 T\/2<\/b><\/span><\/span>2N – 2<\/b><\/span><\/span><\/sup>\u00a0= 0\u00a0 —\u00a0\u00a0 T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>.[1 + 1\/2 + 1\/4 +… + 1\/2<\/b><\/span><\/span>N-1<\/b><\/span><\/span><\/sup>] = T.[1 + 1\/4 + 1\/16 + … + 1\/2<\/b><\/span><\/span>2N \u2013 2<\/b><\/span><\/span><\/sup>] = 0\u00a0 —\u00a0 aplicando a soma dos termos de uma PG\u00a0\"\"\u00a0S = a<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.(q<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sup>\u00a0\u2013 1)\/(q \u2013 1) \u00a0—\u00a0 T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>.2.(1 \u2013 1\/2<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sup>) = (4T\/3).(1 \u2013 1\/4<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0 T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>.(1 \u2013 1\/2<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sup>) = (2T\/3).(1 \u2013 1\/4<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0 T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0= (2T\/3).(1 \u2013 1\/4<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sup>)\/(1 \u2013 1\/2<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sup>)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 b) Para N tendendo a infinito\u00a0 —\u00a0\u00a0 T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0= (2T\/3). (1 \u2013 0)\/(1 \u2013 0) = 2T\/3<\/b><\/span><\/span><\/p>\n

46-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>L\u00edquido\u00a0 —\u00a0 Q<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>.(t<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 entre 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C e 120<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 300.000= m<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>.(120 \u2013 0)\u00a0 —\u00a0 m<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>=2.500\u00a0 —\u00a0 Q<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>.(t<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 entre 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C e 100<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 100.000= m<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>.(100 \u2013 0)\u00a0 —\u00a0 m<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>=1.000\u00a0 —\u00a0 misturando-os\u00a0 —\u00a0 m<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>.(t<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) + m<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>.(t<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) = 0\u00a0 —\u00a0 2.500.(t<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u2013 10) + 1.000.(t<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 115) = 0\u00a0 —\u00a0 3.500t<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0= 140.000\u00a0 —\u00a0 t<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=40<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

47-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>I. Q=m.c.\u0394\u03b8\u00a0 —\u00a0 300=m.0,3.10\u00a0 —\u00a0 m=100g\u00a0 —\u00a0 II. Q=m.c.\u0394\u03b8\u00a0 —\u00a0 400=m.0,2.10\u00a0 —\u00a0 m=200g\u00a0 —\u00a0 III. Q=m.c.\u0394\u03b8450=150.c.10\u00a0 —\u00a0 c=0,3 cal\/g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 IV. Q=m.c.\u0394\u03b8\u00a0 —\u00a0 Q=150.0,4.10\u00a0 —\u00a0 Q= 600cal\u00a0 —\u00a0 V. Q=m.c.\u0394\u03b8=100.0,5.10\u00a0 —\u00a0 Q=500cal\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

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

49-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O metal tem menor calor espec\u00edfico que a madeira e tem maior facilidade em perder calor, se resfriando mais r\u00e1pido e liberando menos calor, pois, supondo que o bloco de madeira tenha calor espec\u00edfico dez vezes\u00a0 maior, a capacidade t\u00e9rmica do bloco de madeira\u00a0 ser\u00e1 dez\u00a0 maior que a do bloco de chumbo de mesma massa. Como os blocos sofrem a mesma queda de temperatura, o bloco de madeira ir\u00e1 fornecer dez vezes mais calor que o bloco de chumbo e assim sendo ir\u00e1 derreter mais gelo do que o bloco de chumbo\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

50-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Para um aquecimento de 12.000cal ser\u00e1 necess\u00e1rio um aumento de temperatura \u0394\u03b8 de\u00a0 —\u00a0 Q=m.c. \u0394\u03b8\u00a0 —\u00a0 12.000=500.1.\u0394\u03b8\u00a0 —\u00a0 \u0394\u03b8=24<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 acima de 24<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C, pelo gr\u00e1fico\u00a0 —\u00a0 F, G, H, J e K\u00a0 —\u00a0<\/b><\/span><\/span>\u00a05 capitais<\/b><\/span><\/span><\/p>\n

51-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como o calor espec\u00edfico c \u00e9 o mesmo (\u00e1gua) e ambas recebem a mesma quantidade de calor Q aquele de menor massa sofrer\u00e1 maior eleva\u00e7\u00e3o de temperatura\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

52-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>01. Falsa\u00a0 —\u00a0 depende tamb\u00e9m da massa dos corpos.<\/b><\/span><\/span><\/p>\n

02. Correta\u00a0 —\u00a0 veja teoria.<\/b><\/span><\/span><\/p>\n

04. Correta\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

08- Falsa\u00a0 —\u00a0 \u00e9 maior nas subst\u00e2ncias l\u00edquidas.<\/b><\/span><\/span><\/p>\n

R- (02 + 04)=06<\/b><\/span><\/span><\/p>\n

53-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0,9=m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/4\u00a0 —\u00a0 m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=3,6kg\u00a0 —\u00a0 Q<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=3,6.0,5.(200 \u2013 20)\u00a0 —\u00a0 Q<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=324kcal\u00a0 —\u00a0 80% – 324kcal\u00a0 —\u00a0 100% – Q\u00a0 — Q=405kcal em 12min=1\/5h\u00a0 —\u00a0 \u03a6=Q\/\u0394t=405\/(1\/5)\u00a0 —\u00a0 \u03a6=2.025kcal\/h\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

54-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Q=mc(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=500.1.(80 \u2013 30)\u00a0 —\u00a0 Q=25.000cal\u00a0 —\u00a0 10.000cal \u2013 1 minuto\u00a0 —\u00a0 25.000cal \u2013 t minutos\u00a0 —\u00a0 t=25.000\/10.000\u00a0 —\u00a0 t=2,5 min\u00a0 —<\/b><\/span><\/span>\u00a0 R- B<\/b><\/span><\/span><\/p>\n

55-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>250.1.(20 \u2013 10) + 50.0,0924.(20 \u2013 80) + 50.c.(20 \u2013 100) = 0 —\u00a0\u00a0 2.500 \u2013 277,2 \u2013 4.000c = 0\u00a0 —\u00a0 c=2.222\/4.000=0,5557 cal\/g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

57-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo o poder calor\u00edfico da gasolina 3,6.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>J\/L\u00a0 —\u00a0 1L de gasolina libera Q=3,6.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>J de energia t\u00e9rmica que \u00e9 utilizada no aquecimento de uma massa m de N<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>N0<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0 Q=mc\u0394\u03b8\u00a0 —\u00a0 3,6.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>=m.1,2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.(550 \u2013 300)\u00a0 —\u00a0 m=120kg\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

59-<\/b><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/span>Dados: massa de \u00e1gua \u00a0\u00a0—\u00a0 \u00a0m = 1 kg; varia\u00e7\u00e3o de temperatura \u00a0\u00a0—\u00a0 \u00a0DT = 80 \u2013 30 = 50 \u00b0C; Tens\u00e3o el\u00e9trica \u00a0\u00a0—\u00a0 \u00a0U = 100 V; calor espec\u00edfico da \u00e1gua \u00a0\u00a0—\u00a0 \u00a0c = 4,2\u00b410<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0J\/kg.\u00b0C e varia\u00e7\u00e3o de tempo \u00a0\u00a0—\u00a0 \u00a0Dt = 10 min = 600 s\u00a0 —\u00a0 Testando cada uma das afirma\u00e7\u00f5es\u00a0 —\u00a0<\/b><\/span><\/span>I. Correta<\/b><\/span><\/span>\u00a0 —\u00a0 Q = mcDT = 1(4,2\u00b410<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>)(50) \u00de Q = 2,1\u00b410<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0J\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u00a0II. Errada\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P =\u00a0\"\"W\u00a0 —<\/b><\/span><\/span>III. Correta<\/b><\/span><\/span>\u00a0 —\u00a0 P =\u00a0\"\"W\u00a0<\/b><\/span><\/span>\u00a0— \u00a0R- C<\/b><\/span><\/span><\/p>\n

60-<\/b><\/span><\/span><\/span> <\/b>Dados\u00a0 —\u00a0 \u00a0Q<\/b><\/span><\/span>Al<\/b><\/span><\/span><\/sub>\u00a0= Q<\/b><\/span><\/span>Fe<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u00a0c<\/b><\/span><\/span>Al<\/b><\/span><\/span><\/sub>\u00a0= 2<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>c<\/b><\/span><\/span>Fe<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 DT<\/b><\/span><\/span>Al<\/b><\/span><\/span><\/sub>\u00a0= DT<\/b><\/span><\/span>Fe<\/b><\/span><\/span><\/sub>\u00a0= DT\u00a0 —\u00a0 Q<\/b><\/span><\/span>Al<\/b><\/span><\/span><\/sub>\u00a0= Q<\/b><\/span><\/span>Fe<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u00a0\u00a0m\u00ad<\/b><\/span><\/span>Al\u00a0<\/b><\/span><\/span><\/sub>c<\/b><\/span><\/span>Al<\/b><\/span><\/span><\/sub>DT = m<\/b><\/span><\/span>Fe\u00a0<\/b><\/span><\/span><\/sub>c<\/b><\/span><\/span>Fe\u00a0<\/b><\/span><\/span><\/sub>DT\u00a0 —\u00a0 \u00a0\u00a0\u00a0m<\/b><\/span><\/span>Al\u00a0<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>c<\/b><\/span><\/span>Fe<\/b><\/span><\/span><\/sub>\u00a0= m<\/b><\/span><\/span>Fe\u00a0<\/b><\/span><\/span><\/sub>c<\/b><\/span><\/span>Fe<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0\u00a0 m<\/b><\/span><\/span>Al<\/b><\/span><\/span><\/sub>\u00a0=\u00a0\"\"\u00a0\u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

62-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span>T<\/b><\/span><\/span><\/sub>\u00a0= 0,5 W\u00a0 —\u00a0 \u00a0h = 50%\u00a0 —\u00a0 \u00a0m = 100 g\u00a0 —\u00a0\u00a0 c = 4,2 J\/g.\u00b0C\u00a0 —\u00a0 quantidade de calor necess\u00e1ria para aquecer a massa de \u00e1gua de 1\u00b0C\u00a0 —\u00a0 Q = m<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>c<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>Dt\u00a0 —\u00a0 \u00a0Q = 100<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>(4,2)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>(1)\u00a0 —\u00a0\u00a0\u00a0 Q = 420 J\u00a0 —\u00a0 pot\u00eancia \u00fatil\u00a0 —\u00a0 P<\/b><\/span><\/span>u<\/b><\/span><\/span><\/sub>\u00a0= h<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>P<\/b><\/span><\/span>T<\/b><\/span><\/span><\/sub>\u00a0= 0,5<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>(0,5) = 0,25 W\u00a0 —\u00a0 P<\/b><\/span><\/span>u<\/b><\/span><\/span><\/sub>\u00a0=\u00a0\"\"\u00a0\u00a0— \u00a0\"\"s\u00a0 —\u00a0 \u00a0\u00a0Dt = 28 min\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span>\u00a0 \u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

63-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>c = 4.200 J\u00a0 —\u00a0\u00a0 DT = (100 \u2013 28) = 72 \u00b0C\u00a0 —\u00a0 Q = DE = 3,6 \u00b4 10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>\u00a0J\u00a0 —\u00a0 Q = m c DT \u00a0\u00a0—\u00a0 m=Q\/c\u0394T=3,6.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>\/4,2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.72\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

m = 1.190 kg\u00a0\u00a0 —\u00a0\u00a0 m @ 1,2 \u00b4 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0kg\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

64<\/b><\/span><\/span><\/span>–\u00a0<\/b><\/span><\/span>Q<\/b><\/span><\/span>\u00e1gua +\u00a0<\/b><\/span><\/span><\/sub>Q<\/b><\/span><\/span>porc<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 m<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>c<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) + m<\/b><\/span><\/span>porc<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>\u00a0porc<\/b><\/span><\/span><\/sub>..(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=0\u00a0 — 150.c<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>(t \u2013 100) + 125.c<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\/5.(t \u2013 9) = 0\u00a0 —\u00a0 15t \u2013 15.000 + 25t \u2013 225 = 0\u00a0 —\u00a0 t=87<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

65- <\/b><\/span><\/span><\/span>(<\/b><\/span><\/span>01) Falsa\u00a0 — a varia\u00e7\u00e3o de temperatura depende da quantidade de calor recebida (Q) e da capacidade t\u00e9rmica do corpo (C), de acordo com a express\u00e3o \u2206t=Q\/C\u00a0 —\u00a0 temos, ent\u00e3o, que se dois corpos apresentarem a mesma raz\u00e3oQ\/C, apresentar\u00e3o a mesma varia\u00e7\u00e3o de temperatura\u00a0 —\u00a0 se estiverem \u00e0 mesma temperatura inicial, apresentar\u00e3o a mesma temperatura final.<\/b><\/span><\/span><\/p>\n

(02) Correta\u00a0 — temperatura \u00e9 a medida do estado de agita\u00e7\u00e3o das part\u00edculas, ou seja, da energia cin\u00e9tica m\u00e9dia das part\u00edculas.<\/b><\/span><\/span><\/p>\n

(04) Correta. A unidade de calor espec\u00edfico \u00e9 cal\/g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 observe nessa unidade que, quanto maior o calor espec\u00edfico, mais energia (calorias) ser\u00e3o necess\u00e1rias para aumentar de 1 \u00b0C uma unidade de massa\u00a0 — o calor espec\u00edfico representa uma esp\u00e9cie de dificuldade da subst\u00e2ncia em variar temperatura.<\/b><\/span><\/span><\/p>\n

(08) Falsa\u00a0 —\u00a0 basta observar aqui que o calor espec\u00edfico da \u00e1gua (l\u00edquida) \u00e9 1 cal\/(g.\u00b0C), enquanto que o calor espec\u00edfico do alum\u00ednio (s\u00f3lido) \u00e9 0,22 cal\/(g.\u00b0C).\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

R- (02 + 04) = 06<\/b><\/span><\/span><\/p>\n

66-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 P = 100 W\u00a0 — \u00a0m = 60 kg\u00a0 —\u00a0 c = 4,2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0J\/kg\u00b0C\u00a0 —\u00a0 \u2206\u03b8 =5<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 express\u00e3o da pot\u00eancia\u00a0 —\u00a0 P=Q\/\u2206t\u00a0 —\u00a0 P=mc\u2206\u03b8\/\u2206t\u00a0 —\u00a0 \u2206t=60.4,2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.5\/100\u00a0 —\u00a0 \u2206t=12.600s=12.600\/3.600\u00a0 —\u00a0 \u2206t=3,5h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

67-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 — \u00a0m = 80 kg = 80.000 g\u00a0 — \u2206t = 40 \u2013 36,5 = 3,5 \u00b0C\u00a0 — \u00a0c = 1 cal\/g\u00b0C\u00a0 —\u00a0 equa\u00e7\u00e3o do calor sens\u00edvel\u00a0 —\u00a0 Q = m c \u2206t \u00a0\u00a0— \u00a0Q = 80.000x1x3,5 = 280.000 cal\u00a0 —\u00a0 Q = 280 kcal\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

68-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 — \u00a0m = 5 g\u00a0 —\u00a0 c = 0,8 J\/g\u00b7\u00b0C\u00a0 —\u00a0 \u2206\u03b8= [880 \u2013 (-20)] = 900 \u00b0C\u00a0 —\u00a0 equa\u00e7\u00e3o fundamental da calorimetria\u00a0 —\u00a0 Q = m c \u2206\u03b8 = (5).(0,8).(900)\u00a0 —\u00a0 Q = 3.600 J\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

 <\/p>\n

70- <\/b><\/span><\/span><\/span>\u00a0a) Quantidade de calor absorvido pelo calor\u00edmetro\u00a0 —\u00a0 temperatura inicial do calor\u00edmetro=20<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 temperatura final do calor\u00edmetro=30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>.\u2206\u03b8<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 mas m<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=C<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0 Q<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=C<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>. \u2206\u03b8<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=10.(30 \u2013 20)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=100 cal<\/b><\/span><\/span>\u00a0 —\u00a0 quantidade de calor absorvido pela \u00e1gua\u00a0 —\u00a0 as temperaturas inicial e final da \u00e1gua s\u00e3o as mesmas que do calor\u00edmetro\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>. \u2206\u03b8<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=500.1.(30 \u2013 20)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=5 000 cal.<\/b><\/span><\/span><\/p>\n

b) A temperatura final de todos os elementos do sistema \u00e9 30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 n\u00e3o havendo perdas de calor para o meio externo, os tr\u00eas elementos trocam calor apenas entre si\u00a0 —\u00a0 a soma das quantidades de calor trocadas por eles deve ser nula\u00a0 —\u00a0 Q<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0+ Q<\/b><\/span><\/span>calor\u00edmetro\u00a0<\/b><\/span><\/span><\/sub>+ Q<\/b><\/span><\/span>barra<\/b><\/span><\/span><\/sub>\u00a0= 0\u00a0 —\u00a0 5000 + 100 + m<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>. \u2206\u03b8<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0= 0\u00a0 —\u00a0 5100 + 200.c<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>.(30 \u2013 80) = 0\u00a0 —\u00a0 c<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=5100\/10000\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>c<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=0,51cal\/g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

71- <\/b><\/span><\/span><\/span>Ocorre trocas de calor apenas entre o leite, o caf\u00e9 e o ado\u00e7ante, pois a capacidade t\u00e9rmica da x\u00edcara, segundo o enunciado, \u00e9 desprez\u00edvel\u00a0 —\u00a0 sendo as densidades as mesmas d=m\/V\u00a0 —\u00a0 m=d.V\u00a0 —\u00a0 sendo t a temperatura final da mistura e d a densidade de cada elemento que participa das trocas de calor\u00a0 —\u00a0 n\u00e3o havendo perda de calor para o ambiente, a soma das quantidades de calor trocadas \u00e9 nula\u00a0 —\u00a0 Qleite\u00a0+ Qcaf\u00e9 +\u00a0Qado\u00e7ante\u00a0= 0\u00a0 —\u00a0 mleite.cleite.(t \u2013 to) + mcaf\u00e9.ccaf\u00e9.(t \u2013 to) + mado\u00e7ante.cado\u00e7ante.(t \u2013 to) = 0\u00a0 —\u00a0 d.Vleite.cleite.(t \u2013 to) + d.Vcaf\u00e9.ccaf\u00e9.(t \u2013 to) + d.Vado\u00e7ante.cado\u00e7ante.(t \u2013 to) = 0\u00a0 —\u00a0 as densidades se cancelam e as unidades mL tamb\u00e9m podem ser mantidas, pois tamb\u00e9m se cancelam\u00a0 —\u00a0\u00a0 100.0,9.(t \u2013 50) + 50.1.(t \u2013 80) + 2.2.(t \u2013 20) = 0\u00a0 —\u00a0 90t \u2013 4 500 + 50t \u2013 4 000 + 4t \u2013 80 = 0\u00a0 —\u00a0 144t = 8 580\u00a0 —\u00a0 t = 59,58oC\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/p>\n

 <\/p>\n

72- <\/b><\/span><\/span><\/span>Define-se capacidade t\u00e9rmica (C) ou capacidade calor\u00edfica de um corpo como sendo o produto da massa desse corpo pelo calor espec\u00edfico da subst\u00e2ncia de que ele \u00e9 constitu\u00eddo, ou seja,\u00a0 —\u00a0\u00a0 C=m.c\u00a0 —\u00a0 como Q=m.c.\u0394t\u00a0 —\u00a0 Q=C.\u0394t\u00a0 —\u00a0 ou Q=C\/(t \u2013 to)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Se, por exemplo, a capacidade t\u00e9rmica de um corpo \u00e9 C=40cal\/oC, isto significa que, quando esse corpo receber ou ceder 40 calorias, sua temperatura aumentar\u00e1 ou diminuir\u00e1 de 1\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C.<\/b><\/span><\/span><\/p>\n

Observe na express\u00e3o C=Q\/\u2206t que, de acordo com o enunciado, como Q e \u2206t s\u00e3o os mesmos para os dois blocos eles devem possuir a mesma capacidade t\u00e9rmica\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Calor espec\u00edfico (sens\u00edvel) e trocas de calor sem mudan\u00e7a de estado \u00a0 01–\u00a0R- A\u00a0\u00a0—\u00a0 veja teoria 02-\u00a0R- A\u00a0 —\u00a0 veja teoria 03-\u00a0Observe os dados do gr\u00e1fico\u00a0 —\u00a0R- C 04- R- A\u00a0\u00a0—\u00a0 veja teoria 05-\u00a0Observe que eles possuem a mesma massa (m) e recebem a mesma quantidade de calor (Q)\u00a0\u00a0 —\u00a0 Q=m.c.\u0394\u03b8\u00a0 —\u00a0 \u0394\u03b8=Q\/m.c\u00a0<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2032,"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-2036","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2036","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=2036"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2036\/revisions"}],"predecessor-version":[{"id":10734,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2036\/revisions\/10734"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2032"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2036"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}