{"id":2046,"date":"2016-03-08T00:46:14","date_gmt":"2016-03-08T00:46:14","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2046"},"modified":"2024-08-21T14:34:15","modified_gmt":"2024-08-21T14:34:15","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-calor-latente-trocas-de-calor-com-mudanca-de-estado","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/fisica-termica\/calorimetria\/calor-latente-trocas-de-calor-com-mudanca-de-estado\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-calor-latente-trocas-de-calor-com-mudanca-de-estado\/","title":{"rendered":"Calor latente \u2013 trocas de calor com mudan\u00e7a de estado – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Calor latente \u2013 trocas de calor com mudan\u00e7a de estado<\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Aquecimento do gelo\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=m.c.(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=100.0,5.(0 \u2013 (-10))\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=500cal<\/b><\/span><\/span><\/p>\n

Fus\u00e3o do gelo\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=m.L<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=100.80\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=8.000cal<\/b><\/span><\/span><\/p>\n

Aquecimento da \u00e1gua\u00a0 —\u00a0 Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=m.c.(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=100.1.(100 \u2013 0)\u00a0 —\u00a0 Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=10.000cal<\/b><\/span><\/span><\/p>\n

Vaporiza\u00e7\u00e3o da \u00e1gua\u00a0 —\u00a0 Q<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>=m.L<\/b><\/span><\/span>V<\/b><\/span><\/span><\/sub>=100.540\u00a0 —\u00a0 Q<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>=54.000cal<\/b><\/span><\/span><\/p>\n

Aquecimento do vapor\u00a0 —\u00a0 Q<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sub>=m.c.(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=100.0,5.(110 \u2013 100)\u00a0 —\u00a0 Q<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sub>=500cal<\/b><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span>T<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0+ Q<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>\u00a0+ Q<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sub>=73.000cal ou<\/b><\/span><\/span>\u00a0Q<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sub>=73kcal<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

02-<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span>a) Q=m.L\u00a0 —\u00a0 (200 \u2013 100)=5.L\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>L=20cal\/g<\/b><\/span><\/span><\/span><\/p>\n

b) Q=m.c.(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>) \u00a0—\u00a0 (100 \u2013 0)=5.c.(100 \u2013 0)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>c=0,2cal\/g<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span> <\/b>R- C\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 observe no gr\u00e1fico<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) O primeiro trecho de gr\u00e1fico corresponde ao aquecimento do gelo. Pela equa\u00e7\u00e3o fundamental do calor sens\u00edvel, Q = m.c.\u0394T \u00a0\u00a0—\u00a0<\/b><\/span><\/span><\/p>\n

1000 = 100.c.20\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u00a0c = 0,5 cal\/g.\u00b0C<\/b><\/span><\/span><\/p>\n

b) O segundo trecho, uma linha horizontal sobre o eixo das abscissas, corresponde a fus\u00e3o do gelo. Pela equa\u00e7\u00e3o: Q = m.L\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

8000 = 100.L\u00a0 —\u00a0 \u00a0<\/b><\/span><\/span>L = 80 cal\/g<\/b><\/span><\/span><\/p>\n

c) O terceiro e \u00faltimo trecho do diagrama corresponde ao aquecimento da \u00e1gua provinda da fus\u00e3o do gelo. A capacidade t\u00e9rmica \u00e9 dada pelo produto entre a massa e o calor espec\u00edfico sens\u00edvel, ou seja, C = m.c\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>C = 100.1 = 100 cal\/\u00b0C<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>\u00a0= 1,6 \u00d710<\/b><\/span><\/span>22<\/b><\/span><\/span><\/sup>\u00a0J (Energia total acumulada pela Terra)\u00a0 —\u00a0 E<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u00a0= 3,2 \u00d710<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0J (Energia necess\u00e1ria para o derretimento de 1 kg gelo a 0\u00b0C)\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1kg \u2013 3,2.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>J\u00a0 —\u00a0 mkg \u2013 1,6.10<\/b><\/span><\/span>22<\/b><\/span><\/span><\/sup>J\u00a0 —\u00a0 m=1,6.10<\/b><\/span><\/span>22<\/b><\/span><\/span><\/sup>\/3,2.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=0,5.10<\/b><\/span><\/span>17<\/b><\/span><\/span><\/sup>J\u00a0 —\u00a0 m=50.10<\/b><\/span><\/span>15<\/b><\/span><\/span><\/sup>kg\u00a0 —\u00a0 dividindo por 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0passa para tonelada\u00a0 —\u00a0 m=50.10<\/b><\/span><\/span>12<\/b><\/span><\/span><\/sup>ton\u00a0 —\u00a0 m=50 trilh\u00f5es de toneladas\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Durante a fus\u00e3o ele recebe Q=20cal\/s.100s=2.000cal\u00a0 —\u00a0 Q=mL\u00a0 —\u00a0 2.000=100.L\u00a0 —\u00a0 L=20cal\/g\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Aquecendo 2,5.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>g de \u00e1gua de 18<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C a 96<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q=mc(\u03b8 \u2013 \u03b8<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=2,5.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.1.(96 \u2013 18)\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>= 1,95.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>cal\u00a0 —\u00a0 vaporiza\u00e7\u00e3o de 1,0.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>g de \u00e1gua a96<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q=mL=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.540\u00a0 —\u00a0\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=5,4.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>cal\u00a0 —\u00a0 calor total\u00a0 —\u00a0 Q<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=1,95.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 5,4.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Q<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=7,35.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>cal\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=1.000W=1.000J\/s\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=1.000\/4,2cal\/s\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=Q\/\u0394t\u00a0 —\u00a0 1.000\/4,2=7,35.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\/\u0394t\u00a0 —\u00a0<\/b><\/span><\/span>\u0394t=3.087s=51min<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Pelo gr\u00e1fico, em 24h 12kg de massa se fundem\u00a0 —\u00a0 portanto em 1h, se funde m=0,5kg\u00a0 —\u00a0 Q=m.L=0,5.320\u00a0 —\u00a0 Q=160kj\u00a0 —<\/b><\/span><\/span><\/p>\n

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

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Subst\u00e2ncia B durante a fus\u00e3o recebe de calor Q\u00a0 —\u00a0 Q=P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u0394t=20.(90 \u2013 30)\u00a0 —\u00a0 Q= 1.200cal\u00a0 —\u00a0 Q=m.L\u00a0 —\u00a0 1.200=50.L\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>L=24 cal\/g<\/b><\/span><\/span><\/p>\n

b) calor espec\u00edfico da subst\u00e2ncia B\u00a0 —\u00a0 Q=mc\u0394\u03b8\u00a0 —\u00a0 Q=P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u0394t\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u0394t=mc\u0394\u03b8\u00a0 —\u00a0 c<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>. \u0394t\/m\u0394\u03b8=20.30\/50.(80 \u2013 20)\u00a0 —\u00a0 c<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=0,2cal\/g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 energia t\u00e9rmica (calor) liberada pela subst\u00e2ncia B quando sua temperatura diminui de 280<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C a 80<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=m.c.(80 \u2013 280)=50.0,1.(-200)\u00a0 —\u00a0 Q<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>= – 1.000 cal\u00a0 —\u00a0 energia t\u00e9rmica (calor) que a subst\u00e2ncia A deve receber para atingir a temperatura de 80<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=m.c.(80 \u2013 20)=50.0,2.60\u00a0 —\u00a0 Q<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>= 600cal\u00a0 —\u00a0 a fus\u00e3o da subst\u00e2ncia B \u00e9 realizada com Q=1000 \u2013 600=400cal\u00a0 —\u00a0 Q=mL\u00a0 —\u00a0 400=m.24\u00a0 —\u00a0 m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=50\/3g (parte de B que se liquefaz) —\u00a0 no estado s\u00f3lido B fica com m<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>=50 \u2013 50\/3\u00a0 —\u00a0 m<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>=100\/3g\u00a0 —\u00a0 observe que a temperatura de equil\u00edbrio t\u00e9rmico entre A e B \u00e9 de\u00a0<\/b><\/span><\/span>80<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C.<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que, pelo gr\u00e1fico T<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>>T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 como Q=m.L e como Q e m s\u00e3o as mesmas\u00a0 —\u00a0 L<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>>L<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) fus\u00e3o do gelo\u00a0 —\u00a0 Q=m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>L<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 pot\u00eancia da fonte\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=Q\/\u0394t\u00a0 —\u00a0 Q=P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u0394t\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u0394t=m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>.L<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.60=m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>.80\u00a0 —\u00a0 3P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=4m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u00a0(I)\u00a0 —\u00a0 aquecimento da massa (m=m<\/b><\/span><\/span>\u00e1gua\u00a0<\/b><\/span><\/span><\/sub>+ m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>) de 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C a 5<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q\u2019=mc(\u03b8 \u2013 \u03b8<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=(2.400 + m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>).1.(5 \u2013 0)\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u0394t\u2019=(2.400 + m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>).1.(5)\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.20=12.000 + m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 4P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=2.400 + m<\/b><\/span><\/span>gelo\u00a0<\/b><\/span><\/span><\/sub>(II)\u00a0 —\u00a0 dividindo membro a membro (I) e (II)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

3P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/4P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=4m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\/2.400 + m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u2248553,85 g<\/b><\/span><\/span><\/p>\n

b) Substituindo m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>=554,85 em I\u00a0 —\u00a0 3P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=4.553,85\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u2248738,47cal\/min<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span>a) Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>\u00a0+ Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>=0\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>.c<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>.(T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>-T<\/b><\/span><\/span><\/span>o1<\/b><\/span><\/span><\/span><\/sub>) + m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>.c<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>.(T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>-T<\/b><\/span><\/span><\/span>o2<\/b><\/span><\/span><\/span><\/sub>)=0\u00a0 —\u00a0\u00a0 1000 \u00d7 1,0 \u00d7 (T – 55) + 500 \u00d7 1 \u00d7 (T – 25 ) = 0\u00a0 —\u00a0\u00a0 1500 T = 67500\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>T = 45 \u00b0C.<\/b><\/span><\/span><\/span><\/p>\n

b) m=1.000 + 500=1.500kg\u00a0 —\u00a0\u00a0 Q = 1500 \u00d7 1,0 \u00d7 (100 – 45) = 1.500 \u00d7 55 = 82500 cal\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0Q=82,5 kcal<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

c) Para resfriar a \u00e1gua\u00a0 —\u00a0 \u00a0Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= – 1500 \u00d7 1 \u00d7 100\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>= – 150 000 cal\u00a0 —\u00a0\u00a0 Para congelar a \u00e1gua\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= – 80 \u00d7 1500 cal\u00a0\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=- 120 000 cal\u00a0 —\u00a0 portanto o calor total a ser RETIRADO da \u00e1gua ser\u00e1\u00a0 —\u00a0\u00a0 Q<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0= Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= – 270 000 cal\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>= – 270 kcal.<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Calor que a \u00e1gua (m=1kg=1.000g) perde quando fica em superfus\u00e3o, se solidificando de 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C a -5,6<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q=mc\u0394\u03b8=1.000.1.(-5,6 \u2013 0) —\u00a0 Q=-5.600cal\u00a0 —\u00a0 quando a superfus\u00e3o cessa essa quantidade de calor torna s\u00f3lida uma massa de \u00e1gua m com L=-80cal\/g (perde calor ao se solidificar)\u00a0 —\u00a0 Q=mL\u00a0 —\u00a0 -5.600=m.(-80)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m=70g<\/b><\/span><\/span><\/p>\n

b) Quando o bloco met\u00e1lico de massa m=930g \u00e0 temperatura \u03b8<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=91<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C \u00e9 colocado no recipiente onde voc\u00ea tem gelo a 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C, \u00e1gua a 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C, todo o sistema \u00e9 aquecido at\u00e9 \u00e0 temperatura t\u00a0 —\u00a0 Fus\u00e3o do gelo\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=mL=70.80\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=5.600cal\u00a0 —\u00a0 aqucimento da \u00e1gua\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=mc(\u03b8 \u2013 \u03b8<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=1.000.1. (\u03b8 \u2013 0)\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=1.000.\u03b8\u00a0 —\u00a0 resfriamento do bloco met\u00e1lico\u00a0 —\u00a0 Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>= mc(\u03b8 \u2013 \u03b8<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=C.(\u03b8 \u2013 \u03b8<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=400.(\u03b8 \u2013 91)\u00a0 —\u00a0 Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=400.\u03b8 \u2013 36.400\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>3<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 5.600 + 1.000.\u03b8 + 400.\u03b8 \u2013 36.400=0\u00a0 —\u00a0 1.400.\u03b8=30.800\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\u03b8=22<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Pot\u00eancia da fonte t\u00e9rmica\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=energia\/tempo=Q\/\u0394t=m.c.(\u03b8 \u2013 \u03b8<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\/\u0394t=m.1.(100 \u2013 10)\/5\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=18m\u00a0 —\u00a0 intervalo de tempo para a vaporiza\u00e7\u00e3o da \u00e1gua\u00a0 —\u00a0 Q=P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u0394t=m.L<\/b><\/span><\/span>v\u00ad<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 18m.\u0394t=m.540\u00a0 —\u00a0 \u0394t=15min\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Calor necess\u00e1rio para fundir os 2 kg de gelo a 0 \u00b0C\u00a0 —\u00a0 Q = m.L = 2000.80 = 160000 cal\u00a0 —\u00a0 calor necess\u00e1rio para resfriar os 3 kg de \u00e1gua at\u00e9 0 \u00b0C\u00a0 —\u00a0 Q = m.c.T = 3000.1.(0-40) = – 120000 cal\u00a0 —\u00a0 observe que o calor liberado pela \u00e1gua, 120 kcal, n\u00e3o \u00e9 suficiente para fundir todo gelo (visto que s\u00e3o necess\u00e1rios 160 kcal)\u00a0 —\u00a0 portanto o equil\u00edbrio t\u00e9rmico ocorrer\u00e1 a 0 \u00b0C, sendo que os 3 kg de \u00e1gua l\u00edquida original continuar\u00e1 l\u00edquida e teremos uma parte do gelo derretida que vale\u00a0 —\u00a0\u00a0 Q = m.L\u00a0 —\u00a0 120000 = m.80\u00a0 —\u00a0 \u00a0m = 120000\/80 = 1500 g = 1,5 kg\u00a0 —\u00a0 assim a quantidade de \u00e1gua gelada final ser\u00e1 de 3 + 1,5 = 4,5 kg\u00a0 —<\/b><\/span><\/span>\u00a0 R- D<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A quantidade efetiva de energia utilizada foi\u00a0 —\u00a0 Q = m.c.\u2206T + m.L\u00a0 —\u00a0 Q = 1000.1.(100-20) + 500.540\u00a0 —\u00a0 Q = 80000 + 270000\u00a0 —\u00a0 Q = 350000 cal = 1.470.000 J\u00a0 —\u00a0 a quantidade de energia liberada foi\u00a0 —\u00a0 W = P.\u2206t\u00a0 —\u00a0 W = U.i.\u2206t\u00a0 —\u00a0\u00a0 W = 127.12.(20.60)\u00a0 —\u00a0 W = 1.828.800 J\u00a0 —\u00a0 o rendimento (\u03b7) ser\u00e1\u00a0 —\u00a0 \u03b7 =Q\/W=1.470.000\/1.828.800\u00a0 —\u00a0 \u03b7=0,80\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03b7=80%<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Volume que evaporou\u00a0 —\u00a0 V=1000mL \u2013 950mL=50mL=50.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V=0,050.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d=m\/V\u00a0 —\u00a0 m=d.V\u00a0 —\u00a0 Q=m.L=d.V.L\u00a0 —\u00a0 Q=1,0.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.0,050.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.2,3.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Q=115.000J\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) P = U.i = 220.1 = 220 W = 220 J\/s\u00a0 —\u00a0 Q = m.c.<\/b><\/span><\/span>\u0394\u03b8<\/b><\/span><\/span><\/span>=1.4,2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.(45 \u2013 23)=1.4,2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.22\u00a0 —\u00a0 Q=9,24.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>J\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=Q\/\u2206t\u00a0 —\u00a0 220=9,24.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\/\u2206t\u00a0 —<\/b><\/span><\/span>\u00a0\u2206t=420s=7min<\/b><\/span><\/span><\/p>\n

b) Se a \u00e1gua est\u00e1 a 45 \u00b0C e sua temperatura cai para 0 \u00b0C, h\u00e1 libera\u00e7\u00e3o de\u00a0 —\u00a0 Q = m.c. \u2206\u03b8 = 1 . 4,2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.45 = 189.000 J\u00a0 —\u00a0 para fundir todo o gelo s\u00e3o necess\u00e1rios Q = m.L = 0,6.330.000 = 198.000 J\u00a0 —\u00a0 como n\u00e3o ocorrer\u00e1 o derretimento total do gelo, haver\u00e1 gelo residual e, deste modo, a temperatura final desse experimento ser\u00e1<\/b><\/span><\/span>\u00a00 \u00b0C.<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>a)A sublima\u00e7\u00e3o do gelo seco ocorre devido ao calor absorvido pelo isopor\u00a0 —\u00a0 m<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>\u00a0\u2013 massa de gelo seco que sofreu sublima\u00e7\u00e3o entre 18h e 24h\u00a0 —\u00a0 \u0394t=24 \u2013 18=6h\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=Q\/\u0394t\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>.L\/\u0394t\u00a0 —\u00a0 60=m<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>.648.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/6×3.600\u00a0 —\u00a0 m<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>=2kg\u00a0 —\u00a0 se ele vai utilizar 2kg e 2kg foram sublimados, a massa total adquirida foi de\u00a0<\/b><\/span><\/span>m<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=4kg<\/b><\/span><\/span><\/p>\n

b) N\u00famero de mols de gelo secoque sofre sublima\u00e7\u00e3o\u00a0 —\u00a0 n=m\/M=2.000\/44\u00a0 —\u00a0 n=45,4545\u00a0 —\u00a0 n\u224845,5 mol —\u00a0 nas condi\u00e7\u00f5es normais de press\u00e3o e temperatura (CNTP), 1 mol do g\u00e1s ocupariam 22,4L, ent\u00e3o 45,5 mol ocupar\u00e3o um volume de V=45,5.22,4\u00a0 —\u00a0 V\u22481.020L=1.020.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V\u22481,02.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Como cada cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0de g\u00e1s formado carrega 0,01g, 1,02.10<\/b><\/span><\/span>6 cm<\/b><\/span><\/span><\/sup>3 carregar\u00e1\u00a0 —\u00a0 m<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=0,101.1,02.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>=1,02.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>g\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0m<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=10,2kg<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Devido ao alto calor espec\u00edfico da \u00e1gua, ela serve como regulador t\u00e9rmico para os seres vivos. Quando a temperatura do organismo aumenta, ele elimina \u00e1gua na forma de suor. Essa \u00e1gua, ao evaporar, absorve calor desse organismo, regulando sua temperatura. Cada 1 grama que se transforma em vapor absorve 540 cal\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

21-\u00a0<\/b><\/span><\/span>De \u00e1gua a 25<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup>a \u00e1gua a 120<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=m.c.\u0394\u03b8=1.1.95\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=95kcal\u00a0 —\u00a0 vaporizando a \u00e1gua a 25<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=m.L=1.526\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=526kcal\u00a0 —\u00a0 Q<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=95 + 526\u00a0 —\u00a0 Q<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=621kcal<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- Q<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=621 kcal.<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Q(gelo) + Q(agua) + Q(\u00e1gua congelada) = 0\u00a0 —\u00a0 725.0,5.(0-x) + 2500.1.(0-5) + 64.(-80) = 0\u00a0 —\u00a0 -362,5.x – 12500 – 5120 = 0\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

-362,5.x = 17620\u00a0 —\u00a0 x = -17620\/362,5 = -48,6\u00b0C<\/b><\/span><\/span><\/p>\n

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

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quantidade de calor m\u00e1xima cedida pela \u00e1gua ao gelo\u00a0 —\u00a0 Q=m.c.(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=100.1.(0 \u2013 30)\u00a0 —\u00a0 Q= – 3.000cal\u00a0 —\u00a0 essa quantidade de calor consegue fundir uma massa de\u00a0 —\u00a0 Q=m.L\u00a0 —\u00a0 3.000=m.80\u00a0 —\u00a0 m=37,5g\u00a0 —\u00a0 como existiam 100g de gelo e foram derretidos s\u00f3 37,5g, sobraram no copo gelo e \u00e1gua a 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span>cedido\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0+ Q<\/b><\/span><\/span>recebgelo\u00a0<\/b><\/span><\/span><\/sub>+ Q<\/b><\/span><\/span>recfus\u00e3ogelo<\/b><\/span><\/span><\/sub>\u00a0+ Q<\/b><\/span><\/span>gelo+\u00e1gua<\/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>(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) + m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>.(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) + m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>.L<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>gelo-\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>)=0\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

500.1.(30 \u2013 40) + m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>.0,5.(0 \u2013 (-7)) + 80m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>.1<\/b><\/span><\/span>.<\/b><\/span><\/span><\/sub>(30 \u2013 0)=0\u00a0 —\u00a0\u00a0 113,5m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u00a0= 5.000\u00a0 —\u00a0 m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>=44g\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Q= 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.L=0,1×4.200x(0 \u2013 20) + 0,1.3,3.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q=80.570cal ou Q=80,57kca<\/b><\/span><\/span>l<\/b><\/span><\/span><\/p>\n

b) porta fechada\u00a0 —\u00a0 volume V constante\u00a0 —\u00a0 Lei geral dos gases perfeitos\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=PV\/T\u00a0 —\u00a0 1\/293=P\/270\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0P=0,92atm<\/b><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span>a) Q=m.c.(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>) + m.L=25.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/span><\/sup>.410.(30 \u2013 10) + 25.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/span><\/sup>.80.000\u00a0 —\u00a0 Q=205 + 2.000\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>Q=2.205J<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=Q\/\u0394t\u00a0 —\u00a0 750=Q\/60\u00a0 —\u00a0 Q=45.000J\u00a0 —\u00a0 1g \u2013 2.500J\u00a0 —\u00a0 m \u2013 45.000J\u00a0 —\u00a0 m=45.000\/2.500\u00a0 —\u00a0 m=18g=0,018kg\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0a taxa \u00e9 de 0,018kg\/min<\/b><\/span><\/span><\/p>\n

b) massa de \u00e1gua perdida pelo atleta\u00a0 —\u00a0 m=0,018kg\/minx30min \u00a0—\u00a0\u00a0<\/b><\/span><\/span>m=0,54kg=0,54L<\/b><\/span><\/span><\/p>\n

30<\/b><\/span><\/span><\/span>–\u00a0<\/b><\/span><\/span>Q=m.L\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=Q\/\u0394t\u00a0 —\u00a0 Q=P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u0394t\u00a0 —\u00a0 mL= P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u0394t\u00a0 —\u00a0 m.2,26.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=113×3.600\u00a0 —\u00a0<\/b><\/span><\/span>m=180g\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span>suco<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>suco<\/b><\/span><\/span><\/sub>.(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) + m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>.L\u00ad<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0+ 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>)=0\u00a0 —\u00a0 250.1.(t \u2013 20) + 50.80 + 50.1.(t \u2013 0)=0\u00a0 —\u00a0 250t \u2013 5.000 +4.000 + 50t=0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>t=3,3<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span>chumbo<\/b><\/span><\/span><\/sub>.L<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>chumbo<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>chumbo<\/b><\/span><\/span><\/sub>.(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) + m<\/b><\/span><\/span>\u00e1gua.<\/b><\/span><\/span><\/sub>L<\/b><\/span><\/span>v<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 -300.5,5 + 300×0,03(100 \u2013 327) + m.540=0\u00a0 —\u00a0 m=6,8g\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

33<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> <\/b>Observe que a temperatura inicial e final do \u00f3leo \u00e9 a mesma, ent\u00e3o ele pode ser excluido da resolu\u00e7\u00e3o do exerc\u00edcio\u00a0 —\u00a0 fus\u00e3o do gelo a 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=mL=40.80\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=3.200cal\u00a0 —\u00a0 \u00e1gua de 0<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sup>C a 30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=mc\u0394\u03b8=40.1.30\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=1.200cal\u00a0 —\u00a0 massa m de \u00e1gua a 70<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C sendo resfriada at\u00e9 a temperatura \u03b8\u00a0 —\u00a0 Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=mc(\u03b8 \u2013 \u03b8<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=m.1.(30 \u2013 70)\u00a0 —\u00a0 Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=-40m\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>3<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 3.200 + 1.200 \u2013 40m=0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m=110g<\/b><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Em 1 s \u00e9 fornecida a quantidade de calor\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=800W=800J\/s =Q\/\u0394t —\u00a0 Q=800J=800\/4,2cal\u00a0 —\u00a0 esse Q aquece \u00e1gua de 20<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C a 100<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C (Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>) e depois a vaporiza a 100<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C (Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>= mc(\u03b8 \u2013 \u03b8<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=m.1.80=80m\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=mL=m.540\u00a0 —\u00a0 Q=Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 — 800\/4,2=540m + 80m\u00a0 —\u00a0 m=0,31g\u00a0 —\u00a0 d=m\/V\u00a0 —\u00a0 1g\/m\u2113=0,31g\/V\u00a0 —\u00a0 V= 0,31m\u2113\u00a0 —\u00a0 vaz\u00e3o\u00a0 —\u00a0 V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=0,31\/1\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=0,31 m\u2113\/s<\/b><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>De acordo com o gr\u00e1fico no instante inicial a temperatura da \u00e1gua \u00e9 de 10 \u00b0C e no instante 50 s \u00e9 de 30 \u00b0C\u00a0 —\u00a0 a quantidade de calor absorvida pela \u00e1gua foi de Q = m.c.DT = 200.1.(30-10) = 4000 cal\u00a0 —\u00a0\u00a0 o que significa que a fonte de calor possui fluxo de 4000 cal\/50 s = 80 cal\/s\u00a0 —\u00a0 a partir do instante inicial considerando-se 135 s o total de calor fornecido para \u00e1gua, por esta fonte, ser\u00e1 de 135.80 = 10.800 cal\u00a0 —\u00a0 Q = m.c.DT —\u00a0\u00a0 10.800 = 200.1.(T \u2013 10)\u00a0 —<\/b><\/span><\/span>\u00a0\u00a0 T = 64\u00b0C<\/b><\/span><\/span>\u00a0 —\u00a0 Para fundir 200 g de gelo s\u00e3o necess\u00e1rios\u00a0 —\u00a0 Q = m.L = 200.80 = 16000 cal\u00a0 —\u00a0 com um fluxo de 80 cal\/s \u00a0—\u00a0 \u00a016000\/80 =<\/b><\/span><\/span>\u00a0200 s\u00a0 —\u00a0 R- A<\/b><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a)Observe nos gr\u00e1ficos que a mudan\u00e7a de estado f\u00edsico, solidifica\u00e7\u00e3o, ocorreu entre as leituras de calor 16.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0J (veja no gr\u00e1fico maior) e 1.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0J (veja no detalhe gr\u00e1fico a direita)\u00a0 —\u00a0 Q = m.L\u00a0 —\u00a0 (1.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0\u2013 16.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>) = 0,5.L\u00a0 —\u00a0 -15.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0= 0,5.L\u00a0 —\u00a0 L = -30.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0=<\/b><\/span><\/span>L=-3,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0J\/kg\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

b) o resfriamento na fase s\u00f3lida pode ser lido no detalhe gr\u00e1fico, onde se pode ver que este resfriamento ocorreu entre 1.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0J e zero.\u00a0 —\u00a0 Q = m.c.DT\u00a0 —\u00a0 (0 \u2013 1.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>) = 0,5.c.(-10 \u2013 0)\u00a0 —\u00a0 -10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0= -5.c \u00a0—\u00a0\u00a0<\/b><\/span><\/span>\u00a0c = 2000 J\/(kg\u00b0C) = 2 kJ\/(kg\u00b0C)<\/b><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Calor de fus\u00e3o do gelo\u00a0 —\u00a0 \u00a0Qg = mLg = m(80)\u00a0 —\u00a0 calor sens\u00edvel da \u00e1gua: Qa = mcaDT = m(1)(100) = m(100)\u00a0 —\u00a0 calor de vaporiza\u00e7\u00e3o da \u00e1gua: Qv = mLv\u00ad = m(540)\u00a0 —\u00a0 a quantidade de calor at\u00e9 o instante t \u00e9\u00a0 —\u00a0 Q = Qg + Qa + Qv = m(80) + m(100) + m(540) = m(720)\u00a0 —\u00a0 como o calor \u00e9 fornecido \u00e0 taxa constante, at\u00e9 o instante t\/2 a quantidade fornecida \u00e9 Q\/2= m(360)\u00a0 —\u00a0 essa quantidade de calor \u00e9 suficiente para derreter o gelo e aquecer a \u00e1gua at\u00e9 100\u00b0C e vaporizar uma massa m\u2019 de \u00e1gua, que pode ser calculada por\u00a0 —\u00a0 m(360) =m(80) + m(100) + m\u2019(540)\u00a0 —\u00a0 m\u2019(540) = m(360) \u2013 m(180)\u00a0 —\u00a0 \u00a0m\u2019(540) = 180 m \u00de m\u2019 =m\/3\u00a0 —\u00a0 assim, na metade do tempo teremos uma mistura de \u00e1gua e vapor a 100 \u00b0C, restando 2\/3 da massa inicial da \u00e1gua\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

38-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Dado: m = 100 g.\u00a0 —\u00a0 pela leitura do gr\u00e1fico, conclui-se que\u00a0 —\u00a0 de 0\u00b0C at\u00e9 40 \u00b0C a subst\u00e2ncia esteve na fase s\u00f3lida\u00a0 —\u00a0 a fus\u00e3o ocorreu na temperatura de 40 \u00b0C com absor\u00e7\u00e3o de 400 cal\u00a0 —\u00a0 ap\u00f3s a fus\u00e3o, iniciou-se novo aquecimento, de 40 \u00b0C a 80 \u00b0C\u00a0 —\u00a0 na fase l\u00edquida a quantidade de calor absorvida foi\u00a0 —\u00a0 Q<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>\u00a0= 1.000 \u2013 600 = 400 cal\u00a0 —\u00a0 Q<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>\u00a0= m c DT\u00a0 —\u00a0 c=Q\/m\u0394T\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

Q=400\/100×40\u00a0 —\u00a0 \u00a0<\/b><\/span><\/span>c = 0,1 cal\/g.\u00b0C\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0durante a fus\u00e3o, a temperatura se manteve constante, absorvendo nesse processo\u00a0 —\u00a0 Q = 600 \u2013 200 = 400 cal\u00a0 —\u00a0 Q<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=mL\u00a0 —\u00a0 L=400\/100\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>L=4 cal\/g<\/b><\/span><\/span><\/p>\n

b) Dados: mg = 50 g; Lg = 80 cal\/g; cag = 1 cal\/g.\u00b0C\u00a0 —\u00a0 \u00a0quantidade calor necess\u00e1ria para fundir totalmente o gelo\u00a0 —\u00a0 Q<\/b><\/span><\/span>g<\/b><\/span><\/span><\/sub>\u00a0= m L<\/b><\/span><\/span>g\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 Q<\/b><\/span><\/span>g<\/b><\/span><\/span><\/sub>\u00a0= 50 (80) = 4.000 cal\u00a0 —\u00a0 se a subst\u00e2ncia recebeu 1.000 cal para aquecer de 0 \u00b0C at\u00e9 80 \u00b0C, para retornar a 0\u00b0C, fazendo o processo inverso, ela liberou, tamb\u00e9m, 1.000 cal, que foram absorvidas pelo gelo\u00a0 —\u00a0 essa quantidade \u00e9 insuficiente para fundir totalmente o gelo\u00a0 —\u00a0 a massa (m\u2019) de gelo fundida \u00e9, ent\u00e3o, a que recebeu Q\u2019= 1.000 cal\u00a0 —\u00a0 Q\u2019= m\u2019 Lg \u00de 1.000 = m\u2019 (80)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

m\u2019 = 12,5 g.\u00a0<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

40-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>m = 100 g; cgelo = 0,5 cal\/g.\u00b0C; L<\/b><\/span><\/span>fus\u00e3o<\/b><\/span><\/span><\/sub>\u00a0= 80 cal\/g; c<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0= 1 cal\/g.\u00b0C; L<\/b><\/span><\/span>V<\/b><\/span><\/span><\/sub>\u00a0= 540 cal\/g; Dt = 5 min = 300 s e 1 cal = 4,2 J\u00a0 —\u00a0 \u00a0\u00a0a quantidade de calor total \u00e9 igual ao calor sens\u00edvel do gelo de \u201310\u00b0C at\u00e9 0 \u00b0C, mais o calor latente de fus\u00e3o do gelo, mais o calor sens\u00edvel da \u00e1gua de 0 \u00b0C a 100 \u00b0C e mais o calor de vaporiza\u00e7\u00e3o da \u00e1gua\u00a0 —\u00a0 Q = Qgelo + Qfus\u00e3o + Q\u00e1gua + Qvaporiza\u00e7\u00e3o\u00a0 —\u00a0 \u00a0Q = m c<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u00a0DT<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u00a0+ m L<\/b><\/span><\/span>fus\u00e3o<\/b><\/span><\/span><\/sub>\u00a0+ m c<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0DT<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0+ m L<\/b><\/span><\/span>V<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q = 100 (0,5) [0 \u2013 (-10)] + 100 (80) + 100 (1) (100 \u2013 0) + 100 (540) \u00a0—\u00a0<\/b><\/span><\/span><\/p>\n

Q = 500 + 8.000 + 10.000 + 54.000\u00a0 —\u00a0 Q= 72.500 cal=72.500×4,2\u00a0 —\u00a0 Q=304.500J\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=Q\/\u0394t=304.500\/300\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=1.015W\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

41-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0= 200 g; m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u00a0= 150 g; t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= 30 \u00b0C; c<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0= 1 cal\/g.\u00b0C; L<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u00a0= 80 cal\/g\u00a0 —\u00a0 nesse tipo de problema, envolvendo gelo e \u00e1gua, voc\u00ea precisa sempre verificar se, no equil\u00edbrio t\u00e9rmico, sobra gelo ou se h\u00e1 fus\u00e3o total\u00a0 —\u00a0\u00a0 para isso, tem que comparar o calor latente necess\u00e1rio para fus\u00e3o do gelo (Q<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>) com o calor sens\u00edvel liberado pela \u00e1gua (Q<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>) at\u00e9 0 \u00b0C\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u00a0= m<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u00a0L<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u00a0= 150 (80)\u00a0 —\u00a0 Q<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u00a0= 12.000 cal\u00a0 —\u00a0 Q<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0= m<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0c<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0DT = 200 (1) (0 \u2013 30)\u00a0 —\u00a0\u00a0 Q<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0= \u2013 6.000 cal ( o sinal negativo indica apenas que houve libera\u00e7\u00e3o de calor)\u00a0 —\u00a0 Comparando essas quantidades de calor (em m\u00f3dulo), verificamos que a quantidade de calor necess\u00e1ria para fundir o gelo (12.000 cal) \u00e9 menor que a quantidade de calor liberada pela \u00e1gua (6.000 cal\u00a0 —\u00a0 \u00a0apenas metade da necess\u00e1ria). Portanto, apenas metade da massa de gelo se funde e a temperatura de equil\u00edbrio t\u00e9rmico \u00e9 0 \u00b0C.\u00a0<\/b><\/span><\/span><\/p>\n

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

42-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quantidade calor liberada pela \u00e1gua para o resfriamento desejado\u00a0 —\u00a0 Q<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0= m<\/b><\/span><\/span>\u00a0\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0c<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0|DT| = 200 (1) |20 \u2013 24|\u00a0 —\u00a0 Q=-800cal<\/b><\/span><\/span><\/p>\n

—\u00a0 quantidade de calor necess\u00e1ria para fundir um cubo de gelo\u00a0 —\u00a0 Q<\/b><\/span><\/span>cubo<\/b><\/span><\/span><\/sub>\u00a0= m<\/b><\/span><\/span>cubo<\/b><\/span><\/span><\/sub>\u00a0Lfus\u00e3o = 10 (80) = 800 cal\u00a0 — Q<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0+ \u00a0Q<\/b><\/span><\/span>cubo<\/b><\/span><\/span><\/sub>=0<\/b><\/span><\/span><\/p>\n

—\u00a0 sendo a quantidaded de calor cedida igual \u00e0 recebida, voc\u00ea deve observar que basta um cubo de gelo para provocar o resfriamento desejado da \u00e1gua \u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

43-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Fus\u00e3o do gelo Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=mL=50.80\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=4.000cal\u00a0 —\u00a0 resfriando \u00e1gua de 80<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C a 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=mc\u0394t=150.1.80\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=- 12.000cal\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

o calor fornecido pela \u00e1gua derreteu o gelo e ainda sobraram 8.000cal para aquecer as (150 + 50)g restantes at\u00e9 uma temperatura t\u00a0 —\u00a0 Q=mc\u0394t\u00a0 —\u00a0 8.000=200.1.(t \u2013 0)\u00a0 —\u00a0 t=8.000\/200\u00a0 —\u00a0 t=40<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

44-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A quantidade de calor liberada pelo fog\u00e3o \u00e9\u00a0 —\u00a0 Q = P Dt = m L \u00de m = P.\u0394t\/L=300×600\/2,25.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 m=80g\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

45-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00e1gua a 30oC a \u00e1gua a 100oC\u00a0 —\u00a0 Q1=mc\u0394\u03b8=0,5×4.190.(100 \u2013 30)\u00a0 —\u00a0 Q1= 146.650J=146.65kJ\u00a0 —\u00a0 vaporizando \u00e1gua a 100oC\u00a0 —\u00a0 Q2=m.L=0,5.2.246\u00a0 —\u00a0 Q2=1.123kJ\u00a0 —\u00a0 Q=Q1 + Q2=146,65 + 1.123=1.269,65kJ\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

46-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quantidade de calor que deve ser retirada\u00a0 do corpo para que sua temperatura diminua de 37<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C para 30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q=mc\u0394\u03b8=60.1.(30 \u2013 37)\u00a0 —\u00a0 Q= – 420kcal\u00a0 —\u00a0 essa quantidade de calor em m\u00f3dulo deve ser fornecido pelo gelo a 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C e fus\u00e3o\u00a0 —\u00a0 Q=mL\u00a0 —\u00a0 420=m.80\u00a0 —\u00a0 m=5,25kg\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

47-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>No sentido cient\u00edfico, supondo que a \u00e1gua esteja fervendo a c\u00e9u aberto, ao n\u00edvel do mar, haver\u00e1, por conta do recebimento de<\/b><\/span><\/span><\/p>\n

energia (calor), ebuli\u00e7\u00e3o\u00a0 —\u00a0 a temperatura da massa l\u00edquida permanecer\u00e1 constante e sempre de 100<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C, logo, a energia cin\u00e9tica<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

m\u00e9dia por mol\u00e9cula permanecer\u00e1 constante\u00a0 —\u00a0 por\u00e9m, haver\u00e1 cada vez menos mol\u00e9culas na massa l\u00edquida\u00a0 —\u00a0 portanto, a quantidade de energia da massa l\u00edquida diminui, apesar do recebimento de energia\u00a0 — \u00a0assim, se voc\u00ea entendesse a quantidade de calor do corpo como temperatura, como na linguagem corrente, ser\u00eda obrigado a dizer que a temperatura da \u00e1gua diminui durante o tempo em que estiver fervendo\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

48-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 — \u00a0L = 2,2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0J\/g\u00a0 —\u00a0 P = 300 W\u00a0 — \u00a0\u2206t = 10 min = 600 s\u00a0 —\u00a0 quantidade de calor liberada pelo fog\u00e3o\u00a0 —\u00a0 P=Q\/\u2206t\u00a0 —Q=P.\u2206t\u00a0 —\u00a0 Q=m.L\u00a0 —P. \u2206t=m.L\u00a0 —\u00a0 m=P.\u2206t\/L=300.600\/2,25.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 m=80g\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

49-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>01. Correta\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

02) Correta\u00a0 —\u00a0 veja como exemplo a fus\u00e3o da \u00e1gua\u00a0 —\u00a0 na patina\u00e7\u00e3o sobre o gelo, o deslizamento \u00e9 facilitado porque, quando o<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

patinador passa, parte do gelo se \u00a0transforma em \u00e1gua, reduzindo o atrito\u00a0 —\u00a0 estando o gelo a uma temperatura inferior a 0\u00baC, isso ocorre porque a press\u00e3o da l\u00e2mina do patim sobre o gelo faz com que ele derreta\u00a0 —\u00a0 isso ocorre porque um aumento da press\u00e3o diminui a temperatura de fus\u00e3o\u00a0 —\u00a0 outro exemplo, vaporiza\u00e7\u00e3o da \u00e1gua\u00a0 —\u00a0 uma panela de press\u00e3o cozinha alimentos em \u00e1gua em um tempo menor do que as panelas comuns\u00a0 —\u00a0 esse desempenho da panela de<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

press\u00e3o se deve ao fato de que a tampa provoca no interior da panela um aumento da press\u00e3o que, por sua vez provoca tamb\u00e9m um aumento na temperatura de vaporiza\u00e7\u00e3o fazendo a \u00e1gua ferver a uma temperatura acima de 100<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C.<\/b><\/span><\/span><\/p>\n

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

08. Correta\u00a0 —\u00a0 com o aumento de temperatura, a energia cin\u00e9tica m\u00e9dia das mol\u00e9culas aumenta, e assim, mais mol\u00e9culas conseguem se transformar em vapor, aumentando a velocidade de evapora\u00e7\u00e3o.\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

R- (01 + 02 + 04 + 08) = 15<\/b><\/span><\/span><\/p>\n

50-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 — V = 2 L\u00a0 — \u00a0P = 420 W\u00a0 —\u00a0 c = 1 cal\/g.\u00b0C = 4,2 J\/g.\u00b0C\u00a0 —\u00a0\u00a0 L = 540 cal\/g = 2.268 J\/g\u00a0 —\u00a0\u00a0 d = 1 kg\/L\u00a0 —\u00a0 \u00a0\u2206T = (100 \u2013 20) = 80 \u00b0C\u00a0 —\u00a0 massa de \u00e1gua usada\u00a0 —\u00a0 d=m\/V\u00a0 —\u00a0 1=m\/2\u00a0 —\u00a0 m=2kg=2.000g\u00a0 —\u00a0 quantidade de calor necess\u00e1ria para que 20% da massa (0,2 M) de \u00e1gua seja vaporizada\u00a0 —\u00a0 Q = Q<\/b><\/span><\/span>sens\u00edvel<\/b><\/span><\/span><\/sub>\u00a0 +\u00a0 Q<\/b><\/span><\/span>latente<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q = M c \u2206T + (0,2 M) L\u00a0 —\u00a0 Q = 2.000 (4,2) (80) + (0,2×2.000)x2.268 = 67.200 + 907.200\u00a0 —\u00a0 Q = 1.579.200 J\u00a0 —\u00a0 a pot\u00eancia \u00fatil \u00e9 20% da pot\u00eancia total\u00a0 —\u00a0 P<\/b><\/span><\/span>\u00fatil<\/b><\/span><\/span><\/sub>\u00a0= 0,8 P = 0,8 (420)\u00a0 —\u00a0 P<\/b><\/span><\/span>\u00fatil<\/b><\/span><\/span><\/sub>\u00a0= 336 W\u00a0 —\u00a0 P<\/b><\/span><\/span>\u00fatil<\/b><\/span><\/span><\/sub>=Q\/\u2206t\u00a0 —\u00a0 \u2206t=Q\/P<\/b><\/span><\/span>\u00fatil<\/b><\/span><\/span><\/sub>=1.579.200\/336\u00a0 —\u00a0 \u2206t=4.700s\u00a0 —\u00a0 \u2206t=1h, 18min e 20s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

51-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) C\u00e1lculo da \u00e1rea da tampa (40 cm:<\/b><\/span><\/span>\uf020<\/b><\/span><\/span>50 cm)\u00a0 —\u00a0 A = 0,4 m.<\/b><\/span><\/span>\uf020<\/b><\/span><\/span>0,5 m\u00a0 —\u00a0 A=0,2m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u00a0pot\u00eancia solar incidente na Terra (placa e<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0consequentemente na \u00e1gua) (1 kW por m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0 1m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 1 kW\u00a0 —\u00a0 0,2m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0,2.1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0,2 kW\u00a0<\/b><\/span><\/span><\/p>\n

b) A quantidade de energia (E) necess\u00e1ria para aquecer a \u00e1gua no interior da panela \u00e9 dada por\u00a0 —\u00a0 Q=E=mc\u2206t\u00a0 —\u00a0 E=300.4.(100 \u2013 25)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>E=90.000J ou E=90 kJ<\/b><\/span><\/span><\/p>\n

c) A quantidade de energia (E\u2018) para aquecer a \u00e1gua de 25<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C at\u00e9 100\u00b0C \u00e9 de 90.000 J (item anterior)\u00a0 —\u00a0\u00a0\u00a0 para evaporar 1\/3 da \u00e1gua\u00a0 —\u00a0 m=300\/3=100g\u00a0 —\u00a0 Q\u2019=E\u2019=mL=100.2.200\u00a0 — E\u2019 = 220.000 J\u00a0 —\u00a0\u00a0 E<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>\u00a0= =90.000 + 220.000\u00a0 —\u00a0 E<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=310.000 J\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=E<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>\/\u2206t\u00a0 —\u00a0 200=310.000\/\u2206t\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u2206t=1.550 s<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

V=300ml=300.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>l=0,3l=0,3.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d=mV\u00a0 —\u00a0 1=m\/0,3.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 m=0,3.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>g=300g\u00a0 —\u00a0 como a \u00e1gua n\u00e3o evapora, trata-se apenas de calor sens\u00edvel, sem mudan\u00e7a de estado\u00a0 —\u00a0 Q=m.c.(\u03b8 \u2013 \u03b8<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=300.1.(40 \u2013 10)\u00a0 —\u00a0 Q=9.000 cal\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

 <\/p>\n

53-<\/b><\/span><\/span><\/span> <\/b>Como\u00a0 o enunciado pede para considerar o sistema \u00e1gua e gelo isolado haver\u00e1 trocas de calor somente entre eles e, ao final dessas trocas voc\u00ea ter\u00e1 apenas \u00e1gua l\u00edquida \u00e0 temperatura de equil\u00edbrio t\u00e9rmico \u03b8<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q \u2013 quantidade de calor cedido pelos 200g de \u00e1gua a 25<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C para sua temperatura passar de 25<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C a \u03b8<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q=m.c.(\u03b8<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 \u03b8)=200.1.(\u03b8<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 25)\u00a0 —\u00a0 Q=200.\u03b8<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 5000\u00a0 —\u00a0 Q\u2019 \u2013 calor recebido pelos 50g de gelo a 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>para mudar de estado e se transformar em 50g de \u00e1gua a 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q\u2019=m.L=50.80\u00a0 —\u00a0 Q\u2019=4000cal<\/b><\/span><\/span><\/p>\n

Q\u2019\u2019 –\u00a0 calor absorvido pelos 50g de \u00e1gua a 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C a 50g de \u00e1gua a \u03b8<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q\u2019\u2019=m.c.(\u03b8<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 0)=50.1.\u03b8<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q\u2019\u2019=50.\u03b8<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 a soma alg\u00e9brica dessas quantidades de calor trocadas pelo sistema \u00e9 nula\u00a0 —\u00a0 Q + Q\u2019 + Q\u2019\u2019 = 0\u00a0 —\u00a0 200.\u03b8<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 5000 + 4000 + 50.\u03b8<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=0\u00a0 —<\/b><\/span><\/span><\/p>\n

250.\u03b8<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0= 1000\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u03b8<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=4<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

02- Maior\u00a0 —\u00a0 voc\u00ea pode, agora incluindo o copo nas trocas de calor, efetuar o equil\u00edbrio t\u00e9rmico entre toda a \u00e1gua (250g) a 4<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C com o copo que est\u00e1 a 25<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C e, \u00e9 claro que a temperatura final estar\u00e1 entre 4<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C e 25<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C, ou seja, ser\u00e1 maior que 4<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C.<\/b><\/span><\/span><\/p>\n

54-<\/b><\/span><\/span><\/span> Como a fus\u00e3o e a vaporiza\u00e7\u00e3o ocorrem \u00e0 temperatura constante, pelo gr\u00e1fico a fus\u00e3o ocorre \u00e0 temperatura de 40<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C e a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

<\/a> vaporiza\u00e7\u00e3o \u00e0 temperatura de 60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 portanto entre 10<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C e 40<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C o corpo se encontra no estado s\u00f3lido, entre 40<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C e 60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C, no estado l\u00edquido e, a partir de 60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C no estado gasoso\u00a0 —\u00a0 calor espec\u00edfico no estado s\u00f3lido\u00a0 —\u00a0 Q=m.c.(\u03b8 \u2013 \u03b8<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 (600 \u2013 0)=50.c.(40 \u2013 10)\u00a0 —\u00a0 c=0,4cal\/g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —<\/b><\/span><\/span><\/p>\n

A vaporiza\u00e7\u00e3o ocorre na temperatura de 60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C e para isso o corpo recebeu Q=1.800 \u2013 1.200=600cal\u00a0 —\u00a0 Q=m.L\u00a0 —\u00a0 600=50L \u00a0—\u00a0 L=12cal\/g\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

 <\/p>\n

55-<\/b><\/span><\/span><\/span> <\/b>De gelo a -20<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C a gelo a 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 calor sens\u00edvel\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=m.c<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>.\u2206\u03b8=2.0,5.[0 \u2013 (-20)]\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=20kcal\u00a0 —\u00a0 de gelo a 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C a \u00e1gua a 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 calor latente de fus\u00e3o\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=m.L=2.80\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=160kcal\u00a0 —\u00a0 de \u00e1gua a 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C a \u00e1gua a 50<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 calor sens\u00edvel\u00a0 —\u00a0 Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>= m.c<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.\u2206\u03b8=2.1.(50 \u2013 0)\u00a0 —\u00a0 Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=100kcal\u00a0 —\u00a0 Q<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=20 + 160 + 100=280kcal\u00a0 —\u00a0 como a fonte t\u00e9rmica fornece 20kcal por minuto\u00a0 —\u00a0 t=280kcal\/20kcal\/min=14minx60=840s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Calor latente \u2013 trocas de calor com mudan\u00e7a de estado 01-\u00a0Aquecimento do gelo\u00a0 —\u00a0 Q1=m.c.(t \u2013 to)=100.0,5.(0 \u2013 (-10))\u00a0 —\u00a0 Q1=500cal Fus\u00e3o do gelo\u00a0 —\u00a0 Q2=m.LF=100.80\u00a0 —\u00a0 Q2=8.000cal Aquecimento da \u00e1gua\u00a0 —\u00a0 Q3=m.c.(t \u2013 to)=100.1.(100 \u2013 0)\u00a0 —\u00a0 Q3=10.000cal Vaporiza\u00e7\u00e3o da \u00e1gua\u00a0 —\u00a0 Q4=m.LV=100.540\u00a0 —\u00a0 Q4=54.000cal Aquecimento do vapor\u00a0 —\u00a0 Q5=m.c.(t \u2013 to)=100.0,5.(110<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2043,"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-2046","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2046","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=2046"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2046\/revisions"}],"predecessor-version":[{"id":10732,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2046\/revisions\/10732"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2043"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2046"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}