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Exercício de desobrir a massa do gelo
Exercício:
Um pedaço de gelo a 0°C é colocado em 200g de água a 30°C, num recipiente de capacidade térmica desprezível e isolado termicamente. O equilíbrio térmico se estabelece em 20°C. O calor latente de fusão do gelo é 80cal/g e o calor específico da água é 1,0cal/g.°C. A massa do pedaço de gelo, usado no experimento, é:
a) 10g
b) 20g(x)
c) 30g
d) 40g
Informações Adicionais:
tentei fazer mais n entendi alguem poderia me ajudar so sei q a resposta é 20!! XD vlw pela atenção
tem duas equaçoes bases para esse exercicio.
1) Q=m.c.ΔT (calor sensivel)
2) Q=m.L (calor latente)
como existe equilibrio termico na questao, quando estiver em 20ºC a troca de calor sera nula.
Qgelo + Qágua=0.
Pelo calor latente do gelo. Qgelo = m.L –> m*80 = 80m cal
Até atingir o equilibrio termico o gelo se transforma em água entao vamos colocar Qágua, mas nao esqueça que estamos tratando da parte do gelo ainda.
–> Qágua m* 1* (20-0) = 20m cal
e agora calcular a perda de temperatura do recipiente de água até atingir 20ºC
Qágua= m.c.Δt -> 200*1* (20-30) = -2000 cal (o sinal neg. só indica que a energia saiu do sistema)
como Qgelo + Qágua=0,
Qgelo = 80m cal+20m cal = 100m
Qagua = -2000 cal
100m + (- 2000) = 0 –> 100m = 2000 –> m = 2000/100 = 20g
vlw pela ajuda cara!!! XD
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