What is the amount of heat required (in calories) to convert 10 g of ice at -10 ℃ into steam at 100 ℃ ? Given that latent heat of vaporization of water is 540 cal g- 1 , latent heat of fusion of ice is 80 cal g- 1 , the specific heat capacity of water and ice are 1 cal g- 1 ° C- 1 and 0.5 cal g- 1 ° C- 1 respectively.

Question

What is the amount of heat required (in calories) to convert 10 g of ice at -10 ℃ into steam at 100 ℃ ? Given that latent heat of vaporization of water is 540 cal g- 1 , latent heat of fusion of ice is 80 cal g- 1 , the specific heat capacity of water and ice are 1 cal g- 1 ° C- 1 and 0.5 cal g- 1 ° C- 1 respectively. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

10 g of ice at -10℃ to ice at 0 ℃ Q1=cm, Δ θ =0.5× 10× 10=50 cal 10 g of ice 0 ℃ to water at 0 ℃ Q2=mL=10× 80=800 cal 10 g of water at 0 ℃ to water at 100 ℃ Q3=cm, Δ θ =1× 10× 100=1000 cal 10 g water at 100 ℃ to steam at 100 ℃ Q4=mL=10× 540=5400 cal Total heat required, Q=Q1+Q2+Q3+Q4 = 50+800+1000+5400 = 7250 cal