Heat of fusion of ice cal g

Asked by maham237 @ 20/02/2021 in Chemistry viewed by 349 People

Water's heat of fusion is 80. cal/g , its specific heat is 1.0calg⋅∘C, and its heat of vaporization is 540 cal/g . A canister is filled with 330 g of ice and 100. g of liquid water, both at 0 ∘C . The canister is placed in an oven until all the H2O has boiled off and the canister is empty. How much energy in calories was absorbed?

Answered by maham237 @ 20/02/2021

Refer to the diagram shown below.

State 1:
The heat of fusion is required for the ice to become water.
The amount of heat required is
Q₁ = (330 g)*(80 cal/g) = 26,400 cal

State 2:
Sensible heat is required to raise water at 0 C to the boiling point of 100 C.
The heat required is
Q₂ = (430g)*(1 cal/(g-C))*(100 C) = 43,000 cal

State 3:
The heat of vaporization is required to boil off the water.
The heat required is
(430 g)*(540 cal/g) = 232,200 cal

Total heat absorbed is
Q₁+Q₂+Q₃ = 301,600

Answer: 301,600 cal or 301.6 kcal

Heat of fusion of ice cal g

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