Question

How much heat (in kJ) is required to warm 13.0 g of ice, initially at -15.0 ∘C, to steam at 110.0 ∘C? The heat capacity of ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C.

Answer 1

The heat required will be sum of different heats required for different processes

1) Q1 = Heat required to raise the temperature of ice from -15∘C to 0∘C

Q1 = Mass X heat capacity of ice X change in temperature

Q1 = 13 X 2.09 X (0-(-15) = 407.55 Joules

2) Q2 = Heat required to convert ice to liquid water

Q2 = heat of fusion of ice X mass of ice = 334 J / g X 13 = 4342 Joules

3) Q3 = Heat rquired to raise the temperature of water to 100 ∘C

Q3 = specific heat of water X mass of water X change in temperature

= 4.18 J / g X 13 X (100) = 5434 Joules

4) Q4 = Heat required to convert water to steam

Q4 = MAss X heat of vaporization of water = 13 X 2256 J /g = 29328 Joules

5) Q5 = Heat required to raise the temperature of steam to 110∘C

Q5 = Heat capacity of steam X mass X change in temperature = 2.01 X 13 X 10 = 261.3 Joules

Total heat required = Q1+Q2+Q3+Q4+Q5 = 407.55 + 4342 + 5434 + 29328 + 261.3 = 39772.85 Joules

= 39.772 KJ