How many calories are required to raise the temperature of 10 kg of ice from -20 to 50?
(The specific heat of ice is 0.50 kcal/kg-C) The calculation must be done in three steps: (1) raising the temperature of the ice from -20 to 0 (2) converting the solid ice at 0 to a liquid water at 0 (3) raising the temperature of the liquid water from 0 to 50 The total number of calories required is the sum of the three steps. Step 1 raising the temperature of the ice from -20 to 0 H1 = m cT where the ice temperature change is: T = TF – TI = 0 – (-20) = 20 H1 = 10 kg (0.50 kcal/kgC)(20) = 100 kcal Step 2 converting the solid ice at 0 to a liquid water at 0 The latent heat of fusion (LF) of water is 80 kcal/kg. This is found using Table 5.3 in the textbook. H2 = m LF = 10 kg (80 kcal/kg) = 800 kcal Notice that there is no temperature-change term in this calculation. The transition from ice to water takes place at a constant temperature, 0 As long as there is any ice left, the temperature of the ice-water mixture will remain at the melting point temperature, 0 As soon as all of the ice
