How much heat is required to convert 5.53 g of ice at - 12.0 ∘C to water at 24.0 ∘C? (The heat capacity of ice is 2.09 J / (g⋅∘C), ΔHvap (H2O) = 40.7kJ/mol, and ΔHfus (H2O) = 6.02kJ/mol.)

2.55 * 10³ J = 2.55 kJ

Explanation:

Specific heat capacity of ice = 37.8 J / mol °C

Specific heat capacity of water = 76.0 J / mol °C

Ice at - 12 °C is converted to ice at 0 °C by absorbing heat Q₁

Ice at 0°C melts to water at 0 °C. Let Heat absorbed during this phase change be Q₂.

Let heat absorbed to raise the temperature of water from 0 C to 24°C be Q₃.

Total heat = Q = Q₁ + Q₂ + Q₃

Q₁ = (37.8 j/mol C) (5.53 g / 18.01532 g / mol) (0 - (-12)) = 139.23749 j

Q₂ = (5.53 g/18.01532 g H₂O / mol) (6.02 x10³ j) = 1847.905 j

Q₃ = (76 j/mol C) ((5.53 g/18.01532 g H₂O / mol) (24-0) = 559.8968 j

Total Heat required = Q = 139.23749 j + 1847.905 j + 559.8968 j

= 2547.039 j = 2.55 * 10³ J = 2.55 kJ