How many kilojoules are required to convert 115.0 g of ice at 0.0 ∘c to liquid water at 32 ∘c? the heat of fusion of water is 334 j/g, and the heat capacity of water is 4.184 j/g ∘c?

The answer is 53.8 kJ.

Solution:There are two major steps in converting ice to liquid water. It begins with a phase change when ice melts at 0.0°C, and then a temperature change when the liquid water rises in temperature from zero to 32°C.

The amount of heat involved with the phase change melting is given by

q = (mass of water) (ΔHfus)

= (115.0 g) (334 J/g)

= 38410 J = 38.41 kJ

The amount of heat involved with temperature change is

q = mcΔT

= (115.0g) (4.184J/g°C) (32°C - 0.0°C)

= 15397.12 J = 15.39712 kJ

Summing up the two values gives the total heat required to convert ice to liquid water:

q = 38.41 kJ + 15.39712 kJ = 53.8 kJ