Phase Changes : How much heat must be lost to freeze 75.3 g of water at 11°C?

Question

How much heat must be lost to freeze 75.3 g of water at 11°C? 

Answer 1

To freeze water that is originally at 11ºC, one must first lower the temperature from 11 to 0ºC, the freezing point for water. The heat lost during this aspect is q = mC∆T where q = heat, m = mass, C = specific heat of water and ∆T = change in temperature. Thus, we have...

q = (75.3 g)(4.184 J/gº)(11º) = 3466 J heat lost

Next, to freeze the water at 0ºC, all the heat lost goes into the freezing process and there is no change in the temperature. This is the phase change and q = m∆Hf where q = heat, m = mass and ∆Hf = heat of fusion for water. Thus, we have...

q = (75.3 g)(334 J/g) = 25,150 J heat lost

Adding the heat loss together, we get 3466 J + 25,150 J = 28,616 J = 28.6 kJ of heat lost