J.R. S. answered 04/07/22
Ph.D. in Biochemistry--University Professor--Chemistry Tutor
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