What is the quantity of heat (In KJ)

Question

What is the quantity of heat (in kJ) associated with cooling 150.3 g of water from 25.60 °C to ice at -10.70 °C?

Csolid = 2.092 J/g* C°

Cliquid = 4.184 J/g* C°

Tfusion = 0.00C°

deltaHfusion = 6.01 KJ/Mol

Answer 1

To go from liquid water at 25.60ºC to solid ice at -10.70º, we have to go through several steps. We will calculate the heat associated with each step, and add them up at the end to arrive at our answer.

Step 1: cool 150.3 g liquid water from 25.60º to 0º:

q = mC∆T = (150.3 g)(4.184 J/gº)(25.6º) = 16099 J

Step 2: Freeze 150.3 g liquid water @ 0º (this is a phase change, and no change in temperature)

∆Hfusion = 6.01 kJ/mol x 1 mol 18 g = 0.334 kJ/g = 334 J/g

q = m∆Hfusion = (150.3 g)(334 J/g) = 50200 J

Step 3: cool 150.3 g of ice from 0º to -10.70º

q = mC∆T = (150.3g)(2.092 J/gº)(10.70º) = 3364 J

Total heat = 16099 J + 50200 J + 3364 J = 69663 J = 69.66 kJ of heat associated with the change.

The heat is lost so the value would be -69.66 kJ.