How much heat must be lost to freeze 52.0 g of water at 14 degress Celsius?

Question

Please don't use mol in the explanation. Thank you.

Answer 1

First we must reduce the temperature of the water form 14º to 0º

q = mC∆T = (52.0 g)(4.184 J/g/deg)(14 degrees) = 3046 J

Next we have to freeze the water at 0º. This is simply a phase change with no change in temperature.

q = m∆Hfusion = (52.0 g)(334 J/g) = 17,368 J

Total heat = 3046 J + 17,368 J = 20,414 J = 20,400 J must be lost (3 significant figures)