A glass containing 234.5 g of H2O at exactly 20°C was placed in a freezer. The water loses 34.5 kJ as it cools to a constant temperature. What is its new temperature?

Question

(The specific heat of water is 4.184 J/g·°C.)

Answer 1

q = mC∆T

q = heat lost = 34.5 kJ = 34,500 J

m = mass of water = 234.5 g

C = specific heat of water = 4.184 J/gº

∆T = change in temperature = ?

Solving for ∆T ...

∆T = q / mC = 34,500 J / (234.5 g)(2.184 J/gº)

∆T = 35.2º

Since heat is being LOST from the water, we must subtract this from the original temperature of 20º.

Final temperature = 20º - 35.2º = -15.2º

NOTE: This temperature is below the freezing point of water (0ºC) so the water would have had to go through a phase transition from liquid to solid and then the solid would have to have lost additional heat. Without the ∆Hfusion and specific heat of ice, we can't calculate the exact final temperature. So, maybe this was a trick question, or maybe that additional information was omitted. Just sayin'.