A quantity of 19.1 cm^3 of water at 12.4 degree C is placed in a Freezer compartment & allowed to freeze to solid ice @ -7.2 degree C. How money Joules of energy must be withdrawn by me Fridge? V = 19.1 cm^3 T_1 = 12.4 degree C T_2 = -7.2 degree C Solution mass of the water: m = density*volume = 1000* 19.1*10^-6 = 0.0191 kg work done is, W = mcdT + mLf + mcicedT = 0.0191*4186*[0-12.4] + 0.0191*333.55x103 + 0.0191*2108*[-7.2-0] = 5089.5 J.