Question

An ice cube tray of negligible mass contains 0.290 kg of water at 14.0 ∘C. Part A How much joules of heat must be removed to cool the water to 0.00∘C and freeze it? Part B How much calories of heat must be removed?Part C How much Btu of heat must be removed?

Answer 1

m = 0.29 kg

initial temperature = 14 deg C

heat to bring the water to 0 deg C = mass*specific heat*change of temperature = 0.29*4184*14 = 16987.04 J

heat to freeze it completely = mass * latent heat of fusion = 0.29 * 334000 = 96860 J

a)So total heat removed injoules = 96860 + 16987.04 = 113847.04 J

b) 1 calorie = 4.184 J

total heat removed in calories = 113847.04/4.184 = 27210.1 calories

c) 1 BTU = 1055.06 J

so total heat removed in BTU = 113847.04/1055.06 = 107.9 BTU