Question

A Carnot freezer that runs on electricity removes heat from the freezer compartment, which is at -10.0 ∘C and expels it into the room at 28.0 ∘C. You put an ice-cube tray containing 385 g of water at 18.0 ∘C into the freezer.

A. What is the coefficient of performance of this freezer?

B. How much energy is needed to freeze this water?

C. How much electrical energy must be supplied to the freezer to freeze the water?

D. How much heat does the freezer expel into the room while freezing the ice?

Answer 1

A) The efficiency of a carnot cycle heat engine is given by the following formula:

η = 1 – (Tc/Th)

Where Tc is absolute temperature of cold reservoir and Th is absolute temperature of hot reservoir.

So the efficiency, η = 1 – (263/301) = 12.62%

Now as we know that the Coefficient of Performance (COP) of the heat engine is the reciprocal of its efficiency. So the COP = 1/η = 7.92.

B) Total energy needed = energy needed to cool it+ to freeze it + cool the ice to –10 deg C.

The specific heat of water is 4.186 kJ/kgC and that of the ice is 2.06 kJ/kgC. The heat of fusion of ice is 334 kJ/kg.

So Q1 = 4.186*0.385*18 = 29.01 kJ.

Q2 = 334*0.385 = 128.59 kJ.

Q3 = 2.06*0.385*10 = 7.93 kJ

So the total energy = Q1+Q2+Q3 = 165.53 kJ.

C) The electrical energy required = total energy/efficiency = 1311.65 kJ

D) The heat expelled = 1311.65 - 165.53 kJ = 1146.12 kJ.