Question

Problem 16.20

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

we want the energy to cool it, freeze it, and cool the ice to –10º

specific heat of water is 4.186 kJ/kgC
specific heat of ice is 2.06 kJ/kgC
heat of fusion of ice is 334 kJ/kg

E1 = 4.186 kJ/kgC x 0.385 kg x (18–0)C = 29.009 kJ
E2 = 334 kJ/kg x 0.385 kg = 128.59 kJ
E3 = 2.06 kJ/kgC x 0.385 kg x (0– –10)C = 7.931 kJ
E = E1+E2+E3 = 165.53 kJ

Efficiency of a carnot cycle heat engine is
Efficiency η = 1 – (Tc/Th)
Where Tc is absolute temperature of cold reservoir
Where Th is absolute temperature of not reservoir

η = 1 – (263/301) = 12.624%

The Coefficient of Performance (COP) of the heat engine is the reciprocal of its efficiency
= 7.92

Energy input = 165.53/7.92 = 20.9 kJ