Question

In a refrigerator one removes heat from a lower temperature and deposits to the surrounding at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1 kW power, and heat is transferred from −3°C to 27°C, find the heat taken out of the refrigerator per second assuming its efficiency is 50 % of a perfect engine.

Answer 1

Given :

T₁ = 27°C = 27 + 273 = 300 K

T₂ = −3°C = −3 + 273 = 270 K

Efficiency,

η = 1 – \(\frac{T_2}{T_1}\)

= 1 - \(\frac{270}{300}\)

= 1 – 0.9

= 0.1 = \(\frac{1}{10}\)

Efficiency of refrigerator is 50% of percentage engine.

η' = 50% of η = 0.5 ×\(\frac{1}{10}\)=\(\frac{1}{20}\)

∴ Coefficient of performance,

\(β=\frac{Q_2}{W}=\frac{1-η'}{η'}\)

\(β=\frac{1-\frac{1}{20}}{\frac{1}{20}}=\frac{1-0.05}{0.05}\)

= \(\frac{0.95}{0.05}\) = 19

Q₂ = 19% of work done by motor on refrigerator

= 19 × 1

= 19 K J/S