Question

In a refrigerator one removes heat from a lower temperature and deposits to the surroundings 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

Carnot designed a theoretical engine which is free from all the defects of a practical engine. The Carnot engine is the most efficient heat engine operating between two given temperatures. The efficiency of Carnot engine is η = `1 - T_2/T_1`

According to the problem, the efficiency of a perfect engine working between – 3°C to 27°C (i.e., T₂ = – 3 + 273 = 270 K and T₁ = 27 + 273 = 300 K).

Thus according to the Carnot theorem `η_("engine") = 1 - T_2/T_1`

= `1 - (270 K)/(300 K)`

= 0.1

Since the efficiency of the refrigerator (η_ref) is 50% of `η_("engine")`.

∴ η_ref = 0.5,  `η_("engine")` = 0.05

If Q₁ is the heat transferred per second at a higher temperature by doing work W, then

η_ref = `W/Q_1` or `Q_1 = W/η_(ref) = (1 kJ)/0.05` = 20 kJ  ......(As W = 1 K W × 1s = 1 kJ)

Since η_ref is 0.05, heat removed from the refrigerator per second, i.e.,

`Q_2 = Q_1 - η_(ref)  Q_1 = Q_1`  ....(1 – η_ref)

= 20 kJ(1 – 0.05)

= 19 kJ

Therefore, heat is taken out of the refrigerator at a rate of 19 kJ per second.