Question

Consider a Carnot engine with a hot source kept at 500 K. From the hot source, 100 J of energy (heat) is withdrawn at 500 K. The cold sink is kept at 300 K. The efficiency of the Carnot engine is ___________ (rounded off to one decimal place).

Hint:

The efficiency of a Carnot engine is the maximum possible efficiency for any heat engine operating between the same two temperatures. It is independent of the working substance and the amount of heat transferred. Always use absolute temperatures (in Kelvin) when calculating Carnot efficiency.

Answer 1

The efficiency (\( \eta \)) of a Carnot engine is determined solely by the temperatures of the hot and cold reservoirs: \[ \eta = 1 - \frac{T_C}{T_H} \] where \( T_H \) is the absolute temperature of the hot source and \( T_C \) is the absolute temperature of the cold sink. Given: Temperature of the hot source, \( T_H = 500 \) K Temperature of the cold sink, \( T_C = 300 \) K Substitute these values into the formula: \[ \eta = 1 - \frac{300 \, \text{K}}{500 \, \text{K}} = 1 - 0.6 = 0.4 \] The efficiency of the Carnot engine is 0.4. Rounded off to one decimal place, the answer remains 0.4.