A Carnot engine operating between two reservoirs has efficiency 1/3. When the temperature of cold reservoir raised by x, its efficiency decreases to 1/6. The value of x, if the temperature of hot reservoir is 99°C, will be :

Solution

Physics sir · 3 steps

Convert 99 °C to kelvin first. The drop in efficiency times the hot temperature then gives x directly.

1. Hot reservoir in kelvin. Carnot efficiency only works with absolute temperatures. So T_H = 99 + 273 = 372 K, and it stays fixed.

   η = 1 - T_C/T_H ⇒ T_C = (1-η) T_H

2. Before and after. With η = 1/3, T_C = 2/3 × 372 = 248 K. With η = 1/6, the new cold temperature is 5/6 × 372 = 310 K.
3. So. x = 310 - 248 = 62 K, which is option (b). As a check, x = (1/3 - 1/6) × 372 = 62 K.

The idea to keep. In a Carnot engine with fixed T_H, any change in the cold temperature changes efficiency by Δ T_C / T_H. Always put temperatures in kelvin before using 1 - T_C/T_H.

Answer from the NTA final answer key. Solution by PrepWiser.