JEE MainPhysicsThermodynamics
A Carnot engine performs 2 kJ of work per cycle and rejects 6 kJ of heat to the sink. If the temperature of the source is 327^ C , what is the temperature of the sink?
Options
- A1527^ C
- B177^ C
- C450^ C
- D527^ C
Correct answer
B. 177^ C
Step-by-step solution
The work done by the engine per cycle is W = 2 kJ . The heat rejected to the sink per cycle is Q_ out = 6 kJ . According to the first law of thermodynamics, the total heat absorbed from the source is: Q_ in = W + Q_ out = 2 kJ + 6 kJ = 8 kJ The temperature of the source in Kelvin is: T_H = 327 + 273 = 600 K For a Carnot engine, the ratio of heat rejected to heat absorbed is equal to the ratio of their absolute temperatures: Q_ out Q_ in = T_C T_H Substituting the known values: 6 8 = T_C 600 T_C = 600 3 4 = 450 K Co