Exams › JEE Main › Physics
A Carnot engine, whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500K. It is desired to have an engine of efficiency 60%. Then, the intake temperature for the same exhaust (sink) temperature must be:
- efficiency of Carnot engine cannot be made larger than 50%
- 1200 K
- 750 K
- 600 K
Correct answer: 750 K
Solution
To achieve a higher efficiency in a Carnot engine, the temperature of the heat source must be increased. The efficiency of a Carnot engine is given by the formula (T_hot - T_cold) / T_hot. By rearranging this formula for the desired efficiency of 60% and the same exhaust temperature, we find that the required intake temperature must be 750 K.
Related JEE Main Physics questions
⚔️ Practice JEE Main Physics free + battle 1v1 →