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⚛️ Physics  ·  Thermodynamics  ·  NEET & JEE

A Carnot engine has 60% efficiency. If cold reservoir is at 300 K, hot reservoir temperature:

Answer: 750 K.

  • A 500 K
  • B 600 K
  • C 750 K
  • D 900 K

Correct answer: C. 750 K

Explanation: eta = 1 - T<sub>cold</sub>/T<sub>hot</sub>. 0.6 = 1 - 300/T<sub>hot</sub>. 300/T<sub>hot</sub> = 0.4. T<sub>hot</sub> = 750 K.

Carnot Cycle (P-V Diagram)VP1→2: isothermal expansion (T_H)2→3: adiabatic expansion3→4: isothermal compression (T_C)4→1: adiabatic compression1234Net work done by the gas = area enclosed by the loop 1→2→3→4→1

The Carnot cycle alternates two isothermal steps (heat absorbed at TH, released at TC) with two adiabatic steps (no heat exchange); the enclosed loop area equals the net work output, and η = 1−TC/TH is the maximum possible efficiency between those two temperatures.

Concept context

Laws of thermodynamics, heat engines, entropy, and gas processes.

Read the full Thermodynamics notes →