Zaymiey

⚛️ Physics  ·  Thermodynamics  ·  NEET & JEE

A heat engine performs 1000 J of work and rejects 1500 J to the cold reservoir. Efficiency:

Answer: 40%.

  • A 25%
  • B 33%
  • C 40%
  • D 67%

Correct answer: C. 40%

Explanation: Q<sub>hot</sub> = W + Q<sub>cold</sub> = 1000 + 1500 = 2500 J. eta = W/Q<sub>hot</sub> = 1000/2500 = 0.4 = 40%.

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 →