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⚛️ Physics  ·  Thermodynamics  ·  NEET & JEE

A heat engine absorbs 800 J per cycle and rejects 320 J to the cold reservoir. Its efficiency is:

Answer: 60%.

  • A 60%
  • B 40%
  • C 50%
  • D 30%

Correct answer: A. 60%

Explanation: &eta; = 1 - Q<sub>c</sub>/Q<sub>h</sub> = 1 - 320/800 = 0.60 = 60%.

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 →