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⚛️ Physics  ·  Thermodynamics  ·  NEET & JEE

Entropy change when 0.5 kg of ice melts at 0 degrees C. Latent heat = 336 kJ/kg:

Answer: 616 J/K.

  • A 168 J/K
  • B 336 J/K
  • C 616 J/K
  • D 168000 J/K

Correct answer: C. 616 J/K

Explanation: Q = mL = 0.5 x 336000 = 168000 J. T = 273 K. delta S = Q/T = 168000/273 = 615.4 J/K approximately 616 J/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 →