Q. When 1 kg of ice at 0°C melts to water at 0°C, the resulting change in its entropy, taking latent heat of ice to be 80 cal/g, is
A) 293 cal/K
B) 273 cal/K
C) 8 × 10⁴ cal/K
D) 80 cal/K
Explanation
[Solution from source] Heat absorbed Q = mL = 1000 g × 80 cal/g = 8 × 10⁴ cal at constant T = 273 K. ΔS = Q/T = 80000/273 ≈ 293 cal/K. Giving 8 × 10⁴ cal/K forgets to divide by the temperature. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.