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Heat and Thermodynamics question

2025 · 3 Apr · Shift 1 · Q60
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Heat and Thermodynamics question

2025 · 3 Apr · Shift 1 · Q60

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
During the melting of a slab of ice at 273 K at atmospheric pressure :
  1. A
    Internal energy of ice-water system remains unchanged.
  2. B
    Positive work is done by the ice-water system on the atmosphere.
  3. C
    Positive work is done on the ice-water system by the atmosphere.
  4. D
    Internal energy of the ice-water system decreases.
View written solutionFree

Correct answer: C

  1. Process identification

A slab of ice melts at T=273 KT=273\,\text{K}T=273K and at atmospheric pressure. This is a phase change from ice to water at constant temperature and constant external pressure.

  1. Use the first law of thermodynamics

For the system (ice + water), ΔQ=ΔU+W\Delta Q = \Delta U + WΔQ=ΔU+W where WWW is the work done by the system.

During melting, heat is absorbed as latent heat of fusion: ΔQ>0\Delta Q > 0ΔQ>0

  1. Change in internal energy

Melting requires energy to weaken/break part of the intermolecular bonding structure of ice. Hence the internal energy increases: ΔU>0\Delta U > 0ΔU>0 So:

  • Option A (internal energy unchanged) is false.
  • Option D (internal energy decreases) is false.
  1. Work done during melting

Ice has a larger volume than the same mass of water near 273 K273\,\text{K}273K. So when ice melts, the volume decreases: ΔV<0\Delta V < 0ΔV<0

At constant atmospheric pressure, work done by the system is W=PΔVW = P\Delta VW=PΔV Since ΔV<0\Delta V<0ΔV<0, W<0W<0W<0 This means the system does negative work on the atmosphere, or equivalently, the atmosphere does positive work on the system.

Therefore:

  • Option B is false.
  • Option C is true.
  1. Final conclusion

The correct option is: C\boxed{\text{C}}C​

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