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MHT CET 2026 (16 Apr, Shift 2) · Heat, Work & First Law of Thermodynamics

Heat, Work & First Law of Thermodynamics question from MHT CET 2026 (16 Apr, Shift 2)

Q5MHT CET 2026 · 16 APR · SHIFT 2Medium

Heat, Work & First Law of Thermodynamics

One gram of a liquid is converted to vapour at 3×1053 \times 10^{5} Pa pressure. If 10% of the heat supplied is used for increasing the volume by 16001600 cm33 during this phase change, then the increase in internal energy in the process will be

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Answer: (c)

  1. Work done in expansion: W=PΔV=3×105×1600×10−6=480 JW = P\Delta V = 3\times10^{5} \times 1600\times10^{-6} = 480\text{ J}
  2. This is 10% of heat supplied: Q=4800 JQ = 4800\text{ J}
  3. ΔU=Q−W=4800−480=4320 J\Delta U = Q - W = 4800 - 480 = 4320\text{ J}
  4. So the answer is (c) 4320 J.
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