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NEET 2018 · Heat, Work & First Law of Thermodynamics

Heat, Work & First Law of Thermodynamics question from NEET 2018

Q19NEET 2018Medium

Heat, Work & First Law of Thermodynamics

A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 105 N m–2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is

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Answer: (b)

  1. Heat: Q=54 cal=54×4.18≈225.7 JQ = 54\text{ cal} = 54 \times 4.18 \approx 225.7\text{ J}
  2. Volume of 0.1 g water =0.1 cc= 0.1\text{ cc}, so ΔV=167.1−0.1=167 cc=167×10−6 m3\Delta V = 167.1 - 0.1 = 167\text{ cc} = 167\times10^{-6}\text{ m}^3
  3. W=PΔV=1.013×105×167×10−6≈16.9 JW = P\Delta V = 1.013\times10^{5} \times 167\times10^{-6} \approx 16.9\text{ J}
  4. ΔU=225.7−16.9≈208.8 J\Delta U = 225.7 - 16.9 \approx 208.8\text{ J}
  5. So the answer is (b) 208.7 J.
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