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NEET 2011 (AIPMT, Prelims) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from NEET 2011 (AIPMT, Prelims)

Q76NEET 2011 · AIPMT · PRELIMSMedium

Heat Engines, Refrigerators & Second Law

When 1 kg of ice at 0oC melts to water at 0oC, the resulting change in its entropy, taking latent heat of ice to be 80 cal/oC, is

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Answer: (d)

  1. Melting is at constant temperature, so ΔS=QT=mLT\Delta S = \frac{Q}{T} = \frac{mL}{T}.
  2. Q=1000 g×80 cal/g=8×104 calQ = 1000\text{ g} \times 80\text{ cal/g} = 8 \times 10^4\text{ cal}, T=273 KT = 273\text{ K}.
  3. ΔS=8×104273≈293 cal/K\Delta S = \frac{8 \times 10^4}{273} \approx 293\text{ cal/K}
  4. So the answer is (d) 293 cal/K.
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