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Heat and Thermodynamics question

2011 · Q158
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Heat and Thermodynamics question

2011 · Q158

NEETPhysicsHeat and ThermodynamicsMCQ+4 / −1
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
  1. A
    273 cal/K
  2. B
    8 ×\times× 104 cal/K
  3. C
    80 cal/K
  4. D
    293 cal/K
View written solutionFree

Correct answer: D

Change in entropy is given by

dS=dQT⇒ΔS=ΔQT=mLf273dS = {{dQ} \over T} \Rightarrow \Delta S = {{\Delta Q} \over T} = {{m{L_f}} \over {273}}dS=TdQ​⇒ΔS=TΔQ​=273mLf​​

ΔS=1000×80273=293 cal/K\Delta S = {{1000 \times 80} \over {273}} = 293\,cal/KΔS=2731000×80​=293cal/K

Note : In the question paper unit of latent heat of ice is given to be cal/°C. It is wrong. The unit of latent heat of ice is cal/g.

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