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JEE Main & Advanced Physics Thermodynamical Processes Question Bank Self Evaluation Test - Thermodynamics

  • question_answer
    When 1 kg of ice at \[0{}^\circ C\] melts to water at \[0{}^\circ C,\]the resulting change in its entropy, taking latent heat of ice to be 80 cal/\[{}^\circ C\], is

    A) 273 cal/K

    B) 253 cal/K

    C) 263 cal/K

    D) 293 cal/K

    Correct Answer: D

    Solution :

    [d] Change in entropy is given by \[dS=\frac{dQ}{T}\text{ or }\Delta S=\frac{\Delta Q}{T}=\frac{m{{L}_{f}}}{273}\] \[\Delta S=\frac{1000\times 80}{273}=293\text{ cal/K}\text{.}\]


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