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JEE Main & Advanced Physics Thermodynamical Processes Question Bank Critical Thinking

  • question_answer
    Find the change in the entropy in the following process 100 gm of ice at 0°C melts when dropped in a bucket of water at 50°C (Assume temperature of water does not change)             [BHU (Med.) 2000]

    A)            ? 4.5 cal/K                             

    B)            + 4.5 cal/K

    C)            +5.4 cal/K                              

    D)            ? 5.4 cal/K 

    Correct Answer: B

    Solution :

                       Gain of entropy of ice                    \[{{S}_{1}}=\frac{\Delta Q}{T}=\frac{mL}{T}=\frac{80\times 100}{(0+273)}=\frac{8\times {{10}^{3}}}{273}cal/K\]                    Loss of entropy of water \[={{S}_{2}}=-\frac{\Delta Q}{T}=-\frac{mL}{T}\]                    \[=\frac{80\times 100}{(273+50)}=\frac{8\times {{10}^{3}}}{323}cal/K\]                    Total change of entropy            \[{{S}_{1}}+{{S}_{2}}=\frac{8\times {{10}^{3}}}{273}-\frac{8\times {{10}^{3}}}{323}=+4.5\ cal/K\]


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