JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Question Bank Calorimetry

  • question_answer
    A block of mass 100 gm slides on a rough horizontal surface. If the speed of the block decreases from 10 m/s to   5 m/s, the thermal energy developed in the process is [UPSEAT 2002]

    A)            3.75 J                                       

    B)            37.5 J

    C)            0.375 J                                     

    D)            0.75 J

    Correct Answer: A

    Solution :

                       According to energy conservation, change in kinetic energy appears in the form of heat (thermal energy).                    Þ i.e. Thermal energy \[=\frac{1}{2}m(v_{1}^{2}-v_{2}^{2})\]                 \[\left[ \because \,\underset{\text{(Joule)}}{\mathop{W}}\,=\underset{\text{(Joule)}}{\mathop{Q}}\, \right]\]            \[=\frac{1}{2}(100\times {{10}^{-3}})({{10}^{2}}-{{5}^{2}})=3.75\,J\]


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