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JEE Main & Advanced Physics Transmission of Heat Question Bank Radiation Newton's Law of Cooling

  • question_answer
    In Newton's experiment of cooling, the water equivalent of two similar calorimeters is 10 gm each. They are filled with 350 gm of water and 300 gm of a liquid (equal volumes) separately. The time taken by water and liquid to cool from \[{{70}^{o}}C\] to \[{{60}^{o}}C\] is 3 min and 95 sec respectively. The specific heat of the liquid will be

    A)            0.3 Cal/gm ´°C                    

    B)            0.5 Cal/gm ´°C

    C)            0.6 Cal/gm ´°C                    

    D)            0.8 Cal/gm ´°C

    Correct Answer: C

    Solution :

                       \[{{S}_{l}}=\frac{1}{{{m}_{l}}}\left[ \frac{{{t}_{l}}}{{{t}_{W}}}({{m}_{W}}{{C}_{W}}+W)-W \right]\] \[=\frac{1}{300}\left[ \frac{95}{3\times 60}(350\times 1+10)-10 \right]\]=0.6 Cal/gm´°C


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