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WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    A copper block of mass 4 kg is heated in a fumance to a temperature \[\text{425}{{\,}^{\text{o}}}\text{C}\]and then placed on a large ice block. The mass of ice that will melt in this process will be (Specific heat of copper\[=500\,J\,k{{g}^{-1}}-{{\,}^{o}}{{C}^{-1}}\] and heat of fusion of ice\[=336\,kJ\,k{{g}^{-1}}\]):

    A)  0.5 kg                                  

    B)  1 kg

    C)  1.5 kg                                  

    D)  2.5 kg

    Correct Answer: D

    Solution :

                     Fall in temperature of copper block when it is placed on the ice block \[=\Delta t=425-0=425{{\,}^{o}}C\]Heat lost by copper block when it is placed on the ice block. \[{{Q}_{1}}={{m}_{1}}s\Delta T\] \[=4\times 500\times 425=850\,kJ\] Heat gained by ice in melting into\[{{m}_{2}}\,kg\] of water. \[{{Q}_{2}}={{m}_{2}}L\] \[={{m}_{2}}\times 336\] \[=336\,{{m}_{2}}\,kJ\] According to calorimetry principle, heat lost = heat gained i.e.,       \[850=336\,{{m}_{2}}\] \[\therefore \]  \[{{m}_{2}}=\frac{850}{336}=2.5\,kg\]


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