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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    A 2 kg copper block is heated to\[500{}^\circ C\]and then it is placed on a large block of ice at\[0{}^\circ C\]. If the specific   heat  capacity  of  copper  is \[400J\,k{{g}^{-1}}{}^\circ {{C}^{-1}}\]and latent heat of fusion of water is\[3.5\times {{10}^{5}}J\text{ }k{{g}^{-1}},\]the amount of ice that can melt is:

    A)  7/8 kg                                  

    B)  7/5 kg

    C)  8/7 kg                  

    D)         5/7 kg

    E)  7/3 kg.

    Correct Answer: C

    Solution :

    Heat emitted by copper = Heat gained by ice \[mc\Delta \theta =mL\] \[\Rightarrow \]               \[m=\frac{mc\Delta \theta }{L}\] Given, \[m=2kg,c=400J\,k{{g}^{-1}}{{C}^{-1}},\]                 \[\Delta \theta =500,L=3.5\times {{10}^{5}}\,J\,k{{g}^{-1}}\] \[\therefore \]  \[m=\frac{2\times 400\times 500}{3.5\times {{10}^{5}}}=\frac{8}{7}kg\]


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