CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer
    1 g of steam at\[100{}^\circ C\]and equal mass of ice at \[0{}^\circ C\]are mixed. The temperature of the mixture in steady state will be (latent heat of steam\[=540\text{ }cal/g,\]latent heat of ice = 80 \[cal/g\]):

    A)  \[50{}^\circ C\]                

    B)         \[100{}^\circ C\]             

    C)         \[67{}^\circ C\]                

    D)         \[33{}^\circ C\]

    E)  \[0{}^\circ C\]

    Correct Answer: B

    Solution :

    Heat taken by ice to raise its temperature, to \[100{}^\circ C\] \[{{Q}_{1}}=1\times 80+1\times 1\times 100\] \[=180\text{ }cal\] Heat given by steam when condensed \[{{Q}_{2}}={{m}_{2}}{{L}_{2}}\] \[=1\times 540=540\text{ }cal\] As\[{{Q}_{2}}>{{Q}_{1}},\]hence, temperature of mixture will remain\[100{}^\circ C\].


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