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WB JEE Medical WB JEE Medical Solved Paper-2008

  • question_answer
    An experiment takes 10 min to raise v temperature of water from \[\text{0}{{\,}^{\text{o}}}\text{C}\]and \[\text{100}{{\,}^{\text{o}}}\text{C}\]and another 55 min to convert it totally into steam by a stabilized heater. The latent heat of vaporization comes out to be

    A)  530 cal/g                            

    B)  540 cal/g

    C)  550 cal/g                            

    D)  560 cal/g

    Correct Answer: C

    Solution :

                     Heat given for raising the temperature of Wg  of water from \[\text{0}{{\,}^{\text{o}}}\text{C}\]to \[\text{100}{{\,}^{\text{o}}}\text{C=W}\times \text{1}\times 100\,\text{cal}\] Time taken \[=10\times 60\,s.\] \[\therefore \]Heat given per second \[=\frac{W\times 1\times 100}{10\times 60}\,\text{cal}\] Heat given out to convert W g to steam \[=W\times L\] This is the heat supplied in\[55\times 60\,s.\] \[\therefore \]  Heat given \[=100\times W\times \frac{55\times 60}{10\times 60}=WL\] \[\therefore \]  \[L=\frac{100\times 55\times 60}{10\times 60}=100\times 5.5\] \[L=550\,cal/g\]


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