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BHU PMT BHU PMT (Screening) Solved Paper-2010

  • question_answer
    An electric immersion heater of 1.08 kW is immersed in water. After the water has reached a temperature of\[100{}^\circ C,\]how much time will be required to produce 100 g of steam?

    A)  50s                                       

    B)  420s

    C)  105s                                     

    D)  210s

    Correct Answer: D

    Solution :

                     L is the latent heat of vaporization of water; the heat required for producing 1 g of steam,\[L=540\text{ }cal=540\times 4.2=2268\text{ }J\] Energy supplied\[=1080\text{ }J/s\] Time required to boil 100 g of water \[=540\times 4.2\times 100/1080=210\text{ }s\]


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