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Manipal Medical Manipal Medical Solved Paper-2003

  • 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)  210 s          

    B)  105 s

    C)  420 s          

    D)  50 s

    Correct Answer: A

    Solution :

     The heat required for producing 1 g of steam = 540 cal \[=540\times 4.2\text{ }J=2268\text{ }J\] Energy given by immersion heater is \[=1.08\text{ }kW=1080\text{ }W\] Now time taken in boiling 100 g of water \[=\frac{2268\times 100}{1080}=210\,\sec \]


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