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BVP Medical BVP Medical Solved Paper-2005

  • question_answer
    An electric immersion heater of 1.08 kW is immersed in water. After the water has reached a temperature of 100°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 \[=540cal=540\times 4.2J=2268J\] Energy given by immersion heater is \[=1.08\,kW=1080W\] Now time taken in boiling 100 g of water                 \[=\frac{2268\times 100}{1080}=210\sec \]


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