RAJASTHAN PMT Rajasthan - PMT Solved Paper-2005

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

    A)  210 s                    

    B)         105 s

    C)  420 s                    

    D)         50 s              

    Correct Answer: A

    Solution :

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


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