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  4. 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?

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Q. 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?

ManipalManipal 2003Current Electricity

Solution:

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