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It takes 10 in for an electric kettle to theat a certain quantity of water from `0^@C` to `100^@C`. It takes 54 min to convert this water `100^@C` into steam Then latent heat of steam isA. 80 cal/gB. 540 cal/kgC. 540 cal/gD. 80 cal/kg |
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Answer» Correct Answer - C Let m be the mass of water. Quantity of heat absorbed by water in 10 min. `=msDeltaT=mxx1xx100=100m` (in calories) Quantity of heat absorbed by water in 54 min. `=(100mxx54)/(10)` Quantity of heat required to convert water into steam`=mL` Hence, `=(100mxx54)/(10)=mL` or `L=540 cal//g`. |
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