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 is

Options

A. 80 cal/g
B. 540 cal/kg
C. 540 cal/g
D. 80 cal/kg

Answer

C

Explanation

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`.