An experiment takes `10` minutes to raise the temperature of water in a container from `0^(@)C "to" 100^(@)C` and another `55` minutes to convert it totally into steam by a heater supplying heat at a uniform rate. Neglecting the specific heat of the container and taking specific heat of water to be `1 "cal//g ^(@)C`, the heat of vapourization according to this experiment will come out to be:-

Options

A. 560 cal/g
B. 550 cal/g
C. 540 cal/g
D. 530 cal/g

Answer

B

Explanation

As `Pt=mC DeltaT`
So, `Pxx10xx60=mC 100…(i)`
and `Pxx55xx60=mL" " …(ii)`
Dividing equation (i) by (ii) we get
`(10)/(55)=(C xx 100)/(L)`
`therefore L=550 cal.//g`.