Question

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 1cal/g C, the heat of vapourization according to this experiment will come out to be

Answer 1

Explanation:

Assuming heater is supplying heat at uniform rate of

Q

˙

cal/s

in 10 minutes total heat transferred =600×

Q

˙

This amount should be equal to heat gained by water =mcΔT where m is mass of water and c is specific heat capacity of water.

mcΔT=600

Q

˙

m×1calg

−1

C

∘

−1

×(100−0)=600

Q

˙

⇒

Q

˙

=

6

m

...(i)

The heater supplies heat at constant temperature to convert water into steam, the amount of heat is given by

55minutes×60×sec×

Q

˙

..(i)

this amount should be equal to heat gained by water to convert into steam at constant temperature =mL where L is latent heat of vaporization

3300

Q

˙

=mL...(ii)

substituting value of

Q

˙

from equation (i)

3300

6

m

=mL

L=550cal/g

Hence correct answer is option B