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
Answers
Answered by
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
Similar questions