Q. An experiment takes 10 min to raise temperature of water from 0$℃$ and 100$℃$ and another 55 min to convert it totally into steam by a stabilized heater. The latent heat of vaporization comes out to be
A) ${530\ calg}^{- 1}$
B) ${540\ calg}^{- 1}$
C) ${550\ calg}^{- 1}$
D) ${560\ calg}^{- 1}$
Explanation
(c) Heat given for raising the temperature of W g of water from
$0℃\ to\ 100℃ = W \times 1 \times 100\ cal$
Time taken = $10 \times 60s$.
$\therefore\ \ \ Heat\ given\ per\ second = \ \frac{W \times 1 \times 100}{10 \times 60}$ cal
Heat given out to convert W g to steam = $W \times L$
This is the heat supplied in $55 \times 60s$
$\therefore\ \ \ Heat\ given = 100 \times W \times \frac{55 \times 60}{10 \times 60} = WL$
$\therefore\ \ \ \ L = \ \frac{100 \times 55 \times 60}{10 \times 60} = 100 \times 5.5$
$L = 550\ calg^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.