Question

An experiment takes $10$ min to raise the temperature of water from $0^{\circ }C$ and ${100}^{\circ }C$ and another $55$ min to convert it totally into steam by a stabilized heater. The latent heat of vaporization comes out to be

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

Answer 1

Answer: (C)

$550cal/g$

Heat given for raising the temperature of Wg of water from $0^{\circ }Cto{100}^{\circ }C=W\times 1\times 100cal$
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 $=100\times W\times \frac{55\times 60}{10\times 60}=WL$
$\therefore L=\frac{100\times 55\times 60}{10\times 60}=100\times 5.5$
$L=550cal/g$