A calorimeter of water equivalent 15 g contains 165 g of water at $25^{\circ} \mathrm{C}$. Steam at

Question

A calorimeter of water equivalent 15 g contains 165 g of water at $25^{\circ} \mathrm{C}$. Steam at $100^{\circ} \mathrm{C}$ is passed through the water for some time. The temperature is increased to $30^{\circ} \mathrm{C}$ and the mass of the calorimeter and its contents is increased by 1.5 g . Calculate the specific latent heat of vaporization of water(in cal/g). Specific heat capacity of water is $1 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$.

Professor PC Thomas Classes · Answer

Let $L$ be the specific latent heat of vaporization of water. The mass of the steam condensed is 1.5 g . Heat lost in condensation of steam is
$
Q_1=(1 \cdot 5 \mathrm{~g}) L .
$
The condensed water cools from $100^{\circ} \mathrm{C}$ to $30^{\circ} \mathrm{C}$. Heat lost in this process is
$
Q_2=(1 \cdot 5 \mathrm{~g})\left(1 \mathrm{cal} \mathrm{~g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\right)\left(70^{\circ} \mathrm{C}\right)=105 \mathrm{cal} .
$
Heat supplied to the calorimeter and to the cold water during the rise in temperature from $25^{\circ} \mathrm{C}$ to $30^{\circ} \mathrm{C}$ is
$
Q_3=(15 \mathrm{~g}+165 \mathrm{~g})\left(1 \mathrm{cal} \mathrm{~g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\right)\left(5^{\circ} \mathrm{C}\right)=900 \mathrm{cal} .
$
If no heat is lost to the surrounding,
$
(1 \cdot 5 \mathrm{~g}) L+105 \mathrm{cal}=900 \mathrm{cal}
$
$
L=530 \mathrm{cal} \mathrm{~g}^{-1} .
$