A steam at $100^{\circ} \mathrm{C}$ is passed into $1 \mathrm{~kg}$ of water contained in a calorimeter of…
A steam at $100^{\circ} \mathrm{C}$ is passed into $1 \mathrm{~kg}$ of water contained in a calorimeter of water equivalent $0.2 \mathrm{~kg}$ at $9^{\circ} \mathrm{C}$ till the temperature of the calorimeter and water in it is increased to $90^{\circ} \mathrm{C}$. The mass of steam condensed in $\mathrm{kg}$ is nearly (specific heat of water $=1 \mathrm{cal} / \mathrm{g}{ }^{\circ} \mathrm{C}$, latent heat of vaporisation $=540 \mathrm{cal} / \mathrm{g}$ )
$0.81$
$0.18$
$0.27$
$0.54$
Solution
By calorimetry,
Heat lost $=$ Heat gained
$
\begin{aligned}
m \times L+ & m \times C \times\left(T_1-T_2\right) \\
& =\left(m_1+m_2\right) \times C \times\left(T_2-T_3\right) \\
m \times 540 & +m \times 1 \times(100-90) \\
& =(1000 \mathrm{~g}+200 \mathrm{~g}) \times 1 \times(90-9) \\
& =m \times 540+10 m=1200 \times 81 \\
550 m & =1200 \times 81 \\
m & =\frac{1200 \times 81}{550}=\frac{24 \times 81}{11} \mathrm{~g} \\
& =\frac{1944}{11} \mathrm{~g}=176 \mathrm{~g}=0.176 \mathrm{~kg}
\end{aligned}
$
or $m=0.18 \mathrm{~kg}$