Steam at $100^{\circ} \mathrm{C}$ is passed into $1 \mathrm{~kg}$ of water contained in a calorimeter at…
Steam at $100^{\circ} \mathrm{C}$ is passed into $1 \mathrm{~kg}$ of water contained in a calorimeter at $9^{\circ} \mathrm{C}$ till the temperature of water and calorimeter is increased to $90^{\circ} \mathrm{C}$. The mass of the steam condensed is nearly
(water equivalent of calorimeter $=0.1 \mathrm{~kg}$, specific heat of water $=1 \mathrm{calg}^{-1}{ }^{\circ} \mathrm{C}^{-1}$
and latent heat of vaporisation $=540 \mathrm{calg}^{-1}$ )
$81 \mathrm{~g}$
$162 \mathrm{~g}$
$243 \mathrm{~g}$
$486 \mathrm{~g}$
Solution
Let mass of the steam condensed is $x$.
Heat released $=$ Heat gained by water
$
\begin{aligned}
& \Rightarrow x \times 540+x \times 1 \times(100-90) \\
& \quad=1 \times 1 \times(90-9)+0.1 \times 1 \times(90-9) \\
& \Rightarrow 540 x+10 x=81+81 \\
& \Rightarrow \quad x=\frac{891}{550}=0 \cdot 162 \mathrm{~kg}, x=162 \mathrm{~g}
\end{aligned}
$