Question

1 kg of ice at $0^{\circ }C$ is mixed with 1 kg of steam at ${100}^{\circ }C$. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = $3.36\times {10}^{5}Jk{g}^{-1}$ and latent heat of vaporization of water = $2.26\times {10}^{6}Jk{g}^{-1}$

• A. 2 kg steam
• B. 2 kg ice
• C. 665 gm water, 1.335 kg steam
• D. 665 gm steam, 1.335 kg water

Answer 1

The correct option is D

665 gm steam, 1.335 kg water

As the mass of ice and steam is the same, and the latent heat of fusion Latent heat of vaporizing, let us start by assuming that all the ice converts to water at 0°C, then heats up to water at ${100}^{\circ }C$, in the process converting some of the steam to water at ${100}^{\circ }C$. Don't worry, if our assumption is incorrect, the mass of water will come out negative and we can then consider other possibilities!
So, Heat gained during ice at $0^{\circ }C$ $\to$ water at ${100}^{\circ }C$

$Q_1$ = $m{L}_1+ms\Delta T$

$m$ = $1kg$

$L_1$ $\to$ Latent heat of fusion
$\Delta T$ = ${100}^{\circ }C$
$S\to$ specific heat of water
$\Rightarrow$ $Q_1$ = $L_1+100S$

Now, heat released by $x$ kg steam converted to water at $\to$ is

$Q_2$ = $x{L}_2$

Where, $L_2$$\to$ latent heat of vaporization

But, $Q_1$ = $Q_2$

$L_1+100S$ = $x{L}_2$
$\Rightarrow$ $x$ = $\frac{L_1+100s}{L_2}$

$\Rightarrow$ $x$ = $0.335kg$ steam converts to water

$\therefore$ Amount of steam left = $(1-0.335)kg$ = $0.665kg$