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 = $3.36\times {10}^{5}Jk{g}^{-1}$, specific heat capacity of water = 4200 J kg ⁻¹ K⁻¹ and latent heat of vaporization of water = $2.26\times {10}^{6}Jk{g}^{-1}$

• A. 2kg steam
• B. 2 kg ice
• C. 665gm 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, 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 = $(1-0.335)kg$ = $665gm$