Question

1 kg of ice at 0$^\circ$ C is mixed with 1 kg 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 $\times {10^5}$ J kg$^{ - 1}$ and latent heat of vaporization of water = 2.26 $\times {10^6}$ J kg$^{ - 1}$.

Answer 1

The specific heat capacity of water, which is the heat required to increase the temperature of unit mass of water by 1 K or 1$^\circ$ C, is approximately 4200 J kg$^{ - 1}$. To find the final composition of the system, we find out the heat required to convert ice to water and steam to water and the heat required to increase the temperature of water to 100$^\circ$ C.

Complete step by step solution:
The system comprises 1 kg ice at 0$^\circ$ C and 1 kg steam at 100$^\circ$ C. Because of the huge temperature gradient, the system is not in thermal equilibrium. To attain thermal equilibrium, there is a change in phase from ice to water and steam to water. But, we also need to find out what mass of the steam and ice will change their phases and what will be the amount of water present when the system reaches thermal equilibrium.

We assume that the system is isolated, i.e. there is no energy interaction between the system and the surrounding. We also assume that there is no energy lost during phase change.

To find out the composition of the system after it has reached thermal equilibrium, we first need to find out the energy required for the phase change of ice and steam to water and the heat required to change the temperature of the water.

Let us denote the heat required for phase change of ice to water as ${Q_1}$, the heat released by phase change of steam to water as ${Q_2}$ and the heat required to change the temperature of water as ${Q_3}$.
The heat required to change the phase of ice to water, ${Q_1}$, can be obtained by multiplying the mass of ice with the latent heat of fusion of ice.
${Q_1} = m \times {L_f} = 1 \times 3.36 \times {10^5} = 3.36 \times {10^5}$ J
The heat released by changing the phase of steam to water, ${Q_2}$, is obtained by multiplying the mass of steam with the latent heat of vaporization of water.
${Q_2} = m \times {L_v} = 1 \times 2.26 \times {10^6} = 2.26 \times {10^6}$ J

Since, the heat released by the phase change of steam to water is more than the heat required to change the phase of ice to water. So, all the ice melts into water and its temperature increases from 0$^\circ$ C to 100$^\circ$ C. But, all the steam will not get converted to water. Let us assume that the weight of steam that gets converted to water is M. On balancing the energy released by the conversion of steam to water and the energy absorbed by the ice to get converted to water and the increase in temperature of water, we obtain,
$M \times 2260000 = 336000 + 4200 = 340200$
$\Rightarrow M = \dfrac{{304200}}{{2260000}} =$ 150.5 g

Therefore, the mass of steam remaining in the system is (1000 – 150.5) g = 849.5 g and the mass of water in the system is the sum of the ice converted into water and the water obtained by the condensation of the steam, which is (1000 + 150.5) g = 1150.5 g.

Hence, the system on obtaining thermal equilibrium will have 849.5 g of steam and 150.5 g of water both at 100$^\circ$ C.

Note: In questions of this kind, we first find out how much energy is required or released during all the different processes and then we compare energy required or released during which process is more. On the basis of that, we find out the composition of the system.