Question

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

Answer 1

Hint: Specific heat of water is the heat required to increase the temperature of 1 kg by 1 degree centigrade. Its value is approximately equal to $4200J\text{ }k{{g}^{-1}}$. To calculate the final composition of the system, we calculate the heat needed to convert ice to water and steam to water. Then, we finally find out the heat required to increase the temperature of water to ${{100}^{o}}C$.

Formula used: $Q=mL$

Complete step by step answer:
Consider a system consisting of 1kg of ice at ${{0}^{o}}C$and 1 kg of steam at ${{100}^{o}}C$. There is a temperature difference in the system, due to which the system is not in thermal equilibrium.
Now, the system will try to attain equilibrium as a consequence of ice will change to water and steam will change to water.
Assuming that the system is isolated, there will be no energy interaction between the system and surrounding. We also assume that no energy is lost during change of phase.
We will first find out the energy required for phase change of ice and steam to water .Also, we need to find the heat energy needed to change the temperature of water.
Now, Let ${{Q}_{1}}$be the heat needed for phase change of ice to water.
${{Q}_{2}}$ be the heat released by phase change of steam to water.
${{Q}_{3}}$be the heat needed to change the temperature of water.
Here , we can say that
${{Q}_{1}}=m{{L}_{1}}\text{ }...........\text{(1)}$
Where m is the mass of ice and ${{L}_{1}}$is the latent heat of fusion of ice.
$\begin{align}
  & \Rightarrow {{Q}_{1}}=1\times 3.36\times {{10}^{5}} \\
 & \therefore {{Q}_{1}}=3.36\times {{10}^{5}}J \\
\end{align}$
Also,
${{Q}_{2}}=m\times {{L}_{2}}$
Where ‘m’ is the mass of steam and ${{L}_{2}}$will be the latent heat of vaporization of water.
$\begin{align}
  & \Rightarrow {{Q}_{2}}=1\times 2.26\times {{10}^{6}} \\
 & \therefore {{Q}_{2}}=2.26\times {{10}^{6}}Joules \\
\end{align}$
Since, ${{Q}_{1}}\text{ }$is greater than ${{Q}_{2}}$, ice melts to water and its temperature increases from ${{0}^{o}}C$to ${{100}^{o}}C$. But all the steam is not converted to water. So, we will assume that the amount of steam that gets converted to water will be ‘M’.
Let us balance the energy released by the conversion of steam to water and the energy absorbed by ice to get converted to water and increase in temperature of water, we will get
$\begin{align}
  & M\times 2260000=336000+4200 \\
 & \Rightarrow M=\dfrac{340200}{2260000} \\
 & \therefore M=150.5g \\
\end{align}$
Therefore the mass of the steam which will remain in the system will be
$1000-150.5=849.5g$
And the mass of water in the system will be the sum of ice converted to water and water which will be obtained by the condensation of the steam,
$1000+150.5g=1150.5g$

Hence the system when attaining thermal equilibrium it will have 849.5g of steam and 150.5g of water both at a temperature of ${{100}^{o}}C$.

Note: For solving questions of these types, first find out how much energy is needed or released in the process. Then we compare the energy required during which process is more.
Using this method we can find out the composition of the system.