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 $=3.36\times {{10}^{5}}J/g$ and latent heat of vaporization of water \[=2.26\times {{10}^{6}}J/g\].

Answer 1

Hint: It is said that ice and steam is mixed together. We are given the mass and temperature of both. We are also given the latent heat of fusion of ice and latent heat of vaporization of water. We know that when ice and steam is mixed water will be formed; hence we can find the heat required to change ice into water and heat absorbed by water to raise its temperature. By calculating the total heat and heat produced by steam we can find the remaining heat in the system. Using this we can find the composition of steam and water in the system at thermal equilibrium.

Formula used:
Heat, $Q=mL$
Heat required for raising the temperature, $Q=ms\Delta T$

Complete step by step answer:
In the question it is said that 1 kg of ice at ${{0}^{o}}C$ is mixed with 1 kg of steam at ${{100}^{o}}C$.
From this we know the amount of ice at ${{0}^{o}}C$,
${{m}_{i}}=1kg$
Amount of steam at ${{100}^{o}}C$,
${{m}_{s}}=1kg$
Latent heat of fusion of ice and latent heat of vaporization of water is given to us
${{L}_{i}}=3.36\times {{10}^{5}}J/g$, ‘${{L}_{i}}$’ is latent heat of fusion of ice
\[{{L}_{s}}=2.26\times {{10}^{6}}J/g\], ‘${{L}_{s}}$’ is latent heat of vaporization of water
From this we can see that the latent heat of fusion of ice is less than the latent heat of vaporization of water, i.e.
${{L}_{i}}<{{L}_{s}}$
Therefore we can say that ice will first melt into water because it requires less heat. And there will be equilibrium between water and steam.
Now we can find the heat required by ice to change into water, ${{Q}_{1}}$
${{Q}_{1}}={{m}_{i}}{{L}_{i}}$
$\begin{align}
  & \Rightarrow {{Q}_{1}}=1\times 3.36\times {{10}^{5}} \\
 & \Rightarrow {{Q}_{1}}=3.36\times {{10}^{5}}J \\
\end{align}$
Now heat absorbed by water to change its temperature from ${{0}^{o}}C$ to ${{100}^{o}}C$, ${{Q}_{2}}$
${{Q}_{2}}={{m}_{i}}{{S}_{w}}\left( 100-0 \right)$
Here ${{S}_{w}}$ is specific heat of water.
We know that specific heat of water, ${{S}_{w}}=4200J$
Therefore, heat absorbed by water to change its temperature from ${{0}^{o}}C$ to ${{100}^{o}}C$
$\begin{align}
  & {{Q}_{2}}=1\times 4200\times 100 \\
 & \Rightarrow {{Q}_{2}}=4.2\times {{10}^{5}}J \\
\end{align}$
Now the total heat absorbed by ice to raise it temperature to ${{100}^{o}}C$
$\begin{align}
  & Q={{Q}_{1}}+{{Q}_{2}} \\
 & \Rightarrow Q=3.36\times {{10}^{5}}+4.2\times {{10}^{5}} \\
\end{align}$
$\Rightarrow Q=7.56\times {{10}^{5}}J$
We know that the heat required to change ice into water is provided by steam. The steam will release heat and this heat released will convert ice into water.
If all steam gets converted into water, we can find the heat released by steam,
$\begin{align}
  & Q'={{m}_{s}}{{L}_{s}} \\
 & \Rightarrow Q'=1\times 2.26\times {{10}^{6}} \\
\end{align}$
$\Rightarrow Q'=2.26\times {{10}^{6}}J$
By comparing this with ‘Q’ we can see that the heat released by steam is greater than the heat required by ice to rise to ${{100}^{o}}C$, i.e.
$Q' > Q$
Therefore the remaining heat after the conversion of ice into water will be
$\begin{align}
  & Q'-Q \\
 & \Rightarrow 2.26\times {{10}^{6}}-7.56\times {{10}^{5}} \\
\end{align}$
$\Rightarrow 1.504\times {{10}^{6}}J$
Let us assume that the mass of steam gets condensed in water, let this be ‘m’ then
$m=\dfrac{Q}{{{L}_{s}}}$
$\Rightarrow m=\dfrac{7.56\times {{10}^{5}}}{2.26\times {{10}^{6}}}$
$\Rightarrow m=0.335kg=335g$
Therefore total amount steam left will be,
$\begin{align}
  & {{m}_{s}}-m \\
 & \Rightarrow 1-0.335 \\
 & \Rightarrow 0.665kg=665g \\
\end{align}$
And the total amount of water at ${{100}^{o}}C$ will be,
$\begin{align}
  & {{m}_{i}}+m \\
 & \Rightarrow 1+0.335 \\
 & \Rightarrow 1.335kg=1335g \\
\end{align}$
Hence at thermal equilibrium we have, $1.335kg$ of water and $0.665kg$ of steam.

Note:
Thermal equilibrium is a state in which there is no exchange of heat energy between the two substances which are in physical contact with each other. Hence we can say that when two substances are in thermal equilibrium, they will be at the same temperature.
Specific heat capacity of a substance is the measure of the amount of heat required to raise the temperature of 1 kg of that substance to 1 Kelvin.