Question

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

Answer 1

Hint: Here in this question ice and steam are mixed so there would be an exchange of heat between the ice and steam and both of them change their state in which ice by absorbing the heat from the steam will convert into the water and simultaneously the steam by releasing the heat will also get converted into water. Hence we have to find the amount of steam and water at thermal equilibrium.

Complete step by step solution:
$ {Q_{ice}} = {m_{ice}} \times {L_{ice}}$
${Q_{water}} = {m_{ice}} \times {S_w} \times \Delta T$
Therefore we can say that,
$ \Rightarrow {Q_{total}} = {m_{cond.}} \times {L_{steam}}$
The latent heat of ice and steam at their respective temperature is given. On comparing them we can deduce that the latent heat of ice is less than the latent heat of steam.
Let, latent heat of fusion (ice) be ${L_{ice}}$ and latent heat of vaporization (steam) be ${L_{steam}}$ .
$\therefore {L_{ice}} < {L_{steam}}$
So the ice will get converted into the water first because it requires less heat as given ${L_{ice}} = 3.36 \times {10^5}\dfrac{J}{{kg}}$
Hence the heat absorbed by the ice to get converted into water will be given by
$ \Rightarrow {Q_{ice}} = {m_{ice}} \times {L_{ice}}$
Where the mass of water
$ \Rightarrow {Q_{ice}} = 1kg \times 3.36 \times {10^5}\dfrac{J}{{kg}}$
$\therefore {Q_{ice}} = 3.36 \times {10^5}J$${m_{ice}} = 1kg$
Now to get converted into steam water I had to absorb some heat. Hence total heat absorbed due to ice is given by (changing to $100^\circ {\text{C}}$)
$ \Rightarrow {Q_{water}} = {m_{ice}} \times {S_w} \times \Delta T$
where ${S_w} = 4200$ (specific heat of water) and $\Delta T = 100^\circ {\text{ - 0}}^\circ $ , hence
$ \Rightarrow {Q_{water}} = 1kg \times 4200 \times 100$
$\therefore {Q_{water}} = 4.2 \times {10^5}J$
Now the total heat absorbed by ice to raise its temperature from $0^\circ {\text{C}}$to $100^\circ {\text{C}}$
$\; \Rightarrow {Q_{total}} = \;{Q_{ice}} + \;{Q_{water}}$
Substituting the values in the above equation we get,
$\; \Rightarrow {Q_{total}} = \;3.36 \times {10^5}J + 4.2 \times {10^5}J$
$\therefore {Q_{total}} = 7.56 \times {10^5}J$
Now as the ice is converting into the water by absorbing the heat that is given off by steam. By doing so steam is also losing heat, as a result, it will start condensing and converting into water.
Let all the steam converted into water, hence heat released by the steam will be
$ \Rightarrow {Q_{steam}} = {m_{steam}} \times {L_{steam}}$
$ \Rightarrow {Q_{steam}} = 1kg \times 2.26 \times {10^6}\dfrac{J}{{kg}}$
Where ${m_{steam}} = 1kg$the mass of steam
$\therefore {Q_{steam}} = 2.26 \times {10^6}J$
Now we find the mass of steam which is converted into water by condensation
$ \Rightarrow {Q_{total}} = {m_{cond.}} \times {L_{steam}}$
$ \Rightarrow {m_{cond.}} = \dfrac{{{Q_{total}}}}{{{L_{steam}}}}$
Therefore on putting the values in the above equation,
$ \Rightarrow {m_{cond.}} = \dfrac{{7.56 \times {{10}^5}J}}{{2.26 \times {{10}^6}\dfrac{J}{{kg}}}}$
$\therefore {m_{cond.}} = 0.335kg$
Similarly, the total amount of water can be obtained as a sum of the amount of water at $100^\circ {\text{C}}$and steam that gets condensed into water
$ \Rightarrow {M_{water}} = 1kg + 0.335kg$
$\therefore {M_{water}} = 1.335kg$
Hence the amount of steam left at thermal equilibrium will be
$ \Rightarrow {M_{steam}} = {m_{steam}} - {m_{cond.}}$
Therefore on putting the value in the above equation,
$ \Rightarrow {M_{steam}} = 1kg - 0.335kg$
$\therefore {M_{steam}} = 0.665kg$
 Hence at thermal equilibrium, we would get $1.335{\text{ }}kg$ water as all the ice is converted into water and $0.665{\text{ }}kg$ steam left.

Note: In the given problem we have used the formula of heat energy formula and we have also used the specific heat formula in the above question in which we have taken specific heat of the water as 4200 which is not available in the above question.