Question

Steam at $100{}^\circ C$ is passed into $20\;g$ of water at $10{}^\circ C$ . When water acquires a temperature of $80{}^\circ C$ , the mass of water present will be [Take specific heat of water = $1cal{{g}^{-1}}{}^\circ {{C}^{-1}}$ , latent heat of steam = $540cal{{g}^{-1}}$ ]
(A) $31.5\;g$
(B) $42.5\;g$
(C) $22.5\;g$
(D) $24\;g$

Answer 1

Hint : Here, the steam that is at a higher temperature is brought in contact with water that is at a lower temperature. Hence, the heat lost by the steam is equal to the heat gained by the water. By equating we can find the total amount of water present after the process.

Complete Step By Step Answer:
Let us note down the given data;
Initial temperature of steam ${{T}_{si}}=100{}^\circ C$
Initial temperature of water ${{T}_{wi}}=10{}^\circ C$
Final temperature of water ${{T}_{wf}}=80{}^\circ C$
Initial mass of water ${{m}_{w}}=20g$
Specific heat of water ${{C}_{pw}}=1cal{{g}^{-1}}{}^\circ {{C}^{-1}}$
Latent heat of steam ${{L}_{s}}=540cal{{g}^{-1}}$
Now, here the steam at a higher temperature is passed through the water at a lower temperature.
By the laws of thermodynamics, we know that the heat flows from the hotter body (here steam) to the colder body (here water).
Hence, the heat flows from the steam whose temperature decreases, to the water whose temperature increases.
This heat flows continues till the temperature of both bodies is equal. Thus the final temperature of the steam is equal to the final temperature of the water.
$\therefore$ Final temperature of steam ${{T}_{sf}}=80{}^\circ C$
But, we know that steam cannot exist below $100{}^\circ C$ , hence at the final temperature, steam is present in the form of water.
Now, the heat gained by water can be calculated as
${{Q}_{w}}={{m}_{w}}{{C}_{pw}}\Delta {{T}_{w}}$
Where, $\Delta {{T}_{w}}$ is the difference in the initial and the final temperature
$\therefore {{Q}_{w}}={{m}_{w}}{{C}_{pw}}({{T}_{fw}}-{{T}_{iw}})$
Substituting the given values,
$\therefore {{Q}_{w}}=(20g)(1cal{{g}^{-1}}{}^\circ {{C}^{-1}})(80{}^\circ C-10{}^\circ C)$
$\therefore {{Q}_{w}}=(20g)(1cal{{g}^{-1}}{}^\circ {{C}^{-1}})(70{}^\circ C)$
Without considering the units,
$\therefore {{Q}_{w}}=20\times 1\times 70$
$\therefore {{Q}_{w}}=1400cal$ …… $(1)$
Now, the heat lost by the steam takes place in two steps. The steam converts from steam to water and then the temperature of the water decreases.
Now, for the phase change from steam to water, we know that a phase change always takes place at a constant temperature.
Hence, the heat lost by steam at $100{}^\circ C$ to convert to water at $100{}^\circ C$ is
${{Q}_{1}}={{m}_{s}}{{L}_{s}}$
Substituting the given values,
$\therefore {{Q}_{1}}={{m}_{s}}(540cal{{g}^{-1}})$
Without considering the units,
$\therefore {{Q}_{1}}=540{{m}_{s}}$
After converting to water, it loses the heat by decrease of temperature up to $80{}^\circ C$
${{Q}_{2}}={{m}_{s}}{{C}_{pw}}({{T}_{is}}-{{T}_{fs}})$
Substituting the given values,
${{Q}_{2}}={{m}_{s}}(1cal{{g}^{-1}}{}^\circ {{C}^{-1}})(100{}^\circ C-80{}^\circ C)$
$\therefore {{Q}_{2}}={{m}_{s}}(1cal{{g}^{-1}}{}^\circ {{C}^{-1}})(20{}^\circ C)$
Without considering the units,
$\therefore {{Q}_{2}}=20{{m}_{s}}$
Adding both the heats, we get the total heat lost by steam
${{Q}_{s}}={{Q}_{1}}+{{Q}_{2}}$
Substituting the obtained values,
$\therefore {{Q}_{s}}=540{{m}_{s}}+20{{m}_{s}}$
$\therefore {{Q}_{s}}=(560{{m}_{s}})cal$ …… $(2)$
Now, as the heat gained by water is equal to the heat lost by steam,
${{Q}_{w}}={{Q}_{s}}$
Substituting the calculated values,
$\therefore 1400=560{{m}_{s}}$
$\therefore {{m}_{s}}=2.5g$
Hence, $2.5\;g$ of steam is converted to water.
Hence, the total water present at the end is = $20+2.5=22.5g$
Hence, the correct answer is Option $(C)$ .

Note :
As per the laws of thermodynamics, heat exchange takes place till the temperature of both bodies is the same. Hence, whenever the final temperature of one body is given, the final temperature of the other body is also the same. Also, the phase change always takes place at a constant temperature.