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Question: Steam at \(100{}^\circ C\) is passed into \(20g\) of water acquires a temperature of \(80{}^\circ C\...

Steam at 100C100{}^\circ C is passed into 20g20g of water acquires a temperature of 80C80{}^\circ C, the mass of water presents will be [Take specific heat of water 1calg1c11ca{{\lg }^{-1}}{{c}^{-1}} and also latent heat of steam is 540g1540{{g}^{-1}}
A.31.5g B.42.5g C.22.5g D.24g \begin{aligned} & A.31.5g \\\ & B.42.5g \\\ & C.22.5g \\\ & D.24g \\\ \end{aligned}

Explanation

Solution

- Hint: First find the heat required to convert 100C100{}^\circ C water to 100C100{}^\circ C vapour. Then calculate the total heat required for the process and from that. After this calculate the mass of water present at this occasion.

Formula used:
H=mSTH=mST
Where mm is the mass of water, SS is the specific heat of water and TT is the temperature increase taking place.
H=mLH=mL
Where LL be the latent heat of steam.LL

Complete step-by-step solution
As per the question, it is mentioned that, the water acquires a temperature up to 80C80{}^\circ C when
Steam at 100C100{}^\circ C is passed through it. The mass of water present at this time is 20g20g .
Therefore firstly let us calculate the heat energy required to increase the temperature in to 100C100{}^\circ C will be,
H=mSTH=mST
Where mm is the mass of water, SS is the specific heat of water and TT is the temperature increase taking place.
As per the equation,
S=1calg1c1S=1ca{{\lg }^{-1}}{{c}^{-1}}
T=10080=20CT=100-80=20{}^\circ C
Substitute these all in the equation of heat will give,
H=20×1×20=400calH=20\times 1\times 20=400cal
Now the temperature of water has reached 100C100{}^\circ C , it is still in a liquid state.
Now let us convert the liquid state in to vapour state,
Therefore we can write that,
The heat consumption during the process will be,
H=mLH=mL
In the question it is mentioned that the latent heat of steam is,
L=540g1L=540{{g}^{-1}}
Substituting the value in it will results in,
H=20×540=10800calH=20\times 540=10800cal
Therefore the total heat produced or developed will be
L=10800cal+400cal=11200calL=10800cal+400cal=11200cal
From this we can calculate the mass of the water.

& m=\dfrac{11200}{400}g \\\ & m=22.5g \\\ \end{aligned}$$ Therefore the mass of water will be $$22.5g$$ . **Hence the correct answer is given as option C.** **Note:** Two general forms of latent heat will be latent heat of fusion in the case of melting and latent heat of vaporization in the case of boiling. These names explain the direction of energy flow if varying from one phase to the next. That means from solid to liquid, and liquid to gas.