Question
Question: Steam at \(100{}^\circ C\) is passed into \(20g\) of water acquires a temperature of \(80{}^\circ C\...
Steam at 100∘C is passed into 20g of water acquires a temperature of 80∘C, the mass of water presents will be [Take specific heat of water 1calg−1c−1 and also latent heat of steam is 540g−1
A.31.5gB.42.5gC.22.5gD.24g
Solution
- Hint: First find the heat required to convert 100∘C water to 100∘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=mST
Where m is the mass of water, S is the specific heat of water and T is the temperature increase taking place.
H=mL
Where L be the latent heat of steam.L
Complete step-by-step solution
As per the question, it is mentioned that, the water acquires a temperature up to 80∘C when
Steam at 100∘C is passed through it. The mass of water present at this time is 20g .
Therefore firstly let us calculate the heat energy required to increase the temperature in to 100∘C will be,
H=mST
Where m is the mass of water, S is the specific heat of water and T is the temperature increase taking place.
As per the equation,
S=1calg−1c−1
T=100−80=20∘C
Substitute these all in the equation of heat will give,
H=20×1×20=400cal
Now the temperature of water has reached 100∘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=mL
In the question it is mentioned that the latent heat of steam is,
L=540g−1
Substituting the value in it will results in,
H=20×540=10800cal
Therefore the total heat produced or developed will be
L=10800cal+400cal=11200cal
From this we can calculate the mass of the water.