Steam at 100∘C is passed into 1.1kg of water contained in a calorimeter of water equivalent 0.02kg at 15∘C till the temperature of the calorimeter and its contents reaches 80∘C. Considering the latent heat of steam =540calg−1 and specific heat capacity of water =4183Jkg−1K−1 or =1calkg−1K−1,
calculate the mass of the steam condensed.
0.130
Given:
Mass of water and calorimeter =1.10+0.02=1.12kg
Specific heat capacity of water =4.184×103Jkg−1K−1=1calkg−1K−1
Let the mass of steam=mkg
Latent heat of vaporization of water at 100∘C=540calg−1
We know that,
Heat lost by steam=Heat gained by water+calorimeter
or, mL+mc(100−80)=1.2×c×(80−15)
where, m is the mass of the steam, L is the latent heat of vaporization and c is the specific heat of water.
or, m[540+(1×20)]=1.2×1×65
or, m=1.2×1×65560=72.8560kg
or, m=0.139kg