Given: Specific heat of water {S}_{w}=1\mathrm{cal}{g}^{-1}^{\circ}{C}^{-1}, Latent heat of steam Ls=540calg−1
Let the mass of steam =mg
So heat lost in change of state from steam to water equal to sum of latent heat and heat use in raise of the temperature of water.
Q1=mLs+mswΔT
Q1=m×540+m×1×(100−80)
Q1=540m+20m=560m
Heat gained by water to change its temperature
Q2=mwSwΔT
Q2=20×1×(80−10)=1400.
Now we know from the principle of calorimetry is that total heat lost by the hot body is equal to total heat gained by cold body i.e it follows law of conservation energy. Hence
Heat lost Q1= Heat gained Q2
560m=1400m=5601400=2.5g
Therefore, total mass of water at 80∘C
m=(20+2.5)g=22.5g