Let T be the final temperature of the mixture.
Let’s calculate the amount of heat released by steam to be converted into water at ToC.
First, steam at 120o will be converted into 110o
Heat relesed during this process Q1=mS∂T=20g×0.46cal g-1 oC-1 × 10oC = 92cal.
Heat released during the conversion of steam in water Q2=mLv=20g × 540calg-1 = 10780cal
Heat released during the conversion of water at 110oC to water at ToC = mS∂T = 20g×1calg-1oC-1(110 – T)
Now Qsteam=Q1 +Q2 + Q3 = (11092 + 20T)
Now let’s calculate heat absorbed by water to get convert to ToC
Qwater = mS∂T = 200g × 1cal g-1 oC-1× (T-20) = 200T – 4000.
By principle of Calorimetry, Qsteam = Qwater.
Hence, 11092 + 20T = 200T – 4000
So, T = 63.84oC.
