In a container of negligible mass 30g of steam at `100^@C` is added to 200g of water that has a temperature of `40^@C` If no heat is lost to the surroundings, what is the final temperature of the system? Also find masses of water and steam in equilibrium. Take `L_v = 539 cal//g and c_(water) = 1 cal//g.^@C.`

Explanation

Given, mass =m =200 g, `m_(1)` = 30g
Temperature `T_(1)` = `100^(@)`C, `T_(2)=40^(@)`C
Latent heat `L_(V))=539calgu^(-1)`
Specific heat `c_(w)` = `1calg^(-1)^(@)C^(-1)`
Let Q be the heat requireid to convert 200g of water at `40^(@)`C into `100^(@)`C, then
Q =`mcDeltaT`
`=(200)(1.0)(100-40)=12000`cal
Now, suppose `m_(0)` mass of clean of steam converts into water to liberate this much amount of het, then
`m_(0)=Q/L=12000/539`=22.26g
Since, it is less than 30g, the temperature of the mixture is `100^(@)`C.
Mass of the water in the mixture = `m+m_(0)`
=200 +22.26=222.26g