`0.1 m^(3)` of water at `80^(@)C` is mixed with `0.3 m^(3)` of water at `60^(@)C`. What is the final temperature of the mixture ?

Explanation

Volume of hot water `= 0.1 m^(3) = 0.1 xx 10^(6)c.C`
`= 10^(5) c.C`
mass of hot water, `m_(1) = 10^(5) xx 1 = 10^(5) g`
mass of hot water, `m_(2) = (0.3 xx 10^(6)) xx 1`
`= 3 xx 10^(5) g`
If `theta` is the final temperature of mixture, the heat loss by hot water = heat gained by cold water
`10^(5) xx 1 (80 - theta) = 3 xx 10^(5) xx 1 xx (0 - 60)`
`(80 - theta) = 3 theta - 180`
`theta = 65^(@)C`