`0.1 m^(3)` of water at `80^(@)C` is mixed with `0.3m^(3)` of water at `60^(@)C`. The finial temparature of the mixture is

Options

A. `65^(@)C`
B. `70^(@)C`
C. `60^(@)C`
D. `75^(@)C`

Answer

A

Explanation

Density of water `=10^(3)kg//m^(3)`
Let the final temperature of the mixture be t.
Assuming no heat transfer to or from container.
Heat lost by water at `80^(@)` C
`=0.1xx10^(3)xxs_("water")xx(80-t)`
heat gained by water at `60^(@)C`
`=0.3xx10^(3)xxs_("water")xx(t-60)`
According to principle of calorimetry
heat lost=heat gain
`0.1xx10^(3)xxs_("water")xx(80-t)=0.3xx10^(3)xxs_("water")xx(t-60)`
`implies1xx(80-t)=3xx(t-60)`
`implies4r=260impliest=260/4""^(@)C=65^(@)C`