`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 = 0.1 xx 10^(3) xx s_("water") xx (80- T)`
Heat gained by water at `60^(@)C = 0.3 xx 10^(3) xx s_("water") xx (T-60)`
According to the principle of calorimetry
heat lost = heat gain
`:. 0.1 xx 10^(3) xx s_("water") xx (80 -T) = 0.3 xx 10^(3) xx s_("water") xx (T-60)`
or `1 xx (80-T) = 3 xx (T-60) or 4T= 260`
or `T= (260)/(4) ""^(@)C = 65^(@)C`