Question

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

Answer 1

Correct Answer - B
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 = `0.1 xx 10^(3) xx S_(water) xx (80-t)`
Heat gained by water = `0.3 xx 10^(3) xx S_(water) (t-60)`
According to 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)`
`rArr 1 xx(80-t)= 3xx (t-60) rArr t=65^(@)`C