Two beaker A and B contain water at different temperatures .When 1 liter of water from beaker A is mixed with 2 litre of water from beaker B the equilibrium temperature is `16^(@)C` and when 2 litre of water from beaker A then the equilibrium temperature is `14^(@)C` .Determine the tempreature of water in the beaker A and B

Explanation

Heat lost = Heat gained
When 1 litre of water from A is mixed with 2 litres of water from B, then the equilibrium temperature attained is `16^@C`.
`Q = msDeltat`. Let `t_A` and `t_B` be the initial temperatures of water present in beaker A and B and let temperature of A be less than temperature of B.
`1 xx S xx (16-t_A)=2xxSxx(t_B-16)`
`(16-t_A)=2(t_B-16)`
`2t_B=t_A=48-(1)`
When 2 litres of water from A is mixed with 2 litres of water from B, then the equilibrium temperature of the mixture is `14^@C`.
`2xxSxx(14-t_A)=2xxSxx(t_B-14)`
`(14-t_A)=(t_B-14)`
`t_B+t_A=28-(2)`
`2t_B+t_A=48-(1)`
`2t_B+2t_A=56-(3)`
`(3)-(1) rArr t_A=8^@C`
`(t_B)+(t_A)=28`
`t_B=28-t_A=28-8=20^@C`