Temperatures of three liquids A, B and C of equal mass are `12^(@)C, 19^(@)C and 28^(@)C` respectively. If A and B are mixed the temperature becomes `16^(@)C` and if we mixed B and C the temperature becomes `23^(@)C`. What will be the temperature of the mixture of A and C?

Explanation

Let mass of each liquid be m and their specific heats be `s_(A), s_(B) and s_(C)` respectively.
By mixing liquids A and B,
heat gained by A = heat lost by B
`"i.e., "m.s_(A)(16-12)=m.s_(B)(19-16)`
`"or, "s_(A).4=s_(B).3`
`"or, "(s_(A))/(s_(B))=(3)/(4)" ...(1)"`
By mixing liquids B and C,
heat gained by B = heat lost by C
`"i.e., "m.s_(B)(23-19)=m.s_(C)(28-23)`
`"or, "s_(B).4=s_(C).5`
`"or, "(s_(B))/(s_(C))=(5)/(4)" ..(2)"`
`therefore" Multiplying (1) and (2),"`
`(s_(A))/(s_(B))xx(s_(B))/(s_(C))=(3)/(4)xx(5)/(4)" or, "(s_(A))/(s_(C))=(15)/(16)`
Let the temperature of the mixture of liquids A and C be `theta`.
Since, heat gained by A = heat lost by C
`m.s_(A)(theta-12)=m.s_(C)(28-theta)`
`therefore" "(s_(A))/(s_(C))=(28-theta)/(theta-12)" or, "(15)/(16)=(28-theta)/(theta-12)`
`"or, "15theta-180=448-16theta" or, "31theta=628`
`therefore" "theta=20.26^(@)`