The temperature of equal masses of three different liquids A, B and C are `12 .^(@)C, 19 .^(@)C` and `28 .^(@)C` respectively. The temperature when A and B are mixed is `16 .^(@)C` and when B and C are mixed is `23^(@)C`. The temperature when A and C are mixed is

Explanation

Let `m=`mass of each liquid, when A and B are mixed,
Heat lost by `B=` heat gained by A
`impliesmS_B(19-16)=mS_A(16-12)implies2S_B=4S_A` …(i)
When B and C are mixed,
Heat lost by `C=` heat gained by B
`impliesmS_C=(28-23)=mS_B(23-19)implies5S_C=4S_B` .(ii)
From eqs. (i) and (ii), we get
`16S_A=12S_B=15S_C` ..(iii)
When A and C are mixed, Let `theta=`final temperature
Heat lost by `C=` heat gained by A
`impliesmS_C=(28-theta)=mS_A(theta-12)`
Using Eq. (iii), we get
`implies15S_C(28-theta)=15S_A(theta-12)`
`implies16S_A(28-theta)=15S_A(theta-12)`
On solving for `theta`, we get `theta=(16xx28+12xx15)/(16+15)`
`impliestheta=20.26^@C`