Equal masses of three liquids A, B  and C have temperatures 10⁰C, 25⁰C and 40⁰C, respectively. If A and B are mixed, the mixture has a temperature of 15⁰C. If B and C are mixed, the mixture has a temperature of 30⁰C. If A and C are mixed, the mixture will have a temperature of

When A and B are mixed,

$m{c}_A(15-10)=m{c}_B(25-15)\text{ or }\frac{c_A}{c_B}=2$                      …(i)

When B and C are mixed,

$m{c}_B(30-25)=m{c}_C(40-30)\text{ or }\frac{c_B}{c_C}=2$                   …(ii)

From Eqs. (i) and (ti), we get 

$\frac{c_A}{c_C}=4$

Now, let  $\theta$ be the equilibrium temperature when A and C are mixed.

Then,   $m{c}_A(\theta -10)=m{c}_C(40-\theta )$ 

or    $4(\theta -10)=40-\theta \left(\mathrm{Using}\frac{c_A}{c_C}=4\right)$

or    $\theta ={16}^{\circ }\mathrm{C}$