The temperature of equal masses of three different liquids x, y and z are ${10}^{0}C,{20}^{0}C$ and  ${30}^{0}C$respectively. The temperature of mixture when X is mixed with y is ${16}^{0}C$ and that when y is mixed with z is ${26}^{0}C$ The temperature of mixture when x and z are mixed will be:

The mass of all three liquids is same, let it be m. Using the concept of specific heat capacity of liquids, i.e. heat lost by the hot liquid is equal to heat gained by the cold liquid.

Heat gained by liquid x=Heat lost by liquid y
 

$\text{ So, }{\mathrm{mS}}_{\mathrm{x}}(16-10)={\mathrm{mS}}_{\mathrm{y}}(20-16)$

$\text{ i.e. }{\mathrm{S}}_{\mathrm{x}}=\frac{2}{3}{\mathrm{S}}_{\mathrm{y}}$

Similarly for the liquids y and z we get the relation between specific heat capacities as,

Heat gained by liquid y=Heat lost by liquid z

$\text{ So, }{\mathrm{mS}}_{\mathrm{y}}(26-20)={\mathrm{mS}}_{\mathrm{z}}(30-26)$

$\text{ i.e. }{\mathrm{S}}_{\mathrm{y}}=\frac{2}{3}{\mathrm{S}}_{\mathrm{z}}$

${\mathrm{S}}_{\mathrm{x}}=\frac{4}{9}{\mathrm{S}}_{\mathrm{z}}$

So, when liquids x and z are mixed we get,

${\mathrm{mS}}_{\mathrm{x}}(\mathrm{T}-10)={\mathrm{mS}}_{\mathrm{z}}(30-\mathrm{T})$

$\mathrm{On} \mathrm{solving} \mathrm{we} \mathrm{get} \mathrm{T}=23.84{}^0\mathrm{C}$