10 g of water at 70°C is mixed with 5 g of water at 30°C. Find the temperature (in degree) of the mixture in equilibrium.

Let t° C be the temperature of the mixture. From energy conservation,

Heat given by 10 g of water = Heat taken by 5 g of water

$\begin{array}{l}\text{ or }{\mathrm{m}}_1{\mathrm{C}}_{\text{water }}\left|{\mathrm{\Delta t}}_1\right|={\mathrm{m}}_2{\mathrm{C}}_{\text{water }}\left|{\mathrm{\Delta t}}_2\right|\\ \therefore \left(10\right)(70-\mathrm{t})=5(\mathrm{t}-30)\\ \therefore \mathrm{t}=56{.67}^{\circ }\mathrm{C}\end{array}$