Question

200 g of hot water at ${80}^{\circ }C$ is added to 300 g of cold water at ${10}^{\circ }C$.Neglecting the heat taken by the container,calculate the final temperature of the mixture of water.Specific heat capacity of water =4200 J k$g^{-1}{K}^{-1}$ .

Answer 1

Let the final temperature be $T^{o}C$.

So, heat lost by hot water = Heat gained by cold water

$\Rightarrow 0.2\times 4200\times (80-T)=0.3\times 4200\times (T-10)$

$\Rightarrow 16-0.2T=0.3T-3$

$\Rightarrow 0.5T=19$

$\therefore T={38}^{o}C$