$100\mathrm{ml}$ of water at $20\mathrm{C}$ and $100\mathrm{ml}$ of water at ${40}^0\mathrm{C}$ are mixed in calorimeter temperature of the mixture is ${28}^0\mathrm{C}$.

Now,

Heat loss = Heat gained

$$\begin{gathered} 100\times (40-28)\times 1=100\times (28-20)\times 1+\text{ water equivalent }\times 8 \\ 1200=800+\text{ water equivalent }\times 8 \\ \text{ water equivalent }\times 8=1200-800 \\ \text{ water equivalent }=\frac{400}{8} \\ \text{ water equivalent }=50 \mathrm{c}\mathrm{a}\mathrm{l}=210\mathrm{J} \end{gathered}$$