This is a problem on 'flow calorimeter' used to measure the specific heat of the liquid.
Amount of heat supplied to the water per second by the heating coil = $Q_{\mathrm{s}}$ $=$ $250 \mathrm{J}$   $=\frac{250}{4186}\text{kcal}$

The volume of liquid flowing out per second = $8.0 {\mathrm{cm}}^3 = 8 \times {10}^{-6} {\mathrm{m}}^3$

Mass of this liquid = $(0.85) \times 1000 \times 8 \times {10}^{-6} \mathrm{kg}$

The temperature rise of this mass of liquid = $15° \mathrm{C}$

Hence,  $=\frac{250}{4186}=\text{mst}=\text{0.85}\times 8\times 10^{-3}\times \text{s}\times 15$

Hence,  $\text{s}=\frac{250\times 10^3}{4186\times \text{0.85}\times 8\times 15}=\text{0.6 }\mathrm{kcal} {\mathrm{kg}}^{-1}{K}^{-1}$