The mass, specific heat capacity and the temperature of a solid are $1000 \mathrm{g}$, $\frac{1}{2}$  $\mathrm{cal} {\left(\mathrm{g}.{}^{\circ }\mathrm{C}\right)}^{-1}$ and $80°\mathrm{C}$ respectively. The mass of the liquid and the calorimeter are $900 \mathrm{g}$ and $200 \mathrm{g}$. Initially, both are at room temperature $20°\mathrm{C}$. Both calorimeter and the solid are made of same material. In the steady state, temperature of mixture is $40°\mathrm{C}$, then find the specific heat capacity of the unknown liquid.

$m_1=mass of solid=1000 \mathrm{g}$

$S_1=specific heat of solid=\frac{1}{2}cal/g-{}^{\circ }\mathrm{C}$

=$S_2$ or specific heat of calorimeter

$m_2$= mass of calorimeter =$200 \mathrm{g}$

$m_3$= mass of unknown liquid = $900 \mathrm{g}$

$S_3$= specific heat of unknown liquid

From law of heat exchange,

Heat given by solid = Heat taken by calorimeter + Heat taken by unknown liquid

$m_1{S}_1\left|\Delta {\theta }_1\right|=m_2{S}_2\left|\Delta {\theta }_2\right|+m_3{S}_3\left|\Delta {\theta }_3\right|$

$1000\times \frac{1}{2}\times (80-40)=200\times \frac{1}{2}(40-20)+900\times S_3(40-20)$

Solving this equation we get, $S_3=1\mathrm{cal}/\mathrm{g}·{}^{\circ }\mathrm{C}$