Question

100 g of water at ${70}^{\circ }C$ is added to 120 g of water at ${30}^{\circ }C$ contained in a vessel. The final temperature of mixture is ${40}^{\circ }C$. Calculate the thermal capacity of the vessel.

Answer 1

Heat lost by water to change temperature from ${70}^{o}C$ to ${40}^{o}C=mC△t=(100\times 4.2\times 30)J=12600J$

Heat gained by $120g$ of water = $(120\times 4.2\times 10)=5040J$

Heat gained by vessel = Heat capacity $\times △t$ = Heat capacity $\times 10$

According to principle of calorimetry,

Heat lost by hot water = heat gained by cold water + heat gained by vessel

$\Rightarrow 12600=5040+($ Heat capacity $\times 10)$

$\Rightarrow$Heat capacity $=756J^o{C}^{-1}$