Question

If A vessel of thermal capacity $30$ calories contains $170g$ of water at ${30}^{\circ }C$. If on dropping a solid at ${93}^{\circ }C$ into the calorimeter, then temperature of mixture becomes ${33}^{\circ }C$, find the thermal capacity of solid.

Answer 1

Please note the difference between heat capacity '$\omega$' and specific heat capacity '$\eta$'.

$\eta \times mass=\omega$

Remember one principle of calorimetry.

Heat lost by hot body = heat gained by cold body unless any heat is lost to the surroundings.

Heat lost by solid = change in temperature$\times$heat capacity of solid

Heat gained by water and vessel= change in temperature$\times$[heat capacity of vessel $+$ (specific heat capacity of water$\times$ mass of water)]

$(33-30)\times \left[\right(170\times {\eta }_{{}_W})+{\omega }_{{}_W}]=(93-33)\times {\omega }_{{}_S}$

$3\times (170+30)=60\times {\omega }_{{}_S}$

$\therefore {\omega }_{{}_S}=10$

Answer:- $10cal{/}^{o}C$