Question

When 0.4 kg of brass of ${100}^{\circ }C$ is dropped into 1 kg of water at ${20}^{\circ }C$, the final temperature is ${23}^{\circ }C$. Find the specific heat of brass.

Answer 1

Given , $m_B=0.4Kg,t_B={100}^{o}C,c_B=?$ ,

$m_w=1kg,t_w={20}^{o}C,c_w=4200J/kg{-}^{o}C$

Temperature of mixture ,$t={23}^{o}C$.

By principle of mixtures ,

Heat given by Brass = Heat taken by water

$m_B{c}_B(100-23)=m_w{c}_w(23-20)$

or $0.4\times c_B\times (100-23)=1\times 4200\times (23-20)$

or $c_B=\frac{4200\times 3}{0.4\times 77}=409J/kg{-}^{o}C$