Question

A piece of copper weighing 500 g is heated to ${100}^{\circ }C$ and dropped into 200 g of water at ${25}^{\circ }C$. Find the temperature of the mixture. The specific heat of Cu is 0.42 J $g^{-1}{}^{\circ }{C}^{-1}$.

Answer 1

Given , $m_{Cu}=500g,t_{Cu}={100}^{o}C,c_{Cu}=0.42J/g{-}^{o}C$ ,

$m_w=200g,t_w={25}^{o}C,c_w=4.2J/g{-}^{o}C$

Let $t^{o}C$ be the temperature of mixture .

By principle of mixtures ,

Heat given by copper=Heat taken by water

$m_{Cu}{c}_{Cu}(100-t)=m_w{c}_w(t-25)$

or $500\times 0.42\times (100-t)=200\times 4.2\times (t-25)$

or $500\times 0.42\times (100-t)=200\times 4.2\times (t-25)$

or $(100-t)=4\times (t-25)$

or $t=200/5={40}^{o}C$