Question

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

• A. ${30}^{o}C$
• B. ${40}^{o}C$
• C. ${50}^{o}C$
• D. ${60}^{o}C$

Answer 1

The correct option is B ${40}^{o}C$

Given, mass of copper, $m_1=500g$
Mass of water, $m_2=200g$
Initial temperature of copper, $t_1={100}^{o}C$
Initial temperature of water, $t_2={25}^{o}C$
Sp. heat of copper, $C_1=0.42J{g}^{-1}{}^o{C}^{-1}$
Sp. heat of water, $C_2=4.2J{g}^{-1}{}^o{C}^{-1}$
Final temperature of the mixture $=t^{o}C$
Then,
Heat lost by the copper piece
$=m_1{C}_1(t_1-t)$
Heat gained by water $=m_2{C}_2(t-t_2)$
We know, Heat lost $=$ Heat gained
$\Rightarrow m_1{C}_1(t_1-t)=m_2{C}_2(t-t_2)$
$\Rightarrow 500\times 0.42\times (100-t)$
$=200\times 4.2\times (t-25)$
$\Rightarrow (100-t)=\frac{200\times 4.2}{500\times 0.42}\times (t-25)$
$=4(t-25)$
This given, $5t=200$
$\Rightarrow t={\frac{200}{5}}^{o}C={40}^{o}C$
Thus, the final temperature of the mixture is ${40}^{o}C$.