Question

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

• A. $307Jk{g}^{-1}{}^o{C}^{-1}$
• B. $407Jk{g}^{-1}{}^o{C}^{-1}$
• C. $507Jk{g}^{-1}{}^o{C}^{-1}$
• D. $607Jk{g}^{-1}{}^o{C}^{-1}$

Answer 1

The correct option is B $407Jk{g}^{-1}{}^o{C}^{-1}$

Here, mass of brass, $m_1=0.4kg$
Temperature of brass, $T_1={100}^{o}C$
Mass of water, $m_2=1kg$
Temperature of water, $T_2={20}^{o}C$
Final temperature, $T={23}^{o}C$
Specific heat of brass, $C_1=?$
Specific heat of water,
$C_2=4180Jk{g}^{-1}{}^o{C}^{-1}$
From relation, Heat lost $=$ Heat gained
$m_1{C}_1(T_1-T)=m_2{C}_2(T-T_2)$
Putting values we get,
$0.4\times C_1\times (100-23)=1\times 4180\times (23-20)$

$\Rightarrow C_1=\frac{1\times 4180\times 3}{0.4\times 77}=407Jk{g}^{-1}{}^o{C}^{-1}$

Specific heat of brass $=407Jk{g}^{-1}{}^o{C}^{-1}$.