Question

A metal bar of mass $1.5kg$ is heated at atmospheric pressure. Its temperature is increased from $30$ to $60$. Then the work done in the process is :
(Volume expansion coefficient of the metal $=5\times {10}^{-5}{}^o{C}^{-1}$
Density of the metal $=9\times {10}^{3}kg{m}^{-3}$
Atmospheric pressure $=1\times {10}^{5}Pa$)

• A. $25\times {10}^{-3}J$
• B. $2.5\times {10}^{-3}J$
• C. $12.5\times {10}^{-3}J$
• D. $1.25\times {10}^{-3}J$

Answer 1

The correct option is A $25\times {10}^{-3}J$

Given - $\gamma =5\times {10}^{-5}{/}^{o}C$,

$d=9\times {10}^{3}kg/m^3$,

$P=1\times {10}^{5}Pa$,

$m=1.5kg$,

${\theta }_1={30}^{o}C$,

${\theta }_2={60}^{o}C$

Change in temperature, $\Delta \theta ={\theta }_2-{\theta }_1=60-30={30}^{o}C$

Volume expansion coefficient of metal is given by,

$\gamma =\frac{\Delta V}{V\times \Delta \theta }$

or $\Delta V=V\times \Delta \theta \times \gamma$

or $\Delta V=\left(m/d\right)\times \Delta \theta \times \gamma$

or $\Delta V=\frac{1.5}{9\times {10}^3}\times 30\times 5\times {10}^{-5}{m}^3$

Now , work done in this process is

$W=P\Delta V$

or $W={10}^5\times \frac{1.5}{9\times {10}^3}\times 30\times 5\times {10}^{-5}$

or $W=25\times {10}^{-3}J$