Q. A cylinder of mass $1kg$ is given heat of $20000\ J$ at atmospheric pressure. If initially temperature of cylinder is $20℃$, then work done by the cylinder will be (Given that Specific heat of cylinder $= 400\ J\ kg^{- 1}$, Coefficient of volume expansion $= 9 \times 10^{- 5}\ ℃^{- 1}$, Atmospheric pressure $= 10^{5}N/m^{2}$ and density of cylinder $9000\ kg/m^{3}$)
A) $0.02\ J$
B) $0.05\ J$
C) $0.08\ J$
D) $0.1\ J$
Explanation
(b) $\Delta Q = mc\Delta T \Rightarrow \Delta T = \frac{20000J}{1kg \times (400J/kg℃)} = 50℃$
$\Rightarrow T_{Final} = 70℃$
Hence $W = P_{atm}\mathrm{\Delta}V = P_{atm}V_{0}\gamma\ \Delta T$
$= (10^{5}N/m^{2})\left( \frac{1}{9 \times 10^{3}}m^{3} \right)(9 \times 10^{- 5}/℃)(50℃) = 0.05J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.