Gujarati
11.Thermodynamics
hard

A cylinder of mass $1\,kg$ is given heat of $20000\, J$ at atmospheric pressure. If initially temperature of cylinder is $20\,^oC$, then work done by the cylinder will be .......$J$ (Given that Specific heat of cylinder $= 400 \,J\, kg^{-1}$, Coefficient of volume expansion $= 9 \times {10^{-5}}\,^o C^{-1}$, Atmospheric pressure $= 10^5 \,N/m^2$ and density of cylinder $9000\,kg/m^3$)

A

$0.02 $

B

$0.05 $

C

$0.08 $

D

$0.1 $

Solution

(b) $\Delta Q = mc\Delta T$ $⇒$ $\Delta T = \frac{{20000J}}{{1kg \times (400J/kg^\circ C)}} = 50^\circ C$

$⇒$ $T$ Final $= 70°C$

Hence $W = {P_{atm}}\Delta V = {P_{atm}}{V_0}\gamma \,\Delta T$

$ = ({10^5}N/{m^2})\,\left( {\frac{1}{{9 \times {{10}^3}}}{m^3}} \right)\,(9 \times {10^{ – 5}}/^\circ C)\,(50^\circ C) = 0.05J$

Standard 11
Physics

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