Heat And Thermodynamics Ques 166
- A metal of mass $1 kg$ at constant atmospheric pressure and at initial temperature $20^{\circ} C$ is given a heat of $20000 J$. Find the following :
$(2005,6 M)$
(a) change in temperature,
(b) work done and
(c) change in internal energy.
(Given, Specific heat $=400 J / kg /{ }^{\circ} C$, coefficient of cubical expansion, $\quad \gamma=9 \times 10^{-5} /{ }^{\circ} C$, density $\rho=9000 kg / m^{3}$, atmospheric pressure $=10^{5} N / m^{2}$ )
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Answer:
Correct Answer: 166.$\text { (a) } 50^{\circ} \mathrm{C} \text { (b) } 0.05 \mathrm{~J} \text { (c) } 19999.95 \mathrm{~J}$
Solution:
Formula:
- (a) From $\Delta Q=m s \Delta T$
$$ \Delta T=\frac{\Delta Q}{m s}=\frac{20000}{1 \times 400}=50^{\circ} C $$
(b) $\Delta V=V \gamma \Delta T=\left(\frac{1}{9000}\right)\left(9 \times 10^{-5}\right)(50)$
$$ =5 \times 10^{-7} m^{3} $$
$\therefore W=p \cdot \Delta V=\left(10^{5}\right)\left(5 \times 10^{-7}\right)=0.05 J$
(c) $\Delta U=\Delta Q-W=(20000-0.05) J$
$$ =19999.95 J $$
BETA
