Question 162: Thermodynamics & KTG

Question

A cylinder of mass $$1 \mathrm{~kg}$$ is given heat of $$20000 \mathrm{~J}$$ at atmospheric pressure. If initially temperature of cylinder is $$20^{\circ} \mathrm{C}$$, find
(A) The final temperature of the cylinder;
(B) The work done by the cylinder;
(C) The change in internal energy of the cylinder.
Given :
The specific heat of cylinder
$$=400 \mathrm{~J} \mathrm{~kg}^{-1 \circ} \mathrm{C}^{-1}$$
Coefficient of volume expansion
$$=9 	imes 10^{-5}{ }^{\circ} \mathrm{C}^{-1} 	ext {; }$$
Atmospheric pressure $$=10^{5} \mathrm{~N} / \mathrm{m}^{2}$$ Density of cylinder $$=9000 \mathrm{~kg} / \mathrm{m}^{3}$$ )

Accepted Answer

(A) Heat supplied to the cylinder is used up in if raising temperature of cylinder $$=m s \Delta \mathrm{T}$$
(i) $$\mathrm{Q}_{1}=m s \Delta \mathrm{T}=1 	imes 400(\mathrm{~T}-20) 	ext {...(i) }$$
S = 400 J kg$$^{-10}$$C$$^{-1}$$
(ii) doing work $$=\mathrm{P}_{	ext {atom }}\left(\mathrm{V}_{2}-\mathrm{V}_{1}ight)$$
Work done $$=\mathrm{P}_{\mathrm{atm}}\left(\mathrm{V}_{2}-\mathrm{V}_{1}ight)$$
$$\mathrm{Q}_{2}=10^{5} 	imes\left(\mathrm{V}_{1} \gamma \Delta \mathrm{T}ight)$$
$$\mathrm{Q}_{2}=10^{5} 	imes \frac{	ext { mass }}{	ext { density }} 	imes \gamma(\mathrm{T}-20)$$
$$\mathrm{Q}_{2}=10^{5} 	imes \frac{1}{900} 	imes 9 	imes 10^{-5} 	imes(\mathrm{T}-20)$$
$$\mathrm{Q}_{2}=0.001(\mathrm{~T}-20)$$ ...... (ii)
$$	herefore \quad \mathrm{Q}_{1}+\mathrm{Q}_{2}=\mathrm{Q}$$ Heat supplied.
(Given $$\mathrm{Q}=20000 \mathrm{~J}$$ )
400.$$(T-20)+0.001(T-20)=20000$$
$$400.001(\mathrm{~T}-20)=2000$$
$$\mathrm{T}-20=\frac{20000}{400.001}=50$$ (approx)
$$\mathrm{T}=50+20=70^{\circ} \mathrm{C}$$ ...... (iii)
(B) Workdone by cylinder
$$\begin{aligned}
\mathrm{W} & =\mathrm{Q}_{2}=0.001(70-2 a) \
& =0.001(50) \
\mathrm{W} & =0.05 \mathrm{~J} .\end{aligned}$$
(C) Change in internal energy
$$\begin{aligned}
\Delta \mathrm{U} & =20000-0.05 \
& =19999.95 \mathrm{~J}=1.9 	imes 10^{4} \mathrm{~J}
\end{aligned}$$

Answer

$$(\mathrm{A}) 70^{\circ} \mathrm{C}$$, (B) $$0.05 \mathrm{~J}$$ (C) $$\Delta \mathrm{U}=1.9 	imes 10^{4} \mathrm{~J}$$

Answer

$$(\mathrm{A}) 70^{\circ} \mathrm{C}$$, (B) $$0.05 \mathrm{~J}$$ (C) $$\Delta \mathrm{U}=2.9 	imes 10^{4} \mathrm{~J}$$

Answer

$$(\mathrm{A}) 60^{\circ} \mathrm{C}$$, (B) $$0.05 \mathrm{~J}$$ (C) $$\Delta \mathrm{U}=1.9 	imes 10^{4} \mathrm{~J}$$

Answer

$$(\mathrm{A}) 70^{\circ} \mathrm{C}$$, (B) $$0.07 \mathrm{~J}$$ (C) $$\Delta \mathrm{U}=1.9 	imes 10^{4} \mathrm{~J}$$

Thermodynamics & KTG

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