Heat and Thermodynamics

Since the radiation is continuously falling on the black body, the quantity of radiation absorbed per second will increase.

The reason given in question is self explanatory. With an increase in temperature, the entire Plank's curve shifts upwards and hence radiation in any spectrum will increase. Reflected energy per unit time will be zero since black body has zero reflectivity and hence it will remain constant.

The bodies which radiate energy in the form of photons can be determined using Kirchoff's

law of radiation. When these photons reach another surface, they may be absorbed, reflected or transmitted. Since the radiation is continuously falling on the black body, the quantity of radiation absorbed per second will increase.

Therefore, the quantity of radiation absorbed by the black body in unit time will increase, since emissivity = absorptivity, hence the quantity of radiation emitted by black body in unit time will increases, Black body radiates more energy in unit time in the visible spectrum, the reflected energy in unit time by the black body remains same.

Hence, the option (A), (B), (C) and (D) are the correct answer.

$ \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W}, \mathrm{~W}=\int \mathrm{P} d \mathrm{~V} $

Along JK : V is constant, P is decreasing and

$\mathrm{P} \propto \mathrm{T}, \mathrm{T}$ should also decrease.

As $\mathrm{PV}=n \mathrm{RT}$.

$ \begin{aligned} \therefore & & \mathrm{W} & =P \Delta V=0 \\ & & \Delta \mathrm{U} & =\Delta \mathrm{Q}<0 \quad \text { (negative) } \end{aligned} $

$\Rightarrow$ Along KL : $\mathrm{P}=$ constant, i.e., isobaric process.

$\therefore n \mathrm{C}_{\mathrm{V}} \Delta \mathrm{T}$ is increasing. So temperature should also increase.

$ \begin{aligned} & \therefore \Delta \mathrm{W}=\mathrm{PdV}>0, \Delta \mathrm{U}=n \mathrm{C}_{\mathrm{V}} \Delta \mathrm{~T}>0 \text { and } \\ & \mathrm{Q}=m c \Delta T>0 . \end{aligned} $

$\therefore$ Volume increases, $\Delta \mathrm{W}>0$

$\Rightarrow$ Along LM : MV = constant, $\Delta \mathrm{V}=0$ [isochoric process]

both P increases and T increases.

$ \therefore W=0, \Delta U>0 \text { and } Q>0 \quad[\Delta Q=\Delta U] $

$\Rightarrow$ Along $\mathrm{MJ}: \mathrm{V}$ is decreasing, $\therefore \Delta \mathrm{W}<0$

T is also decreasing, $\Delta \mathrm{V}<0$ and $\Delta \mathrm{Q}<0$

P increases.

(A) Heat supplied to the cylinder is used up in if raising temperature of cylinder $=m s \Delta \mathrm{T}$

(i) $\mathrm{Q}_{1}=m s \Delta \mathrm{T}=1 \times 400(\mathrm{~T}-20) \text {...(i) }$

S = 400 J kg$^{-10}$C$^{-1}$

(ii) doing work $=\mathrm{P}_{\text {atom }}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$

Work done $=\mathrm{P}_{\mathrm{atm}}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$

$\mathrm{Q}_{2}=10^{5} \times\left(\mathrm{V}_{1} \gamma \Delta \mathrm{T}\right)$

$\mathrm{Q}_{2}=10^{5} \times \frac{\text { mass }}{\text { density }} \times \gamma(\mathrm{T}-20)$

$\mathrm{Q}_{2}=10^{5} \times \frac{1}{900} \times 9 \times 10^{-5} \times(\mathrm{T}-20)$

$\mathrm{Q}_{2}=0.001(\mathrm{~T}-20)$ ...... (ii)

$\therefore \quad \mathrm{Q}_{1}+\mathrm{Q}_{2}=\mathrm{Q}$ 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

$\begin{aligned} \Delta \mathrm{U} & =20000-0.05 \\ & =19999.95 \mathrm{~J}=1.9 \times 10^{4} \mathrm{~J} \end{aligned}$

${{{n_1} + {n_2}} \over {\gamma - 1}} = {{{n_1}} \over {{\gamma _1} - 1}} + {{{n_2}} \over {{\gamma _2} - 1}}$
${{1 + 1} \over {\gamma - 1}} = {1 \over {{7 \over 5} - 1}} + {1 \over {{5 \over 3} - 1}}$

$\therefore$ ${2 \over {\gamma - 1}} = {5 \over 2} + {3 \over 2} = 4$

$\therefore$ $2 = 4\gamma - 4 \Rightarrow \gamma = {6 \over 4} = {3 \over 2}$