Radiation MCQ Questions & Answers in Heat and Thermodynamics | Physics

According to Wien's law, $$\lambda T = {\text{constant}}$$
From graph $${\lambda _1} < {\lambda _3} < {\lambda _2}$$
$$\therefore \,\,{T_1} > {T_3} > {T_2}.$$

According to Newton's law of cooling, $$\Delta T = \Delta {T_0}{e^{ - \lambda t}}$$
$$ \Rightarrow 3T - 2T = \left( {3T - T} \right){e^{ - \lambda \times 10}}\,......\left( {\text{i}} \right)$$
Again for next 10 minutes
$$T' - T = \left( {2T} \right) \times {e^{ - \lambda \left( {20} \right)}}\,......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii), we get
$$\eqalign{ & T' - T = \left( {2T} \right){\left( {{e^{ - \lambda \times 10}}} \right)^2} = \left( {2T} \right){\left( {\frac{1}{2}} \right)^2} = \frac{T}{2} \cr & \therefore T' = T + \frac{T}{2} = \frac{{3T}}{2} \cr} $$

According to Stefan’s law, $$E \propto {T^4}\,\,{\text{or}}\,\,E = \sigma {T^4}$$
Here, $$\sigma $$ is proportionality constant called the Stefan’s constant.
The unit of Stefan's constant is watt
$$metr{e^{ - 2}}kelvi{n^{ - 4}}\,\,{\text{or}}\,\,\frac{W}{{{m^2} - {K^4}}}.$$

Boltzmann corrected Stefan’s law and stated that the amount of radiations emitted by the body, not only depends upon the temperature of the body but also on the temperature of the surrounding. The radiated power by the body is given by
$$P = \sigma \left( {{T^4} - T_0^4} \right)\,......\left( {\text{i}} \right)$$
where $${T_0}$$ is the absolute temperature of the surrounding and $$T$$ is the temperature of body. So for two different cases ratio of radiation power is given by
$$\therefore \frac{{{P_2}}}{{{P_1}}} = \left( {\frac{{T_2^4 - T_0^4}}{{T_1^4 - T_0^4}}} \right)\,......\left( {{\text{ii}}} \right)$$
Here, $${P_1} = 60\,W,{T_1} = {727^ \circ }C = 1000\,K$$
$$\eqalign{ & {T_0} = {227^ \circ }C = 500\,K \cr & {T_2} = {1227^ \circ }C = 1500\,K \cr} $$
Substituting in Eq. (ii), we get
$$\eqalign{ & {P_2} = \frac{{{{\left( {1500} \right)}^4} - {{\left( {500} \right)}^4}}}{{{{\left( {1000} \right)}^4} - {{\left( {500} \right)}^4}}} \times 60 \cr & = \frac{{{{\left( {500} \right)}^4}}}{{{{\left( {500} \right)}^4}}} \times \left[ {\frac{{{3^4} - 1}}{{{2^4} - 1}}} \right] \times 60 \cr & = \frac{{80}}{{15}} \times 60 \cr & = 320\,W \cr} $$

According to Wien’s displacement law, the wavelength $$\left( {{\lambda _m}} \right)$$  of maximum intensity of emission of black body radiation is inversely proportional to absolute temperature $$\left( T \right)$$  of the black body.
i.e. $${\lambda _m}\,T = {\text{constant}}$$
or $${\lambda _m} = \frac{{{\text{constant}}}}{T}$$
or $${\lambda _m} \propto \frac{1}{T}$$
or $${\lambda _m} \propto {T^{ - 1}}$$

From Stefan's law, the rate at which energy is radiated by sun at its surface is $$P = \sigma \times 4\pi {r^2}{T^4}$$
[Sun is a perfectly black body as it emits radiations of all wavelengths and so for it $$e = 1$$ ]
The intensity of this power at the earth’s surface (under the assumption $$R > > {r_0}$$  ) is

$$\eqalign{ & I = \frac{P}{{4\pi {R^2}}} = \frac{{\sigma \times 4\pi {r^2}{T^4}}}{{4\pi {R^2}}} \cr & = \frac{{\sigma {r^2}{T^4}}}{{{R^2}}} = \frac{{\sigma {r^2}{{\left( {t + 273} \right)}^4}}}{{{R^2}}} \cr} $$

When hot water temperature $$\left( T \right)$$ and surrounding temperature $$\left( {{T_0}} \right)$$ readings are noted, and $$\log \left( {T - {T_0}} \right)$$   is plotted versus time, we get a straight line having a negative slope; as a proof of Newton’s law of cooling.

Concept
Apply Newton's law of cooling. Let the temperature of the surrounding be $${t^ \circ }C.$$  For first case,
$$\eqalign{ & \frac{{\left( {70 - 60} \right)}}{{5\,\min }} = K\left( {{{65}^ \circ }C - {t^ \circ }C} \right)\,\,\left( {{{65}^ \circ }{\text{ is average of }}{{70}^ \circ }C{\text{ and }}{{60}^ \circ }C} \right) \cr & \frac{{10}}{{5\min }} = K\left( {{{65}^ \circ }C - {t^ \circ }C} \right)\,......\left( {\text{i}} \right) \cr} $$
For second case, $$\frac{{\left( {60 - 54} \right)}}{{5\,\min }} = K\left( {57 - t} \right)\,......\left( {{\text{ii}}} \right)$$
$$\left( {{{57}^ \circ }C{\text{ is average of }}{{60}^ \circ }C{\text{ and }}{{54}^ \circ }C} \right)$$
From Eqs. (i) and (ii),
$$\frac{{10}}{6} = \frac{{\left( {65 - t} \right)}}{{\left( {57 - t} \right)}}$$
So, $$t = {45^ \circ }C$$

According to Wein's displacement law
$$\eqalign{ & {\lambda _m} \times T = {\text{constant}} \cr & {\text{Here, }}{\lambda _{{m_3}}} < {\lambda _{{m_2}}} < {\lambda _{{m_1}}} \cr & \Rightarrow \,\,{T_3} > {T_2} > {T_1} \cr} $$

The temperature of Sun is higher than that of welding arc which in turn is greater than tungsten filament.

Since no heat flows through the central rod, temperature at the ends of the central rod will be same. So, it will form a wheatstone like network.
So, $$\frac{{{K_1}}}{{{K_2}}} = \frac{{{K_3}}}{{{K_4}}} \Rightarrow {k_1}{k_4} = {k_2}{k_3}$$