Radiation MCQ Questions & Answers in Heat and Thermodynamics | Physics

In this question the given options are wrong as all the four options contain $$e$$ in place of $$\sigma .$$
When a spherical body is kept inside a perfectly block body then the total heat radiated by the body is equal to that of the black body.

According to Stefan’s Law, the rate of loss of heat is $$\frac{Q}{t} = \sigma A\left( {T_1^4 - T_2^4} \right) \times e$$
here $$\sigma = 5.67 \times {10^{ - 8}}J/{m^2} \times \sec .{K^2},$$
$$\eqalign{ & {T_1} = 527 + 273 = 800\,K, \cr & {T_2} = 27 + 273 = 300\,K\& A = 200 \times {10^{ - 4}}{m^2} \cr} $$
So, $$\frac{Q}{t} = 5.67 \times {10^{ - 8}} \times 2 \times {10^{ - 2}}\,\left[ {{{\left( {800} \right)}^4} - {{\left( {300} \right)}^4}} \right] \times 0.4$$
$$ \cong 182\,joule$$

Equation of Wien’s displacement law is given by $${\lambda _m}T = {\text{constant}}$$

Rate of colling $$ = \frac{{ - d\theta }}{{dt}} \propto \left( {\frac{{{\theta _1} + {\theta _2}}}{2} - {\theta _0}} \right)$$
In second case average temperature will be less hence rate of colling will be less. Therefore time taken will be more than 4 minutes.

Rate of heat flow is given by,
$$Q = \frac{{KA\left( {{\theta _1} - {\theta _2}} \right)}}{l}$$
Where, $$K$$ = coefficient of thermal conductivity $$l$$ = length of rod and $$A$$ = Area of cross-section of rod

If the junction temperature is $$T,$$ then
$$\eqalign{ & {Q_{{\text{Copper}}}} = {Q_{{\text{Brass}}}} + {Q_{{\text{Steel}}}} \cr & \frac{{0.92 \times 4\left( {100 - T} \right)}}{{46}} = \frac{{0.26 \times 4 \times \left( {T - 0} \right)}}{{13}} + \frac{{0.12 \times 4 \times \left( {T - 0} \right)}}{{12}} \cr & \Rightarrow \,\,200 - 2\,T = 2\,T + T \cr & \Rightarrow \,\,T = {40^ \circ }C \cr & \therefore \,\,{Q_{{\text{Copper}}}} = \frac{{0.92 \times 4 \times 60}}{{46}} \cr & = 4.8\,\,cal/s \cr} $$

We know from Weins displacement law $${\lambda _m}T = {\text{constant}}$$
So, $$T \propto \frac{1}{{{\lambda _m}}}$$
As, $${\lambda _r} > {\lambda _g} > {\lambda _v}$$
So, $${T_r} < {T_g} < {T_v}$$
Given $$P \to {v_{\max }},Q \to {r_{\max }},R \to {g_{\max }}$$
Hence, $${T_Q} < {T_R} < {T_P}$$
i.e. $${T_P} > {T_R} > {T_Q}$$

$$\eqalign{ & {\text{As, }}P = \frac{1}{3}\left( {\frac{U}{V}} \right) \cr & {\text{But }}\frac{U}{V} = K{T^4} \cr & {\text{So, }}P = \frac{1}{3}K{T^4} \cr & {\text{or, }}\frac{{uRT}}{V} = \frac{1}{3}K{T^4}\,\,\left[ {{\text{As }}PV = uRT} \right] \cr & \frac{4}{3}p{R^3}{T^3} = {\text{constant}} \cr & {\text{Therefore, }}T\,\propto \,\frac{1}{R} \cr} $$

Concept
Apply Stefan's law
According to Stefan's law, the rate at which an object radiates energy is proportional to the fourth power of its absolute temperature, i.e.
$$E = \sigma {T^4}\,{\text{or}}\,\,E \propto {T^4}\,\left( {\sigma = {\text{Stefan's constant}}} \right)$$
so for two different cases, $$\frac{{{E_1}}}{{{E_2}}} = {\left( {\frac{{{T_1}}}{{{T_2}}}} \right)^4}$$
Given, $${T_1} = T,{T_2} = 2T,{E_1} = 20\,kcal/{m^2}\min $$
$$\eqalign{ & \therefore \frac{{20}}{{{E_2}}} = {\left( {\frac{T}{{2T}}} \right)^4}\,{\text{or}}\,\,\frac{{20}}{{{E_2}}} = \frac{1}{{16}} \cr & \therefore {E_2} = 20 \times 16 = 320\,kcal/{m^2}\min \cr} $$

$$E = \sigma \times {\text{area}} \times {T^4};$$     $$T$$ increases by a factor $$\frac{3}{2}.$$  Area increases by a factor $$\frac{1}{4}.$$
$$ \Rightarrow E' = \frac{1}{4} \times {\left( {\frac{3}{2}} \right)^4} \times E = \frac{{81}}{{64}}E$$

The heat radiation emitted by the human body is the infrared radiation. Their wavelength is of the order of $$7.9 \times {10^{ - 7}}m$$   to $${10^{ - 3}}m$$  which is of course the range of infrared region.