Radiation MCQ Questions & Answers in Heat and Thermodynamics | Physics

The energy radiated per second is given by $$E = e\sigma {T^4}A.$$
For same material e is same. $$\sigma $$ is stefan's constant
$$\eqalign{ & \therefore \,\,\frac{{{E_1}}}{{{E_2}}} = \frac{{T_1^4{A_1}}}{{T_2^4{A_2}}} \cr & = \frac{{T_1^4\,4\,\pi r_1^2}}{{T_2^4\,4\,\pi r_2^2}} \cr & = \frac{{{{\left( {4000} \right)}^4} \times {1^2}}}{{{{\left( {2000} \right)}^4} \times {4^2}}} \cr & = \frac{1}{1} \cr} $$

From Stefan’ law, $$E = \sigma {T^4}$$
So, the rate of energy production $$Q = E \times A$$
$$Q = \sigma {T^4} \times 4\pi {R^2}$$
Temperature of star $$T = {\left( {\frac{Q}{{4\pi {R^2}\sigma }}} \right)^{\frac{1}{4}}}$$

$$\eqalign{ & \left[ {\frac{{ - dT}}{{dr}}} \right] = \frac{A}{{mc}}\left[ {T - {T_0}} \right] = \frac{{4\pi {r^2}}}{{\rho \times \frac{4}{3}\pi {r^3}c}}\left[ {T - {T_0}} \right] \cr & \therefore \left[ {\frac{{ - dT}}{{dr}}} \right] \propto \frac{1}{{\rho rc}} \cr} $$

We have $$\theta - {\theta _s} = \left( {{\theta _0} - {\theta _s}} \right){e^{ - kt}}\,......\left( 1 \right)$$
where $${{\theta _0} = }$$  initial temperature of body $$ = {40^ \circ }C$$
$$\theta $$ = temperature of body after time $$t$$
$${\theta _s}$$ = temperature of surrounding
Since body cools from $${40^ \circ }C$$  and $${38^ \circ }C$$  in $$10\,\min ,$$  we have
$${38^ \circ } - {30^ \circ } = \left( {{{40}^ \circ } - {{30}^ \circ }} \right){e^{ - 10k}}\,......\left( 2 \right)$$
Let after 10 min, the body temp. be $$\theta .$$
$$\theta - {30^ \circ } = \left( {{{38}^ \circ } - {{30}^ \circ }} \right){e^{ - 10k}}\,......\left( 3 \right)$$
Dividing equ. (2) by equ. (3) gives,
$$\frac{{{8^ \circ }}}{{\theta - {{30}^ \circ }}} = \frac{{{{10}^ \circ }}}{{{8^ \circ }}} \Rightarrow \theta - {30^ \circ } = {6.4^ \circ } \Rightarrow \theta = {36.4^ \circ }$$

The energy radiated per second by a black body is given by Stefan's Law
$$\frac{E}{t} = \sigma {T^4} \times A,$$    where $$A$$ is the surface area.
$$\frac{E}{t} = \sigma {T^4} \times 4\pi {r^2}\,\,\left( {\because {\text{For}}\,{\text{a}}\,{\text{sphere,}}\,A = 4\pi {r^2}} \right)$$
Dividing (ii) and (i), we get
$$\eqalign{ & \frac{{\frac{E}{t}}}{{450}} = \frac{{{{\left( {1000} \right)}^4}{{\left( {0.06} \right)}^2}}}{{{{\left( {500} \right)}^4}{{\left( {0.12} \right)}^2}}} = \frac{{{2^4}}}{{{2^2}}} = 4 \cr & \Rightarrow \frac{E}{t} = 450 \times 4 = 1800\,W \cr} $$

The energy radiated per second by a black body is given by Stefan's Law
$$\frac{E}{t} = \sigma {T^4} \times A,$$   where $$A$$ is the surface area.
$$\frac{E}{t} = \sigma {T^4} \times 4\pi {r^2}$$   ($$\because $$ For a sphere, $$A = 4\pi {r^2}$$   )
Case (i) : $$\frac{E}{t} = 450,T = 500\,K,r = 0.12\,m$$
$$\therefore \,\,450 = 4\,\pi \sigma {\left( {500} \right)^4}{\left( {0.12} \right)^2}\,\,\,\,.....\left( {\text{i}} \right)$$
Case (ii) : $$\frac{E}{t} = = \,? ,T = 1000\,K,r = 0.06\,m$$
$$\therefore \,\,\frac{E}{t} = 4\,\pi \sigma {\left( {1000} \right)^4}{\left( {0.06} \right)^2}\,\,\,.....\left( {{\text{ii}}} \right)$$
Dividing (ii) and (i), we get
$$\eqalign{ & \frac{{\frac{E}{t}}}{{450}} = \frac{{{{\left( {1000} \right)}^4}{{\left( {0.06} \right)}^2}}}{{{{\left( {500} \right)}^4}{{\left( {0.12} \right)}^2}}} \cr & = \frac{{{2^4}}}{{{2^2}}} \cr & = 4 \cr & \Rightarrow \,\,\frac{E}{t} = 450 \times 4 \cr & = 1800\,W \cr} $$

According to Stefan's law, energy emitted $$E \propto {T^4}$$
$$\eqalign{ & E = \sigma {T^4}\,\left( {\sigma = {\text{stefan's constant}}} \right) \cr & \therefore E \propto {\left( {500} \right)^4} \cr} $$

By Newton's law of cooling, rate of fall of temperature $$ \propto $$ average temperature excess. In each case average temperature excess decreases, so rate of fall of temperature decreases. Hence, $${t_1} < {t_2} < {t_3}.$$   Because more and more time is required to cool, if the average temperature goes $$Q$$ in decreasing.

According to Stefan's law, $$E \propto {T^4}\,\,{\text{or}}\,\,E = \sigma {T^4}$$
where, $$\sigma $$ is constant of proportionality and called Stefan’s constant. Its value is $$5.67 \times {10^{ - 8}}W{m^{ - 2}}{K^{ - 4}}$$
Here, $$E \propto {\left( {727 + 273} \right)^4} \Rightarrow E \propto {\left( {1000} \right)^4}$$
NOTE
If the body at temperature $$T$$ is surrounded by a body at temperature $${T_0},$$ then Stefan’s law is $$E = \sigma \left( {{T^4} - T_0^4} \right)$$
This statement is called Stefan-Boltzmann law.

Total power radiated by Sun $$ = \sigma {T^4} \times 4\,\pi {R^2}$$
The intensity of power at earth's surface $$ = \frac{{\sigma {T^4} \times 4\,\pi {R^2}}}{{4\,\pi {r^2}}}$$
Total power received by Earth $$ = \frac{{\sigma {T^4}{R^2}}}{{{r^2}}}\left( {\pi r_0^2} \right)$$