Question

A black body radiates heat energy at the rate of $2\times 10^{5}J/s{m}^2$ at the temperature of $127^{\circ }C$. Temperature of the black body at which rate of heat radiation $32\times 10^{5}J/s{m}^2$, is

• A. 400 K
• B. 600 K
• C. 800 K
• D. 200 K

Answer 1

Answer: (C)

Here, $E_1=2\times 10^{5}J/s{m}^2$,
$T_1=127^{\circ }C=400K$
and $E_2=32\times 10^{5}J/s{m}^2$
By Stefan's law, the rate of emission of radiated energy per unit area per unit time is $E=\sigma T^4$
$\therefore \frac{E_2}{E_1}=\frac{T_2^4}{T_1^4}$ or
$\frac{T_2}{T_1}={\left(\frac{E_2}{E_1}\right)}^{1/4}={\left(\frac{32\times 10^5}{2\times 10^5}\right)}^{1/4}=2$
or $T_2=2\times T_1=2\times 400=800K$