Question

For a black body at temperature $727{}^{\circ }C$ its radiating power is 60 W and temperature of surrounding is $227{}^{\circ }C$ . If the temperature of the black body is changed to $1227{}^{\circ }C,$ then its radiating power will be

• A. 120 W
• B. 240 W
• C. 304 W
• D. 320 W

Answer 1

Answer: (D)

From Stefans-Boltzmann law $E=\sigma (T^4-T_0^4)$ $\frac{E_2}{E_1}=\frac{(T_2^4-T_0^4)}{(T_1^4-T_0^4)}$ $\Rightarrow$ $E_2=\left(\frac{T_2^4-T_0^4}{T_1^4-T_0^4}\right)E_1$ ?(i) Here $E_1=60W,$ $T_0=227{}^{\circ }C=500K,$ $T_1=727{}^{\circ }C=1000K,$ $T_2=1227{}^{\circ }C=1500K$ From Eq. (i), we get $\therefore$ $E_2=\frac{{(1500)}^4-{(500)}^4}{{(1000)}^4-{(500)}^4}\times 60$ $=\frac{{(500)}^4[3^4-1]}{{(500)}^4[2^4-1]}\times 60$ $=\frac{80}{15}\times 60=320W$