Question

Energy is being emitted from the surface of black body at ${127}^{o}C$ at the rate of $1.0\times {10}^{6}J/s{m}^2$. The temperature of black body at which the rate of energy is $16.0\times {10}^{6}J/s{m}^2$ will be :

• A. ${754}^{o}C$
• B. ${527}^{o}C$
• C. ${254}^{o}C$
• D. ${508}^{o}C$

Answer 1

The correct option is B ${527}^{o}C$

Rate of energy emission is directly proportional to $4^{th}$ power the temperature of the body so $\frac{E_1}{E_2}=\frac{T_1^4}{T_2^4}$

We have $\frac{E_1}{E_2}=\frac{1}{16}$

So we have $\frac{T_1}{T_2}=(\frac{1}{16}{)}^{1/4}=\frac{1}{2}$

so $T_2=2T_1=2(127+273)K=800K$

Or $T_2=800-273={527}^{0}C$

Option B is correct.