Question

An object is at a temperature of ${400}^0$ C. At what temperature would it radiate energy twice as fast? The temperature of the surroundings may be assumed to be negligible

• A. 200°C
• B. 200K
• C. 800°C
• D. 800K

Answer 1

The correct option is D

800K

$\frac{E_2}{E_1}={\left(\frac{T_2}{T_1}\right)}^4\Rightarrow \frac{2}{1}={\left(\frac{400+273}{T}\right)}^4={\left(\frac{673}{T}\right)}^4\Rightarrow T=2^{\frac{1}{4}}\times 673=800K.$