Question

Two spheres of the same material have radii $1 \,m$ and $4\, m$ and temperatures $4000\, K$ and $2000\, K$ respectively., The ratio of energy radiated per second by the first sphere to the second is

• A. $1 : 1$
• B. $16 : 1$
• C. $4 : 1$
• D. $1 : 9$

Answer 1

Answer: (A)

$1 : 1$

Answer 2

Energy radiated per second by a body which has surface area $A$ at temperature $T$ is given by Stefan's law, $E = \sigma AT^4$ Therefore $\frac{E_{1}}{E_{2}} = \left(\frac{r_{1}}{r_{2}}\right)^{2}\left(\frac{T_{1}}{T_{2}}\right)^{4}$ $= \left(\frac{1}{4}\right)^{2}\left(\frac{4000}{2000}\right)^{4}$ $\frac{E_{1}}{E_{2}} = \frac{16}{16} = \frac{1}{1}$