Question

Two bodies of same shape, same size and same radiating power have emissivitys 0.2 and 0.8. The ratio of their temperatures is:

• A. $\sqrt{3}:1$
• B. $\sqrt{2}:1$
• C. $1:\sqrt{5}$
• D. $1:\sqrt{8}$

Answer 1

Answer: (B)

Radiating power of a body of area ${\text{A}}_1,$ emissivity $e_1$ and surface temperature $T_1$ is $P_1=\sigma e_1{T}_1^4{T}_1$ ?(i) Similarly radiating power of a body of area $A_2,$ emissivity $e_2$ and surface temperature $T_2$ is $P_2=\sigma e_2{T}_2^4{A}_2$ ?(ii) Given: $P_1=P_2$ and $A_1=A_2,$ $\sigma e_1{T}_1^4{A}_1=\sigma e_2{T}_2^4{A}_2$ $\Rightarrow$ ${\left(\frac{T_1}{T_2}\right)}^4=\left(\frac{e_2}{e_1}\right)\left(\frac{A_2}{A_1}\right)$ $\frac{T_1}{T_2}={\left(\frac{e_2}{e_1}\right)}^{1/4}.1$ $(\because A_1=A_2)$ Here $e_1=0.2,e_2=0.8$ Hence, $\frac{T_1}{T_2}={\left(\frac{0.8}{0.2}\right)}^{1/4}={\left(\frac{4}{1}\right)}^{1/4}={\left(\frac{2^2}{1^1}\right)}^{1/4}$ $\frac{T_1}{T_1}={\left(\frac{2}{1}\right)}^{1/2}=\sqrt{\frac{2}{1}}$ $T_1:T_2=\sqrt{2}:1$