Q161 / 200 · Manipal (MET) 2010 · Physics

Two bodies $A$ and $B$ having temperatlires $327^{\circ} \mathrm{C}$ and $427^{\circ} \mathrm{C}$ are radiating heat to the surrounding. The surrounding temperature is $27^{\circ} \mathrm{C}$. The ratio of rates of heat radiation of $A$ to that of $B$ is

  • Correct answer
    A
    $0.52$
  • B
    $0.31$
  • C
    $0.81$
  • D
    $0.42$

Solution

If temperature of surrounding is considered, then net loss of energy of a body by radiation $Q=\operatorname{Ae\sigma }\left(T^4-T_0^4\right) \Rightarrow Q \propto\left(T^4-T_0^4\right)$ $\therefore \frac{Q_1}{Q_2}=\frac{T_1^4-T_0^4}{T_2^4-T_0^4}$ $=\frac{(273+327)^4-(273+27)^4}{(273+427)^4-(273+27)^4}$ $=\frac{(600)^4-(300)^4}{(700)^4-(300)^4}=0.52$

Hence, the correct answer is option: 'A'.

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