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Heat of Reaction

A black body ...

Question

A black body at temperature $T$ radiates energy at the rate of $E$ Wm $^{- 2}$ . In reducing its temperature to $T / 2$ the rate of energy radiation will be

E / 16

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E / 4

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E / 2

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E

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Solution

The correct option is D $E / 16$
According to Stefan's law, the ratio of energy emission is directly proportional to fourth power of temperature, by decreasing temperature to half , the energy emitted will be 1/16th times of previous value. Thus the correct answer is A.
$\frac{Δ Q_{B}}{Δ Q_{A}} = \frac{(.5 T)^{4}}{T^{4}} = \frac{1}{16} \Rightarrow Δ Q_{B} = \frac{Δ Q_{A}}{16}$

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