Question

12)A black body is heated from 27°C to 127° C. The ratio of their energies of radiations emitted will be

a)3:4

c)27:64

b)9:16

d)81:256​

Answer 1

Answer:

$81:256$.

Explanation:

Let $T$ denote the absolute temperature of this object.

Calculate the value of $T$ before and after heating:

$T(\text{before}) = 27 + 273 = 300\; \rm K$.

$T(\text{after}) = 127 + 273 = 400\; \rm K$.

By the Stefan-Boltzmann Law, the energy that this object emits (over all frequencies) would be proportional to $T^4$.

Ratio between the absolute temperature of this object before and after heating:

$\displaystyle \frac{T(\text{before})}{T(\text{after})} = \frac{3}{4}$.

Therefore, by the Stefan-Boltzmann Law, the ratio between the energy that this object emits before and after heating would be:

$\displaystyle \left(\frac{T(\text{before})}{T(\text{after})}\right)^{4} = \left(\frac{3}{4}\right)^{4} = \frac{81}{256}$.