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

`81:256`.

Step-by-step 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})} = (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((3)/(4)\right)^(4) = (81)/(256)`.