Question

Two bodies A and B having equal surface areas are maintained at temperature 10°C and 20°C. The thermal radiation emitted in a given time by A and B are in the ratio
(a) 1 1.15
(b) 1 2
(c) 1 4
(d) 1 16

Answer 1

The thermal radiation emitted in a given time by A and B are in the ratio :

• We are given that,

Surface area of A = surface area of B = S

T1 = 10°C and T2 = 20°C

• According to Stefan Boltzmann law, energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by

• U = σAT^4

Here, σ is Stefan Boltzmann constant, T =Temperature in kelvin and A = surface area of body

• UA = σS(T1)^4

• UB = σS(T2)^4

• T1 = 273 + 10 = 283K

• T2 = 273 + 20 = 293K

• UA/UB = (T1)^4/(T2)^4

= (283/293)^4

• UA/UB = 1/1.15

Answer 2

Two bodies A and B having equal surface areas are maintained at temperature 10°C and 20°C. The thermal radiation emitted in a given time by A and B are in the ratio 1 1.15

Explanation:

Step 1:

Stefan’s Boltzmann law states that the total radiant heat power emitted from a surface is proportional to the fourth power of its absolute temperature. This law applies for all the theoretical surfaces that absorb all incident heat radiation.

According to Stefan’s Boltzmann law the energy of thermal radiation is given as

         $\mathrm{U}=\mathrm{e} \sigma \mathrm{A} \mathrm{T}^{4}$

Where e= emissivity

    σ= stefan's constant

           A= area of body

           T= temperature of the body

Step 2:

           For black body e=1  

          $\frac{u_{A}}{u_{B}}=\frac{\sigma \mathrm{AT}_{\mathrm{A}}^{4}}{\sigma \mathrm{AT}_{\mathrm{B}}^{4}}$

               $=\left(\frac{283}{293}\right)^{4}$

          $\frac{u_{A}}{u_{B}}=\frac{1}{1.15}$