Question 750: Thermodynamics & KTG

Question

Two spheres $S_1$ and $S_2$ have same radii but temperatures $T_1$ and $T_2$ respectively. Their emissive power is same and emissivity in the ratio 1:4. Then the ratio $T_1: T_2$ is

Accepted Answer

Given that two spheres, $ S_1 $ and $ S_2 $, have the same radius but different temperatures $ T_1 $ and $ T_2 $, their emissive powers are the same with emissivities in the ratio $ 1:4 $. We need to determine the ratio $ T_1 : T_2 $.
First, recall that emissive power is given by the formula:
$ \frac{Q_1}{A_1 t} = \frac{Q_2}{A_2 t} $
Since the radii of the spheres are the same, it follows that:
$ A_1 = A_2 $
Thus, we have:
$ Q_1 = Q_2 $
Using the Stefan-Boltzmann law, which states that emissive power $ E $ is given by:
$ E = e \sigma A T^4 $
We can write:
$ e_1 \sigma A T_1^4 = e_2 \sigma A T_2^4 $
Given the emissivity ratio $ \frac{e_2}{e_1} = \frac{4}{1} $, we have:
$ \frac{T_1^4}{T_2^4} = \frac{e_2}{e_1} = \frac{4}{1} $
Taking the fourth root on both sides, we find:
$ \frac{T_1}{T_2} = \left( \frac{4}{1} \right)^{\frac{1}{4}} = \frac{\sqrt{2}}{1} $
Hence, the ratio $ T_1 : T_2 $ is $ \sqrt{2} : 1 $.

Answer

$2: 1$

Answer

$\sqrt{2}: 1$

Answer

$1: \sqrt{2}$

Answer

$1: 2$

Thermodynamics & KTG

Two spheres S_1 and S_2 have same radii but temperatures T_1 and T_2 respectively. Their emissive power is same and emissivity in the ratio 1:4. Then the ratio T_1: T_2 is

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