Question 678: Thermodynamics & KTG

Question

Two bodies A and B at temperatures ' $\mathrm{T}_1$ ' K and ' $\mathrm{T}_2$ ' K respectively have the same dimensions. Their emissivities are in the ratio $16: 1$. At $\mathrm{T}_1=\mathrm{xT}_2$, they radiate the same amount of heat per unit area per unit time. The value of $x$ is

Accepted Answer

We are asked:
Two bodies, A and B,
same dimensions
(so same surface area).
Temperatures are $T_1$ and $T_2$.
Emissivities in the ratio $16:1$.
At $T_1 = x T_2$, they radiate the same power per unit area per unit time.
Find $x$.
Step 1: Radiation law
Radiated power per unit area (StefanÃ¢ÂÂBoltzmann law):
$ E = e\sigma T^4 $
where $e$ is emissivity.
Step 2: Apply to both bodies
For body A: emissivity ratio given " $16 : 1$". Let emissivity of A = $16k$, B = $k$.
Then:
$ E_A = (16k)\sigma T_1^4,  $
$ E_B = (k)\sigma T_2^4. $
Step 3: Condition
Same power per unit area:
$ 16k\sigma T_1^4 = k\sigma T_2^4 $
$ 16 T_1^4 = T_2^4 $
$ \left(\frac{T_1}{T_2}ight)^4 = \frac{1}{16}. $
Step 4: Solve
$ \frac{T_1}{T_2} = \left(\frac{1}{16}ight)^{1/4} = \frac{1}{2}. $
So
$ x = 	frac{1}{2} = 0.5. $
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Answer: Option D (0.5)

Answer

8

Answer

4

Answer

2

Answer

0.5

Thermodynamics & KTG

Two bodies A and B at temperatures ' \mathrm{T}_1 ' K and ' \mathrm{T}_2 ' K respectively have the same dimensions. Their emissivities are in the ratio 16: 1. At \mathrm{T}_1=\mathrm{xT}_2, they radiate the same amount of heat per unit area per unit time. The value of x is

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