First slide

Heat transfer-Radiation

Question

The sun having surface temperature $T_{S}$ radiates like a blackbody. The radius of sun is $R_{S}$ and Earth is at a distance R from the centre of sun. Earth absorbs radiation falling on its surface from sun only and is at constant temperature T. If radiation falling on Earth’s surface is almost parallel and Earth also radiates like a blackbody, then

Moderate

A

$T = T_{s} \sqrt{\frac{R_{S}}{2 R}}$
B

$T = T_{S}$
C

$T = \frac{T_{s}}{2} \sqrt{\frac{R_{S}}{R}}$
D

$T = T_{s} \sqrt{\frac{R_{S}}{R}}$

Solution

[Step 1]: Energy absorbed per unit time by earth’s surface, if its radius is r, is
[Step 2]: $Q_{a b s} = \frac{σ 4 π R_{S}^{2} T_{S}^{4}}{4 π R^{2}} \times π r^{2} (∵ s o l a r c o n s \tan t, S = \frac{P}{4 π b^{2}} \Rightarrow S = \frac{4 {πa}^{2} {σT}^{4}_{s}}{4 π b^{2}})$
[Step 3]: Energy radiated per unit time by earth’s surface, $Q_{r a d} = σ (4 π r^{2}) T^{4}$
[Concept]: $Q_{a b s} = Q_{r a d}$
[Step 4]: $\Rightarrow T^{4} = \frac{T_{s}^{4}}{4} \frac{R_{s}^{2}}{R^{2}}$
[Step 5]: $T = T_{s} \sqrt{\frac{R_{s}^{}}{2 R}}$

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