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JIPMER 2015
Exam
JIPMER
Year
2015
Subject
Physics
Chapter
Thermal Properties of Matter

JIPMER 2015 · Physics · Thermal Properties of Matter

The earth receives solar radiation at a rate of 8.2 ~J ~cm^ -2...

The earth receives solar radiation at a rate of 8.2 J cm−2 min−18.2 \mathrm{~J} \mathrm{~cm}^{-2} \mathrm{~min}^{-1}. If the sun radiates as the black bodies, the temperature at the surface of the sun will be (the angle subtended by sun on the earth in suppose 0.53∘0.53^{\circ} and Stefan constant is σ=5.67×10−8Wm−2 K4\sigma=5.67 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^4 )

Options

  1. A5800 K
  2. B6700 K
  3. C8000 K
  4. D7800 K

Correct answer

A. 5800 K

Step-by-step solution

Let diameter of the sun is dd while radius of the earth is rr then D2≈0.53×π180=9.25×10−3\begin{aligned} \frac{D}{2} & \approx 0.53 \times \frac{\pi}{180} \\ & =9.25 \times 10^{-3}\end{aligned} The radiation emitted per unit area of the sun is 4π(d2)2σTA=πD2σTA4 \pi\left(\frac{d}{2}\right)^2 \sigma T^A=\pi D^2 \sigma T^A The radiation received at the earths surface per unit time per unit area is πσdT44πr2=σT44(dR)2\frac{\pi \sigma d T^4}{4 \pi r^2}=\frac{\sigma T^4}{4}\left(\frac{d}{R}\right)^2⇒σT44(DR)2=82Jcm−2min−1 or 14×5.67×10−8×T4×(9.25×10−3)2=8.210−4×60Wm−2⇒T=5794K≅5800K\begin{array}{rlrl}\Rightarrow & & \frac{\sigma T^4}{4}\left(\frac{D}{R}\right)^2 & =82 \mathrm{Jcm}^{-2} \mathrm{ min}^{-1} \\ & \text{ or } & \frac{1}{4} \times 5.67 \times 10^{-8} \times T^4 \times\left(9.25 \times 10^{-3}\right)^2 \\ & = & \frac{8.2}{10^{-4} \times 60} \mathrm{Wm}^{-2} \\ \Rightarrow & & T & =5794 \mathrm{ K} \cong 5800 \mathrm{ K}\end{array}$$
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