logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Ad
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
0 votes

Earth receives solar power at a rate of $\;8.2 Jcm^{-2}min^{-1}\;$ . Assuming that sun behaves like blackbody & the angle it substends on earth is $\;0.50^{0}\;$ , calculate its surface temperature - $\;(T=6 \times 10^{-8} W m^{-2}K^{-4})$

$(a)\;5880K\qquad(b)\;6880K\qquad(c)\;7880K\qquad(d)\;8880K$

Can you answer this question?
 
 

1 Answer

0 votes
Answer : $5880K$
Explanation :
$\large\frac{D}{R} $$\approx 0.5^{0} = 8.72 \times 10^{-3}$
Radiation emitted by sum ( per time unit)= $\;4 \pi (\large\frac{1}{2})$$^2 \sigma T^{4}$
$=\sigma \pi D^{2} T^{4}$
Radiation received by earth per unit area per unit time = $\;\large\frac{\sigma \pi D^{2} T^{4}}{4 \pi R^2}$
$\large\frac{\sigma T^{4}}{4}\;(\large\frac{D}{R})^2=\large\frac{6 \times 10^{-8} \times T^{4}}{4} $$\times (8.72 \times 10^{-3})^2$
$=\large\frac{8.2 \times 10^{4}}{60}$$m^{-2}s^{-1}$
$\large\frac{8.2 \times 10^{4}}{60}=\large\frac{6 \times 10^{-8}}{4} $$\times (8.72)^2 \times 10^{-6} \times T^{4}$
$T=5883 .5 K$
answered Mar 13, 2014 by yamini.v
edited Mar 25, 2014 by balaji.thirumalai
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...