A black sphere I m in radius lies 6 times 10^12 m from the sun which radiates 4

Question

A black sphere I m in radius lies 6 times 10^12 m from the sun which radiates 4 times 10^26 W. This is about at the distance of Pluto. Assume it receives no other significant radiation from its surroundings. How much power does it receive from the sun? (assume it absorbs all sunlight that hits it) If the sphere is at room temperature (300 K) initially, how much power does it initially radiate into space? At what temperature will it eventually reach equilibrium? At what distance from the sun should it be placed so that it would it reach an equilibrium temperature of300 K? (for comparison, the earth is 1.5 times 10^11 m from the sun)

Explanation / Answer

a.According to boltzman stefans law of black body radiation:

L= 4 pi R2 sigma T4

where L1=power radiated by sun=4*10^26

R=Radius of object R2=1m

Radius of sun R1=695700 km=695700*103m

sigma=stefan constant= 5.67*10-8Wm-2 K-4

T1=Temperature of sun=5777 K

Power recieved by object=L2=??

Power radiated by sun= Power recieved by the object

L1=L2

4pi(R1)^2*sigma*(T1)^4=4pi(R2)^2*sigma*(T2)^4

(R1)^2*(T1)^4=(R2)^2*(T2)^4

(695700*10^3)^2*(5777)^4=(1)^2*(T2)^4

(695700*10^3)*2(5777)^4=1*(T2)^4

4.8*10^11**10^6*1.1*10^15=T2^4

5.28*10^32=T2^4

5.28*10^32K

Now, power recived by object=L2

L2=4pi*R2^2sigma*T4

=4*3.14*1*5.67*10^-8*5.28*10^32

=376.01*10^24W

b.Temperature of body=300K

L= power radiated into space=??

R=radius=1m

sigma=5.67*10-8Wm-2K-4

L=4piR*sigma*T^4

L=4*3.14*5.67*10-8*(300)^4

=71.2*10-8* 81*10^8

=5767.2W

c. T=(aI/A sigma)^1/4

where a= cross section area of object=3.14*(1)2=3.14m2

A=surface area=4pi R^2=4*3.14*(1)=12.56m2

I= incident solar energy=1368W/m2

Now T=(3.14*1368)/12.56*5.67*10-8)^1/4

=(4295.52/71.2*10-8)^1/4

=60.3*10-8

=15.075K

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