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Grade: 11
        A copper sphere is suspended in an evacuated chamber maintained at 300 K. The sphere is maintained at a constant temperature of 500 K by heating it electrically. A total of 210 W of electric power is needed to do it. When the surface of the copper sphere is completely blackened, 700 W is needed to maintain the same temperature of the sphere. Calculate the emissivity of copper. 
5 years ago

Answers : (2)

Aditi Chauhan
askIITians Faculty
396 Points
							Sol. Q = eσA(T base 2^4 – T base 1^4)
For any body, 210 = eAσ[(500)^4 – (300)^4]
For black body, 700 = 1 * Aσ[(500)^4 – (300)^4]
Dividing 210/700 = e/1 ⇒ e = 0.3
Q/t = eAe(T base 2^4 – T base 1^4)
⇒ Q/At = 1 * 6 * 10^-8 [(300)^4 – (290)^4] = 6 * 10^-8 (81 * 10^8 – 70.7 * 10^8) = 6 * 10.3
Q/t = KA(θ base 1 – θ base 2)/l
⇒ Q/tA = K(θ base 1 – θ base 2)/l = K * 17/0.5 = 6 * 10.3 = K = 17/0.5 ⇒ K = 6 * 10.3 * 0.5/17 = 1.8

						
5 years ago
L JAIPAL REDDY
13 Points
							
Q1=E1*A* σ(T14-T24)
210=E1*A* σ(5004-3004)...........(1)
700=E2*A* σ(5004-3004)............(2)      {Emissivity of the black body is equal to 1, i.e E2=1}
                                                                          σ= 5.67*10-8 W/M2-K4
SUB VALUES IN EQ 2, WE GET A2=0.226.........(3)
NOE, SUB A IN EQ 1, WE GET, E1=0.30125
10 months ago
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