A spherical ball of surface area 20 cm2 absorbs any radiation that falls on it. It is suspended in a closed box maintained at 57°C. (a) Find the amount of radiation falling on the ball per second. (b) Find the net rate of heat flow to or from the ball at an instant when its temperature is 200°C. Stefan constant = 6.0 × 10−8 W m−2 K−4.
Answers
Thus the rate of heat flow on the ball is t1 = 4.58 W
Explanation:
(a) A = 20 cm^2 = 20×10^−4 m^2
T = 572 °C = 57 + 273 = 330 K
E = AσT^4
E = 20 × 10^−4 × 6 × 10^−8 × (330)^4
E = 1.42 J
(b) (Et) = (Aσe)(T1^4 − T2^4)
[A =20 cm^2 = 20 × 10^−4 m^2
σ = 6 × 10^−8
t1 = 473 K, t2 = 330 K
t1 = 20 × 10^−4 × 6 × 10^−8 × 1 [(473)^4 − (330)^4)]
t1 = 20 × 6 × 10^−12
[5.005×1010−1.185×1010] = 20 × 6 × 3.82 × 10^−2
t1 = 4.58 W
Thus the rate of heat flow on the ball is t1 = 4.58 W
Answer:
Here you go ............
Explanation:
(a) A = 20 cm^2 = 20×10^−4 m^2
T = 572 °C = 57 + 273 = 330 K
E = AσT^4
E = 20 × 10^−4 × 6 × 10^−8 × (330)^4
E = 1.42 J
(b) (Et) = (Aσe)(T1^4 − T2^4)
[A =20 cm^2 = 20 × 10^−4 m^2
σ = 6 × 10^−8
t1 = 473 K, t2 = 330 K
t1 = 20 × 10^−4 × 6 × 10^−8 × 1 [(473)^4 − (330)^4)]
t1 = 20 × 6 × 10^−12
[5.005×1010−1.185×1010] = 20 × 6 × 3.82 × 10^−2
t1 = 4.58 W
Thus the rate of heat flow on the ball is t1 = 4.58 W
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