A spherical light bulb dissipates 100W and is of 5cm diameter. Assume the emissivity is 0.8 and the irradiation is negligible. What is the surface temperature of this spherical …

Question

asked Jul 21, 2022 150k views

1 Answer

Amenzhinsky · Answer 1 · 2022-07-26T18:54:32+0000

Answer:

T=728.9K

Step-by-step explanation:

Power
P=100W

Diameter
d=5

Radius
r=2.5cm=>2.5*10^(-2)m

Emissivity
e=0.8

Generally the equation for Area of Spherical bulb is mathematically given by

$A=4\pi r^2$

$A=4\pi (2.5*10^(-2)m)^2$

A=7.85*10^(-3)m^2

Generally the equation for Emissive Power bulb is mathematically given by

$E=e\mu AT^4$

Where

$\mu=Boltzmann constants\\\\\mu=5.67*10^(-8)$

Therefore

$T^4=(E)/(e\mu A)$

T^4=(100)/(0.8*5.67*10^(-8)*7.85*10^(-3)m^2)

$T=^4\sqrt{2.80*10^(11)}$

T=728.9K