Question

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 light bulb

Answer 1

`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`