Question

A 100 W lightbulb generates 96 W of heat, which is dissipated through a glass bulb that has a radius of 3.1 cm and is 0.8 mm thick. What is the difference in temperature between the inner and outer surfaces of the glass? Answer in ( C°)

Answer 1

given data

power = 100 Wat

H=96W

x=0.0008m

A =4*pi*r^2 = 4*3.14*(3.1*10^-2)^2 = 0.012 m^2

H = k*A*∆T/x

∆T = H*x/(k*A)

k = 1.1 W/(m-ºK)

∆T = (96*0.8*10^-3)/(1.1* 0.012 )=5.818 ºK