A100-w light bulb generates 95w of heat, which is dissipated through a glass bulb that has a radius of 3.0 cm and is 0.50 mm thick. what is the temperature difference between the inner and outer surfaces of the glass? use 'deg c' as your units.

21 °C

Explanation:

Heat generated by bulb in one second, H = 95 W

Radius = 3 cm = 0.03 m, Area = 3.14 x 0.03 x 0.03 = 2.826 x 10^-3 m^2

Thickness, d = 0.5 mm = 0.5 x 10^-3 m

Coefficient of thermal conductivity of glass, K = 0.8 W/mK

Let ΔT be the temperature difference between inner and outer surface.

H = K A ΔT / d

95 = 0.8 x 2.826 x 10^-3 x ΔT / (0.5 x 10^-3)

ΔT = 21 °C

Thus, the temperature difference between the inner and outer surface of the bulb is 21 °C.

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6