Question

A standard filament light bulb has an efficiency rating of 18%. calculate the amount of light energy (useful work) that a 60w light bulb transfers each second. (1w = 1j/s)

Answer 1

If a bulb is rated at 60 W it uses 60 W (60 J/s) for its operation.

If only 18% of this power input provides useful light then the light produced each second is

P = .18 * 60 J / s = 10.8 J / sec

In one sec W = P * t = 10.8 J/s * 1 sec = 10.8 joules