Question

In 1 second, a light bulb transfers 3 joules to light energy and 57 J to heat. What is the energy input in 1 second? What is the efficiency?

Answer 1

energy input per second `=60J`
efficiency `= 0.05`

Explanation:

`3J + 57J = 60J`

total energy input per second `3J + 57J = 60J`

efficiency = efficient energy output / total energy input

efficient energy transfer in a light bulb: electrical energy -> light energy
(wasted energy transfer in a light bulb: electrical energy -> heat energy)

efficient energy output = light energy output `= 3J`
total energy input `= 60J`

efficiency = `(3J)/(60J) = 1/20` or `0.05`