Question

If the efficiency of a 60W light bulb is 9.7%, how much energy is wasted due to heat in 2 hours?

Answer 1

In case of 100% efficiency the wasted energy would be (by definition of the electrical power):

$A_{100\%} = 60W\cdot 2hours = 60W\cdot 7200 sec = 432kJ$

But in case of 9.7% efficiency:

$A_{9.7\%} = A_{100\%}\cdot 0.097 = 432kJ\cdot 0.097 = 41.9kJ$

Answer. A = 41.9 kJ.