Question

A 40W lamp wastes 34 J of energy every second by heating its surroundings.

Answer 1

Answer:

$15\%$.

Explanation:

The efficiency of a machine is the percentage of energy input that was turned into useful energy.

The power rating of this lamp is $40\; \rm W$ (same as $40\; \rm J \cdot s^{-1}$,) meaning that $40\; \rm J$ of energy is supplied to this lamp every second.

The question states that $34\; \rm J$ out of that $40\; \rm J$ of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the $(40\; \rm J - 34\; \rm J) = 6\; \rm J$ of energy supplied to this lamp would be turned into useful energy output.

Thus, every second, this lamp would receive $40\; \rm J$ of energy input and would outputs $6\; \rm J$ of useful work. The efficiency of this lamp would be:

$\begin{aligned}& \text{Efficiency} \\ =\; & \frac{\text{Useful energy out}}{\text{Total energy in}} \times 100\% \\ =\; & \frac{6\; \rm J}{40\; \rm J} \times 100\%\\ =\; &15\% \end{aligned}$.