Chapter 12.3, Problem 21R

To determine

To Find: Efficiency of the fluorescent light.

Answer to Problem 21R

The required efficiency of the bulb is $20\mathrm{\%}$ .

Explanation of Solution

Given:

The power of fluorescent light is $2\text{5}\text{W}$ .

The thermal energy emitted per second is $5.0\text{J}$ .

Formula Used:

  $\mathrm{e}\text{=}\frac{\text{power}\text{emitted}}{\text{total}\text{power}}\times 100\mathrm{\%}$

Here, e is the efficiency of the light.

Substitute $2\text{5}\text{W}$ for total power, $5\text{J}$ for power emitted and solve.

  $\begin{array}{l}\mathrm{e}\text{=}\frac{\text{power}\text{emitted}}{\text{total}\text{power}}\times 100\mathrm{\%}\\ \mathrm{e}=\frac{5}{25}\times 100\mathrm{\%}\\ \mathrm{e}=20\mathrm{\%}\end{array}$

Conclusion:

Hence,the required efficiency of the bulb is $20\mathrm{\%}$ .