A 60 W light bulb is left on for 8.0 h.

Question

(a) How much electrical energy will it use?
E=Pt
8.0h X 3,600= 28800s
E=(60 W)(28800s)=1,728,000 J
The light bulb uses 1.7 X 106 J or 1.7MJ of energy.

(b) If the light bulb is only 15% efficient, how much electrical energy will be converted into radiant energy (light)?

Answer 1

To find the amount of electrical energy converted into radiant energy (light) by the light bulb, we can use the efficiency formula:

\[ \text{Useful energy output} = \text{Efficiency} \times \text{Total energy input} \]

Given that the light bulb is 15% efficient:

\[ \text{Efficiency} = 0.15 \] \[ \text{Total energy input} = 1,728,000 , \text{J} , (\text{from part a}) \]

Now, we can calculate the useful energy output:

\[ \text{Useful energy output} = 0.15 \times 1,728,000 , \text{J} \]

Calculating this gives:

\[ \text{Useful energy output} = 259,200 , \text{J} \]

Therefore, the light bulb converts approximately 259,200 J (or 0.2592 MJ) of electrical energy into radiant energy (light).