Chapter 22, Problem 68A

To determine

Energy used by lightbulb and the thermal energy generated.

Answer to Problem 68A

  $E=108\text{kJ}$ and $Q=95.04\text{kJ}$

Explanation of Solution

Given:

Power rating of light bulb, $P=60\text{W}$

Time, $t=0.5\text{hour}=1800\text{s}$

Percentage of energy converted to light, $\text{12\%}$

Formula Used:

The energy power relation is given as,

  $E=Pt$

Where, $E$ is energy, $P$ is power and $t$ is time.

Calculation:

On putting the given values,

  $\begin{array}{c}E=60\times 1800\\ =108\text{kJ}\end{array}$

As bulb is having efficiency of 12% which means 88% of energy is lost to heat, so

Thermal energy generated in 0.5 hour is,

  $\begin{array}{c}Q=\left(108\text{kJ}\times \frac{88}{100}\right)\\ =95.04\text{kJ}\end{array}$

Conclusion:

Therefore, energy used by light bulb is $E=108\text{kJ}$ and thermal energy generated is $Q=95.04\text{kJ}$ .