Chapter 22.2, Problem 28PP

a.

To determine

Energy transformed by bulb in radiant energy.

Answer to Problem 28PP

The radiant energy is $1.32\text{kJ}$ .

Explanation of Solution

Given:

Power rating of bulb, $P=22\text{W}$

Efficiency of bulb is $\text{22}\text{\%}$

Time, $t=60\text{s}$

Formula used:

The energy power relation is given as,

  $E=Pt$

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

Calculation:

On substituting the values in above relation

  $\begin{array}{c}E=100\times 60\\ =6\times {10}^3\\ =6\text{kJ}\end{array}$

But, the efficiency of the bulb is 22 %

So, the energy transformed by light bulb in radiant energy is

  $\begin{array}{c}E=6\times \frac{22}{100}\\ =\left(\text{6}\times \text{0}\text{.22}\right)\text{kJ}\\ =1.32\text{kJ}\end{array}$

Conclusion:

Therefore, the energy transformed by light bulb in radiant energy is $1.32\text{kJ}$ .

b.

To determine

Thermal energy output per minute.

Answer to Problem 28PP

Thermal energy output is $4.68\text{kJ}$ .

Explanation of Solution

Given:

Power rating of bulb, $P=22\text{W}$

Efficiency of bulb is $\text{22}\text{\%}$

Time, $t=60\text{s}$

Formula used:

The energy power relation is given as,

  $E=Pt$

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

Calculation:

On substituting the values in above relation

  $\begin{array}{c}E=100\times 60\\ =6\text{kJ}\end{array}$

But, the efficiency of the bulb is 22%

So, the remaining 78% will lost as thermal energy.

So the thermal energy output is

  $\begin{array}{c}{E}_T=6\times {10}^3\times \frac{78}{100}\\ =\left(\text{6}\times \text{0}\text{.78}\right)\text{kJ}\\ =4.68\text{kJ}\end{array}$

Conclusion:

Therefore, the thermal energy output per minute is $4.68\text{kJ}$ .