Chapter 22.2, Problem 26PP

a.

To determine

Current through the heater.

Answer to Problem 26PP

  $I=8\text{A}$

Explanation of Solution

Given:

Resistance of heater, $R=15\Omega$

Source voltage, $V=120\text{V}$

Formula used:

From Ohm’s law,

  $V=IR$

Where, $V$ is voltage, $I$ is current and $R$ is resistance

Calculation:

On substituting the values,

  $\begin{array}{c}120=I\times 15\\ I=\frac{120}{15}\\ =8\text{A}\end{array}$

Conclusion:

Therefore, current through the electric heater is $I=8\text{A}$ .

b.

To determine

Energy used by heater.

Answer to Problem 26PP

Energy used by heater is $E=28.8\text{kJ}$

Explanation of Solution

Given:

Resistance of heater, $R=15\Omega$

Source voltage, $V=120\text{V}$

Time, $t=30\text{s}$

Formula used:

The energy consumed by load is given as,

  $E=I^{2}Rt$

Where, $I$ is current through load, $R$ is resistance of load and $t$ is time.

Calculation:

As, $I=8\text{A}$ from previous subpart.

On substituting the values,

  $\begin{array}{c}E=8^2\times 15\times 30\\ =64\times 15\times 30\\ =64\times 450\\ =28.8\text{kJ}\end{array}$

Conclusion:

Therefore, the energy used by the heater is $E=28.8\text{kJ}$ .

c.

To determine

Thermal energy liberated.

Answer to Problem 26PP

  $Q=28.8\text{kJ}$

Explanation of Solution

Given:

Resistance of heater, $R=15\Omega$

Source voltage, $V=120\text{V}$

Time, $t=30\text{s}$

Formula used:

According to law of conservation of energy, total energy of an isolated system remains conserved. It is neither created nor destroyed. There is only a change in form of energy.

Calculation:

Since all the electrical energy is converted into heat. So, the thermal energy liberated in this time is equal to the electrical energy used by the heater.

Hence,

  $Q=E=28.8\text{kJ}$

Conclusion:

Therefore, the thermal energy liberated in given time is $Q=28.8\text{kJ}$ .