Chapter 18, Problem 71GP

(a)

To determine

The resistance of the oven.

Answer to Problem 71GP

  $26\Omega$

Explanation of Solution

Given:

Voltage, $V=240V$

Power, $P=2200W$

Formula used:

Electrical power:

  $P=\frac{V^2}{R}$

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

Calculation:

Rearrange the expression: $P=\frac{V^2}{R}$

So, $R=\frac{V^2}{P}=\frac{{\left(240V\right)}^2}{2200W}=26.18\Omega \approx 26\Omega$

Conclusion: The resistance of the oven is $26\Omega$ .

(b)

To determine

Time that oven will take to bring the given changes.

Answer to Problem 71GP

  $26s$

Explanation of Solution

Given:

Voltage, $V=240V$

Power, $P=2200W$

The resistance of the oven is $26\Omega$ .

Efficiency, $e=75\%=0.75$

Temperature, $\Delta T=100°C-15°C=85°C$

Volume of the water heated, $120mL=0.120L$

Formula used:

Heat energy is $\Delta E=Q=mc\Delta T$ .

Where, $c$ is a specific heat, $\Delta T$ is the temperature change and $m$ is the mass of the object. Therefore,

Calculation:

By using the formula of heat energy:

  $e\left(P_{oven}\right)t=heatabsorbed=mc\Delta T$

Where, $e$ is efficiency, $P_{oven}$ power of oven, $\Delta T$ is the temperature change, $c$ is a specific heat, and $m$ is the mass.

Now, substitute all the given values:

  $0.75\left(P_{oven}\right)t=heatabsorbed=mc\Delta T$

  $t=\frac{mc\Delta T}{0.75\left(P_{oven}\right)}$

  $t=\frac{\left(0.120L\right)\left(\frac{1kg}{1L}\right)\left(4186J/kg\cdot C°\right)\left(85°C\right)}{0.75\left(2200W\right)}$

  $t=25.88s\approx 26s$

Conclusion: The time is around $26s$ .

(c)

To determine

The cost.

Answer to Problem 71GP

  $0.17cents$

Explanation of Solution

Given:

Reference cost is $11cents/kWh$

Time, $t=25.88s$

Formula used:

To calculate the cost of the energy, multiply the kilowatts of power consumed by the number of hours in operation times the cost per $kWh$ .

Calculation:

According to the given information, the cost is

  $Cost=\frac{11cents}{kWh}\left(2.2kW\right)\left(25.88s\right)\frac{1h}{3600s}=0.17cents$

Conclusion: The required cost is $0.17cents$ .