Answer
See answers.
Work Step by Step
a. Use equation 18–6b to calculate the power, and solve for the resistance R.
$$P=\frac{V^2}{R}$$
$$R=\frac{V^2}{P}=\frac{(240V)^2}{2800W}\approx 21\Omega$$
b. Only 65 percent of the oven’s energy makes it into the water as heat.
$$0.65P_{oven}t=Q_{absorbed}=mc\Delta T$$
$$t=\frac{mc\Delta T}{0.65P_{oven}}=\frac{(0.120L)(1kg/L)(4186J/kg\cdot C)(85C)}{0.65(2800W)}\approx 23s$$
c.
$$\frac{11\;cents}{kWh}(2.8kW)(23.46s)\frac{1h}{3600s}=0.20\;cents$$
It only costs one-fifth of a cent.
