a) The rate at which electrical energy is transferred to thermal energy is equal to the power dissipated on the resistor because of the Joule effect:
[tex]P= \frac{V^2}{R}= \frac{(120 V)^2}{17 \Omega}=847 W [/tex]
b) We can find the total energy dissipated in 7 h:
[tex]E=Pt = (847 W)(7 h)=5929 Wh=5.93 kWh[/tex]
The energy costs 0.05 $ per kWh, so the total cost of 5.93 kWh of energy is
[tex]5.93 kW \cdot 0.05 \frac{1}{kWh} =0.30[/tex] $
