Question 26.69: How much electrical energy is transferred to thermal energy ......

We find the rate of energy consumption from Eq. 26-28:

P={\frac{V^{2}}{R}}            (resistive dissipation).                                 (26-28)

P={\frac{V^{2}}{R}}={\frac{(90\,\mathrm{V})^{2}}{400\,\Omega}}=20.3\,\mathrm{W}

Assuming a steady rate, the energy consumed is (20.3 J/s)(2.00 × 3600 s) = 1.46 × 10^{5} J.