Power and energy are often used interchangeably, but they are quite different. Energy is measured in units of joules ( J ) or Calories (Cal), where \( 1 \mathrm{Cal}=4184 \mathrm{~J} \). On the other hand, power is the rate at which energy is used and is measured in watts \( (\mathrm{W} ; 1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}) \). Other useful units of power are kilowatts \( (\mathrm{kW}) \) and megawatts ( \( 1 \mathrm{MW}=10^{6} \mathrm{~W} \) ). If energy is used at a rate of 1 kW for 1 hr , the total amount of energy used is 1 kilowatt-hour \( (\mathrm{kWh}) \), which is \( 3.6 \times 10^{6} \mathrm{~J} \). Suppose the power function of a large city over a 24 -hr period is given by \( \mathrm{P}(\mathrm{t})=\mathrm{E}^{\prime}(\mathrm{t})=345-230 \sin (\pi t / 12) \), where P is measured in MW and \( \mathrm{t}=0 \) corresponds to \( 6: 00 \) p.m. (see accompanying figure). Complete parts (a) through (d) below. Click the icon to view the graph of \( \mathrm{P}(\mathrm{t})=\mathrm{E}^{\prime}(\mathrm{t})=345-230 \sin (\pi \mathrm{t} / 12) \). a. How much energy is consumed by this city in a typical 24 -hr period? Express the answer in MWh and in J . The energy consumed in a typical 24 -hour period in MWh is 8280 MWh . (Simplify your answer.) The energy consumed in a typical 24 -hour period in J is (Simplify your answer. Use scientific notation.) J .

To find the total energy consumed by the city in a typical 24-hour period, we need to integrate the power function \( P(t) = 345 - 230 \sin(\pi t / 12) \) over the time period from 0 to 24 hours. Given: \[ P(t) = 345 - 230 \sin(\pi t / 12) \] The total energy consumed in a 24-hour period is given by the integral of the power function over the time period: \[ \text{Total Energy} = \int_{0}^{24} P(t) dt \] Substitute the power function into the integral: \[ \text{Total Energy} = \int_{0}^{24} (345 - 230 \sin(\pi t / 12)) dt \] Now, we can calculate the total energy consumed by the city in a typical 24-hour period. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{0}^{24} \left(345-230\sin\left(\frac{\pi t}{12}\right)\right) dt\) - step1: Evaluate the integral: \(\int \left(345-230\sin\left(\frac{\pi t}{12}\right)\right) dt\) - step2: Use properties of integrals: \(\int 345 dt+\int -230\sin\left(\frac{\pi t}{12}\right) dt\) - step3: Evaluate the integral: \(345t+\int -230\sin\left(\frac{\pi t}{12}\right) dt\) - step4: Evaluate the integral: \(345t+\frac{2760}{\pi }\times \cos\left(\frac{\pi }{12}t\right)\) - step5: Return the limits: \(\left(345t+\frac{2760}{\pi }\times \cos\left(\frac{\pi }{12}t\right)\right)\bigg |_{0}^{24}\) - step6: Calculate the value: \(8280\) The total energy consumed by the city in a typical 24-hour period is 8280 MWh. To express the answer in Joules, we need to convert MWh to Joules. Given that 1 MWh is equal to \( 3.6 \times 10^6 \) Joules, we can convert 8280 MWh to Joules as follows: \[ 8280 \, \text{MWh} = 8280 \times 3.6 \times 10^6 \, \text{J} \] Now, we can calculate the total energy consumed in Joules. Calculate the value by following steps: - step0: Calculate: \(8280\times 3.6\times 10^{6}\) - step1: Rewrite the expression: \(8280\times 36\times 10^{5}\) - step2: Multiply the terms: \(298080\times 10^{5}\) The total energy consumed by the city in a typical 24-hour period is \( 2.9808 \times 10^{10} \) Joules. Therefore, the answers to parts (a) through (d) are: a. The total energy consumed by the city in a typical 24-hour period is 8280 MWh. b. The total energy consumed by the city in a typical 24-hour period is \( 2.9808 \times 10^{10} \) Joules.