Power and energy are often used interchangeably, but they are quite different. Energy is measured in units of joules ( $$ \mathbf{J} $$ ) 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 ( 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 ( kWh ), which is $$ 3.6 imes 10^{5} \mathrm{~J} $$. Suppose the power function of a large city over a 24 -hr period is given by $$ P(t)=E^{\prime}(t)=345-230 \sin (x t / 12) $$, where $$ P $$ is measured in $$ M W $$ and $$ t=0 $$ corresponds to $$ 6: 00 \mathrm{p} . \mathrm{m} $$. (see accompanying figure). Complete parts (a) through ( d ) below. Click the icon to view the graph of $$ \mathrm{P}(\mathrm{t})=\mathrm{E}^{\prime}(\mathrm{t})=345-230 \sin (\pi / / 12) $$. (Simplify your answer.) c. Fission of 1 gram of uranium-235 (U-235) produces about $$ 16,000 \mathrm{kWh} $$ of energy. How many grams of uranium are needed to meet the energy needs of the city for 1 day? For 1 yr ? The amount of uranium needed in a day is $$ \square $$ g. (Simplify your answer. Round to the nearest integer as needed.)

Question

Power and energy are often used interchangeably, but they are quite different. Energy is measured in units of joules ( $$ \mathbf{J} $$ ) 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 ( 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 ( kWh ), which is $$ 3.6 	imes 10^{5} \mathrm{~J} $$. Suppose the power function of a large city over a 24 -hr period is given by $$ P(t)=E^{\prime}(t)=345-230 \sin (x t / 12) $$, where $$ P $$ is measured in $$ M W $$ and $$ t=0 $$ corresponds to $$ 6: 00 \mathrm{p} . \mathrm{m} $$. (see accompanying figure). Complete parts (a) through ( d ) below. Click the icon to view the graph of $$ \mathrm{P}(\mathrm{t})=\mathrm{E}^{\prime}(\mathrm{t})=345-230 \sin (\pi / / 12) $$. (Simplify your answer.) c. Fission of 1 gram of uranium-235 (U-235) produces about $$ 16,000 \mathrm{kWh} $$ of energy. How many grams of uranium are needed to meet the energy needs of the city for 1 day? For 1 yr ? The amount of uranium needed in a day is $$ \square $$ g. (Simplify your answer. Round to the nearest integer as needed.)

UpStudy AI · Accepted Answer

To determine the amount of uranium-235 (U-235) needed to meet the energy demands of the city for one day, follow these steps:

1. **Calculate the Total Energy Consumed in One Day:**

The power function is given by:
   $$
   P(t) = 345 - 230 \sin\left(\frac{\pi t}{12}\right) \quad \text{(in MW)}
   $$
   
   To find the total energy consumed over a 24-hour period, integrate the power function with respect to time:
   $$
   E = \int_{0}^{24} P(t) \, dt = \int_{0}^{24} \left(345 - 230 \sin\left(\frac{\pi t}{12}\right)\right) dt
   $$
   
   Evaluate the integral:
   $$
   E = 345 \times 24 - 230 \times \left[\frac{12}{\pi} \left(-\cos\left(\frac{\pi t}{12}\right)\right)\right]_{0}^{24}
   $$
   
   Since $$\cos(2\pi) = \cos(0) = 1$$, the sine integral term cancels out:
   $$
   E = 345 \times 24 = 8280 \, \text{MWh}
   $$
   
   Convert MWh to kWh:
   $$
   8280 \, \text{MWh} = 8,280,000 \, \text{kWh}
   $$

2. **Determine the Amount of Uranium-235 Needed:**

Given that fission of 1 gram of U-235 produces 16,000 kWh:
   $$
   \text{Grams of U-235 needed} = \frac{8,280,000 \, \text{kWh}}{16,000 \, \text{kWh/g}} = 517.5 \, \text{g}
   $$
   
   Rounding to the nearest integer:
   $$
   \text{Grams of U-235 needed} = 518 \, \text{g}
   $$

**Answer:**

The amount of uranium needed in a day is 518 g.
The city needs 518 grams of uranium-235 each day to meet its energy requirements.

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