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 \( 3.6 \times 10^{5} \mathrm{~J} \). 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.)
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To determine how many grams of uranium are needed to power the city for one day and for one year, you'll want to first calculate the total energy consumption per day. By integrating the power function \( P(t) \) over 24 hours, you can find the total energy used by the city in \( MWh \), which you can then convert to \( kWh \) and divide by the energy output from 1 gram of U-235. In one day, the city consumes about \( 345 \, \text{MW} \times 24 \, \text{hr} = 8280 \, \text{MWh} \), or \( 8,280,000 \, \text{kWh} \). Each gram of U-235 produces \( 16,000 \, \text{kWh} \), so the amount of uranium needed per day is: \[ \frac{8,280,000 \, \text{kWh}}{16,000 \, \text{kWh/g}} \approx 517.5 \, \text{g} \] Rounding that to the nearest integer gives you approximately \( 518 \) grams of uranium needed for daily energy needs. For a year, simply multiply the daily requirement by 365 days, which gives approximately \( 189,070 \) grams, or about \( 189.1 \) kg of U-235! The city would need around \( 518 \, \text{g} \) of uranium per day and \( 189,070 \, \text{g} \) per year.
