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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 6 - Applications of IntegrationProblem 6.1.53b
Chapter 6, Problem 6.1.53b

Power and energy The terms power and energy are often used interchangeably, but they are quite different. Energy is what makes matter move or heat up and is measured in units of joules (J) or Calories (Cal), where 1 Cal=4184 J. One hour of walking consumes roughly 10⁶ J, or 250 Cal. On the other hand, power is the rate at which energy is used and is measured in watts (W; 1W=1 J/s). Other useful units of power are kilowatts (1 kW=10³ W) and megawatts (1 MW=10⁶ 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 3.6×10⁶ J. Suppose the power function of a large city over a 24-hr period is given by P(t) = E'(t) = 300 - 200 sin πt/12, where P is measured in megawatts and t=0 corresponds to 6:00 P.M. (see figure).


b. Burning 1 kg of coal produces about 450 kWh of energy. How many kilograms of coal are required to meet the energy needs of the city for 1 day? For 1 year? 

Verified step by step guidance
1
Understand that the power function is given by \(P(t) = E'(t) = 300 - 200 \sin\left(\frac{\pi t}{12}\right)\), where \(P(t)\) is in megawatts (MW) and \(t\) is in hours, with \(t=0\) at 6:00 PM.
To find the total energy used by the city over 1 day (24 hours), integrate the power function over the interval from \(t=0\) to \(t=24\): \(E(24) = \int_0^{24} P(t) \, dt = \int_0^{24} \left(300 - 200 \sin\left(\frac{\pi t}{12}\right)\right) dt\) This will give the total energy in megawatt-hours (MWh).
Evaluate the integral step-by-step: - Integrate the constant term \(300\) over 24 hours. - Integrate the sine term \(-200 \sin\left(\frac{\pi t}{12}\right)\) over 24 hours, using the substitution method if needed.
After finding the total energy in megawatt-hours, convert it to kilowatt-hours (kWh) by multiplying by 1000, since 1 MW = 1000 kW.
Use the given information that burning 1 kg of coal produces 450 kWh of energy. Calculate the kilograms of coal required by dividing the total energy in kWh by 450. For the yearly amount, multiply the daily coal consumption by 365.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Power as the Rate of Energy Consumption

Power represents how quickly energy is used or transferred, measured in watts (W), where 1 W equals 1 joule per second. Understanding power as the derivative of energy with respect to time (P(t) = E'(t)) allows us to calculate total energy consumption by integrating power over a time interval.
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Energy Units and Conversion

Energy is measured in joules (J) or kilowatt-hours (kWh), with 1 kWh equal to 3.6×10⁶ J. Converting between these units is essential for practical calculations, such as relating electrical energy consumption to fuel energy content, like the energy produced by burning coal.
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Integration to Find Total Energy from Power Function

Given a power function P(t), the total energy used over a time period is found by integrating P(t) with respect to time. This process sums the instantaneous power usage to find the cumulative energy, which is necessary to determine fuel requirements for a given energy demand.
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