Power and energy The terms power and energy are often used int...

Question

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?

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?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says a city's power consumption over a 24 hour period is modeled by PFT, which is equal to 250 minus 150, multiplied by a sign of the quantity of i T divided by 12 in quantity, where P is in megawatts, T is in hours, and T equal to zero corresponds to midnight. Burning 1 kg of natural gas produces 500 kilowatt hours of energy. How many kilograms of natural gas are needed to supply the city's energy for 1 day and for 1 year? So, the first step in solving this problem is determining how much energy is consumed by the city in one day. So, we call that our definition of power, which will denote as P, and that that definition is equal to the derivative of our energy with respect to time. And so, in order to find our total energy E, it's just gonna be equal to the interval, and here we're gonna be integrating over 1 day, so we're gonna be integrating from 0 to 24 of PFT DT. And we were given in the problem, a function PFT to use, so we'll substitute in that function. And so that means our energy is going to be equal to the interval from 0 to 24 of the quantity of 250. -150. Multiplied by the sign of quantity of IT divided by 12, in quantity DT. So, we just need to integrate this term by term. So doing so, this is going to be equal to the quantity, and here we integrate 250 with respect to T. We get 250 multiplied by T. Then we'll have minus 150, and this gets multiplied by the interval of sine of the quantity of pi T divided by 12 in quantity. And when we integrate this function, we will get minus 12 divided by pi, multiplied by the cosine of quantity of pi T divided by 12 in quantity, and all of this needs to be evaluated from 0 to 24. So, substituting in the limits of integration, we see that this is going to be equal to. 250 Multiplied by 24. Plus 1800 divided by pi, multiplied by the cosine. Of the quantity of i multiplied by 24, divided by 12 in quantity. Minus 250, multiplied by 0. Minus 1800 divided by pi, multiplied by the cosine of the quantity of pi, multiplied by zero divided by 12 in quantity. So simplifying this expression, we see that our energy E is going to be equal to. 6000 Plus 1800 divided by pi, since we'll have the that being multiplied by the cosine of 2 pi, and the cosine of two pi is just 1, then we'll have minus 0, then minus 1800 divided by pi. So again, here, this will be multiplied by the cosine of zero, which is also equal to 1. And so find this expression, we see that this is going to be equal to. 6000 Megawatt hours. And we get the units of megawatts hours, since our power was measured in megawatts, and our time was measured in hours. However, we're going to need to convert this into kilowatt hours to use later. So, the conversion factor is to multiply our our result here by 1000, to convert megawatt hours into kilowatt hours. In doing so, we'll see that this is going to be equal to 6 million. Kilowatt hours. So now we need to use this result here to determine how much natural gas we need to burn. So we're looking for the mass of the natural gas, which we'll note as M, and we're gonna be looking for the mass of natural gas to power the city for one day. So we'll call that M sub D. And this is just going to be equal to the total energy is consumed by the city, which is going to be the 6 million Kill what hours? And this needs to be multiplied by our conversion factor between kilowatt hours and kilograms. Now, we're told that 1 kg, Can produce 500 kilowatt hours of energy. So we're gonna need to multiply our 6 million kilowatt hours here by 1 kg, divided by 500 kilowatt hours. And when we evaluate this expression, we see that our mass is going to be equal to. 12,000 Kilograms. So, that is the amount of natural gas that we need to burn in order to power the city for one day. So that's the first part of our answer to this problem. So the second part, we need to determine. How much natural gas needs to be burned to power the city for an entire year. So, we'll call that mass MY there for a year, and this is just going to be equal to the mass that we have for a single day, which in our case is going to be 12,000. Kilograms. And we need to multiply this by the number of days in the year, which is 365. And when we evaluate this expression, we will get 4 million 380,000. Kilograms. So that is the answer for the second part to this problem. So, these are the answers that we're looking for for this problem. It takes 12,000 kg to power the city for one day, and it takes 4,380,000 kg of natural gas to power the city for an entire year. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Power as the Rate of Energy Consumption

Energy Units and Conversion

Integration to Find Total Energy from Power Function

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