### Video Transcript

The Monroe power plant in Michigan has an installed capacity of 3280 megawatts. If it was to produce this power output constantly for an entire year, how much energy would it produce in that year? Give your answer in standard form in joules to three significant figures. Use a value of 365 for the number of days in a year.

Okay, so, in this question, we’re looking at the power output of a specific power plant, the Monroe power plant in Michigan. We’ve been told that this power plant has an installed capacity of 3280 megawatts. In other words, we can say that the power, which we’ll call 𝑃, that the Monroe power plant can produce is 3280 megawatts.

And we know that the question is talking about power when referring to the installed capacity because megawatts is a unit of power. Now, what we’re being asked to do is to work out how much energy this power plant would produce if it was to run constantly at the full output power, and if it was to do this for an entire year. So, if the power plant was to run at 3280 megawatts constantly for a time, which we’ll call 𝑡, of one year, then how much energy, which we’ll call 𝐸, will it produce?

To answer this question then, we need to find the relationship between power, time, and energy. So, let’s recall for this that power is defined as either energy transferred per unit time or, in the case of a power plant, the energy produced per unit time. In other words, the total amount of energy produced divided by the time taken for that energy to be produced. And in our question, we’ve been given this time taken for the energy to be produced as well as the power itself. And we’re trying to calculate the value of 𝐸, the energy. To do this, we need to rearrange the equation.

We can do this by multiplying both sides of the equation by the time 𝑡 because, this way, on the right-hand side, we’ve got a 𝑡 in the numerator and a 𝑡 in the denominator. Dividing 𝑡 by 𝑡 then just gives us one. And so, on the right-hand side, we’re left with 𝐸 multiplied by one, or simply 𝐸, the energy. And on the left-hand side, we’re left with the amount of time for which the energy is produced multiplied by the power.

Now, we know both quantities on the left-hand side, the time and the power, where the time for which the energy is being produced is one year and the power during this time is a constant 3280 megawatts. So, we could just plug those into this equation and find out the value of the energy produced. However, before we do this, let’s recall that we’re being asked to find the energy 𝐸 in joules.

Now, joules is the base unit of energy. Which means that if we want to find the energy in joules, then we need to convert the time 𝑡 and the power 𝑃 into base units as well. Because we can recall that one watt of power, where watt is the base unit of power, is equal to one joule, which is the base unit of energy, divided by one second, which is the base unit of time.

And so, rearranging this equation in the same way that we rearranged this one to give us 𝑡𝑃 is equal to 𝐸, we’re going to say that one watt multiplied by one second is equal to one joule. Or in other words, we can say that one joule is equal to one watt second. And so, to find the energy 𝐸 in joules, we’re going to need to convert the power so that it’s in watts and convert the time so that it’s in seconds.

So, let’s start by converting the power. Firstly, let’s recall that the prefix mega- means one million. In other words, one megawatt is the same thing as one million watts. And so, the power of 3280 megawatts is the same thing as 3280 multiplied by one million watts. Now, we could simplify this, but for now, let’s keep it as it is. And let’s focus instead on converting the time from years to seconds.

Let’s recall that one year has 365 days in it, which is what we’ve been told in the question. And then, every single day has 24 hours in it. And so, we can multiply this on the right-hand side by 24 hours per day because this fraction that we’re multiplying by is simply equal to one. Because 24 hours is the same thing as a day. And in multiplying by this fraction, we see that the unit of days cancels, leaving us simply with the unit of hours.

And then, we can recall that every single hour has 60 minutes in it. And so, we can multiply by one again, which is the same thing as multiplying by 60 minutes per hour. And then, we see that the unit of hours cancels, leaving us with minutes. And finally, we recall that every single minute has 60 seconds in it. And so, we multiply by 60 seconds per minute, or 60 seconds divided by one minute. And then, the unit of minutes cancels, leaving us finally with the unit of seconds.

And so, we find then that one year is equal to 365 multiplied by 24 multiplied by 60 multiplied by 60 seconds. When we simplify everything on the right-hand side, we find that one year is equal to 31536000 seconds. Which means that we’ve now converted the power into its standard unit and we’ve converted time into its standard unit. And hence, we can say that the energy produced by the power plant 𝐸 is equal to firstly the power, which is 3280 multiplied by one million watts as we saw here, multiplied by the time, which we know to be one year or 31536000 seconds. And so, we put that in here as well.

Now, to make progress, what we’ll do is to multiply all the numbers together and all of the units together. In other words, we’ll compute 3280 multiplied by one million multiplied by 31536000. Wow, those are some big numbers. And we’ll also multiply watts by seconds. But we can recall from earlier that one watt multiplied by one second is equal to one joule. And so, the end result of everything that we multiply here will be a quantity, whatever that numerical value is, in joules.

And so, when we multiply all of the numerical values, we find that the numerical value is so large that it doesn’t make sense to not write it in standard form. And it turns out that the energy produced is 1.0343808 dot dot dot times 10 to the power of 17 joules. And also, let’s recall quickly that standard form is when a number is written in this form, 𝑎 times 10 to the power of 𝑏, where 𝑎 is any value greater than or equal to one and less than 10, and 𝑏 is just an integer, a whole number. And that’s exactly what we see here. It’s in standard form. Because this value 𝑎 is somewhere between one and 10, where it’s larger than or equal to one but less than 10, and the value 𝑏 is an integer. 17 is a whole number.

Another example would be to write the number of seconds in a year in standard form. We can see that, for this number, currently the decimal point is located here after the last zero. Now, in order to write this number as one that’s between one and 10, specifically greater than or equal to one but less than 10, we would have to divide this number by a factor of 10, 100, 1000, 10000, 100000, one million. We’d have to divide this by a factor of 10 million, which is the same thing as 10 to the power of one, two, three, four, five, six, seven.

And so, dividing this number by a factor of 10 million would put the decimal point here. This number would now read 3.1536000 seconds. But to compensate for the fact that we divided by a factor of 10 million, or 10 to the power of seven, we need to multiply by a factor of 10 to the power of seven so that this whole combined number is exactly the same thing as 31536000. And so, tidying everything up a bit, we could write the number of seconds in one year as 3.1536000 times 10 to the power of seven seconds. And that is in standard form.

But anyway, so this is the answer that we found for the energy produced by the Monroe power plant, assuming it works at maximum capacity for an entire year. And we found this energy in joules in standard form. However, we haven’t yet given our answer to three significant figures. So, that’s the only thing that’s left to do out of all the instructions given to us in the question.

So, to round to three significant figures, we can firstly see that the leading numeral is not zero. It’s a one, so it’s significant. And that is the first significant figure. The second significant figure, then, is this zero. And the third significant figure is this three. In other words, if we are to round to three significant figures, we’re going to truncate, or cut off, our value there. But in order to work out what happens to this third significant figure, we need to look at the next value. This number is a four. Four is less than five. And therefore, our third significant figure is going to stay exactly the same. It’s not going to round up.

And so, we can say that, to three significant figures, our energy is 1.03 times 10 to the power of 17 joules. Hence, we found the answer to our question. If the Monroe power plant in Michigan were to be run at full capacity for an entire year, then the amount of energy that it would produce would be 1.03 times 10 to the power of 17 joules in standard form in joules to three significant figures.