Video: Finding the Efficiency of a Device given Its Useful Power Output and Its Total Power Output

A refrigerator has a total power input of 35 W and a useful power output of 7.4 W. What is the efficiency of the refrigerator? Give your answer to 2 significant figures.

03:09

Video Transcript

A refrigerator has a total power input of 35 watts and a useful power output of 7.4 watts. What is the efficiency of the refrigerator? Give your answer to two significant figures.

Okay, so let’s start by underlining all of the important bits here. First of all, we’ve been told that the refrigerator has a total power input of 35 watts. And we’ve also been told that it has a useful power output of 7.4 watts.

We’ve been asked to find the efficiency of the refrigerator. And we’ve been asked to give our answer to two significant figures. So how do we go about doing this? Well, we need to recall the definition of efficiency.

Efficiency is defined as the useful energy output divided by the total energy input. Now, in our case, we’ve got very similar quantities. We’ve been given the total power input and the useful power output. But we want the total energy input and the useful energy output.

So we need to recall the relationship between power and energy. Power is defined as the energy transferred per unit time or in symbols that’s 𝑃 is equal to 𝐸 divided by 𝑡. Power is equal to energy transferred per unit time.

Therefore, if we want to find the efficiency of our refrigerator, we say that this is equal to the useful power output, which is 7.4 watts, multiplied by some period of time, which we’ll call 𝑡. And we divide this by the total energy input, which is 35 watts, again multiplied by that same period of time 𝑡.

Now, what we’ve done here is rearrange this equation by multiplying both sides of the equation by the time 𝑡. This gives us 𝑃 multiplied by 𝑡 is equal to 𝐸. In other words, power multiplied by time is equal to energy.

Then, what we’ve done is we’ve taken our quantities of the useful power output and multiply that by the time 𝑡 to give us the useful energy output in a time 𝑡. And we’ve done the same thing for the denominator. We’ve taken the total power output of 35 watts and multiplied that by the same period of time 𝑡 to give us the total energy input in a time 𝑡. And now, we see that both of the 𝑡s in the numerator and the denominator cancel out.

Therefore, what we’re left with in the end is just our useful power output divided by our total power input all because power and energy are related in this way. But anyway, having done all of that, we just need to evaluate this fraction here. When we do, we find that the efficiency is 0.21142857 and so on and so forth.

But this is not our final answer. Remember we’ve been asked to give our answer to two significant figures. So our first significant figure is the digit two and our second significant figure is this one here next to it. The digit after that is a one. So our second significant figure will not round up. It will stay the same.

And so to two significant figures, our efficiency is 0.21 or as a percentage that’s 21 percent. And that is our final answer.

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