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Ch 20: The Second Law of Thermodynamics

Chapter 20, Problem 20

A certain brand of freezer is advertised to use 730 kW•h of energy per year. (a) Assuming the freezer operates for 5 hours each day, how much power does it require while operating?

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Hey everyone welcome back in this problem. We have a heat pump that consumes 4000 kilowatt hours of electrical energy annually and runs on average for 12 hours a day. Were asked how much power does it need to run? Alright, let's get started. We're thinking about power. Okay. And we're also told some kilowatt hours so that we're thinking work. Okay. So we're thinking about a relationship between work and power. Let's recall that work is equal to power times time. T Alright, we're given the work 4000 kilowatt hours. Okay, let's go ahead and convert this just into watt hours so that we have standard units. So this is going to be 4000 Times 10 to the three. What hours? And on the right hand side we have P. The power, which is what we're looking for times time. Alright, let's think about this. We're talking about how much power it needs to run for a year. Okay? We're talking annually And we're told that it runs an average of 12 hours each day. So we're talking about time. Well, we're gonna need to run for 12 hours per day, Times 365 days. Alright, so the unit of day we will cancel. Okay, we can divide so that we get power P is equal to Times 10 to the three What hours Divided by 12 times 365 is going to be 4380 hours. Okay. The unit of our will cancel when we divide and we are left with the power of 913 watts. Okay. And so the answer to this one is going to be see. The power to run this heat pump for the year is watt. Thanks everyone for watching. See you in the next video.
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