A certain brand of freezer is advertised to use 730 kW•h of energ...

Question

A certain brand of freezer is advertised to use 730 kW•h of energy per year. (c) What is the theoretical maximum amount of ice this freezer could make in an hour, starting with water at 20.0°C?

Accepted Answer

Hey everyone in this problem, we're given the coefficient of performance of an ice making machine is five. And the power input is 800 watts. Were asked how much ice can this ice machine theoretically make in one day, If we start with water at 30°C. Alright, now we're asked how much ice can this machine theoretically make. Okay. So we're talking about how much ice we're making. We're thinking about finding a mass. Okay? So we want to find a mass. Let's look at what we're given. Okay? We're given the coefficient coefficient of performance and let's recall that the coefficient of performance. K. Can be written as the absolute value of Q. C. Divided by the absolute value of w the work. Okay? So the quantity of heat over the work. Now, if we're thinking about finding a mass, recall that we can relate Q. C. The quantity of heat to mass. Okay? So, if we can find QC, we can use our formula to find the mass. All right. So, we know, okay, what about W Do we know the work? We don't know the work, but we do know that the power is 800 watts and we can relate work and power we call that work. It's going to be equal to power times time. Well, in this case the power is 800 watts. Okay? And the time is one day. Okay, so one day We want to convert this into our standard unit. So instead of having days we're gonna multiply by hours per day. And then we're gonna multiply by 3600 seconds per hour. So the unit of day cancels and then the unit of our cancels and we're left with seconds and we're going to get a work of 6.912 times To the seven jewels. Good. Alright so now we have our work we have K. So we can find QC. Okay so K. S. Five coefficient of performance Q. C. We want to find so we can relate to the mass In our work 6.912 Times to the seven jewels. And so the absolute value of QC is going to be equal to when we multiply 3. 56 times 10 to the eight yours. Alright so we found this absolute value of Q. C. We've been saying that we can relate this to the mass. Let's go ahead and do that. Okay but we know in this case the absolute value of Q. C. Is gonna be made up of two things. Okay so the first quantity of heat and the second quantity of heat. Okay the first one is going to be that quantity of heat related to the heat lost by the ice when it's cooling. And the second one is gonna be related to the heat required in order to freeze the ice. Okay. Alright so let's take a look at what these look like. Okay the first one recall that we can write the mass M. Times C. W. Okay. The heat capacity of water. The absolute value of delta T. The change in temperature plus M. Oh a latent heat of fusion and our massive. Okay so these are two quantities on the left hand side. It's just going to be our QC value. Alright so we have Q. C. We want to find em we want to find em we know C. W. And L. F. These are standard values that we can look up in a table in our text worker that our professor provided. Okay so it's just delta T. Okay well what's delta T. Going to be does A. T. Going to be T final minus T. Initial? Okay, well we know that water freezes. So in order to make ice we need water to freeze. Okay, we know that it freezes at zero degrees Celsius. Okay, so we're gonna have zero degrees Celsius plus to 73.15 Calvin. Okay to convert this into Calvin. And then we're going to subtract the initial temperature which we're told is 30 degrees Celsius. 30 plus to 73.15 Calvin. And this gives us a delta t. Negative 30 Calvin. All right, let's go ahead now and substitute what we know. So QC 3.456 times 10 to the eight jewels is equal to let's factor M. The mass K. Times C. W. Which is 190 jewels per kilogram Calvin. Okay. Times Delta T. And we want the absolute value of Delta T. So that's gonna be positive 30 Calvin. Okay. Plus the mass M. Times L. F. Which again is a standard value that you can look up. 334 times 10 to the three. The unit jewel per kilogram. Okay So left hand side, 3.456 times 10. Let's just give ourselves some more room here, Times 10 to the eight jewels is equal to again. So let's go ahead and factor that M. Okay? And do the multiplication, we're gonna have 1. Times 10 to the five jewels per kilogram. Okay, when we multiply the unit of Calvin will cancel. Oops. Not a bracket yet. Plus the second term 334 times 10 to the three jewels per kilogram. Okay. And you can see now in this bracket we have the same unit jewel per kilogram jewel per kilogram so we can add them up. Okay. All right. And we're gonna want to divide by everything on the right hand side to get em by itself we're gonna be left with m is equal to 3. times 10 to the eight jewels divided by when we add these together we get 4. times 10 to the five joules per kilogram. Okay, the unit of jewel will cancel will be left with the unit of kilogram which is the unit of mass we want. So our units check out when we get 751.8 kg. Okay. And so the amount of ice we can make in one day is 751.8 kg back to the top. We're going to look at our answer choices and we're going to see that we have answer. D. Okay. 751.8 kg. Thanks everyone for watching. I hope this video helped see you in the next one.

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