A freezer with a coefficient of performance 30% that of a Carnot refrigerator keeps the inside temperature at -22℃ in a 25℃ room. 3.0 L of water at 20℃ are placed in the freezer. How long does it take for the water to freeze if the freezer's compressor does work at the rate of 200 W while the water is freezing?

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Accepted Answer

Hey, everyone. So this problem is dealing with heat engines. Let's see what it's asking us. We have a refrigerator operating at an efficiency of 60% of the maximum possible value for its temperature range. The inside of this refrigerator is maintained at a temperature of five °C. While the outside environment is at 25 °C. 4 L of ethanol at 25 °C are placed in the freezer section of this refrigerator. If the refrigerator consumes 250 watts of electrical power, we're asked how long it would take to cool the ethanol from an initial temperature of 25 °C to a final temperature of negative 10 °C using the same engine. Ethanol has a specific heat capacity of 2440 joules per kilogram, Kelvin and a density of 0.789 g per milliliter. Our multiple choice answers are a 163.8 seconds. B 1.292 times 10 to the negative three seconds. C 129.2 seconds or D 1.638 times 10 to the negative three seconds. So the first step to this problem is finding out what the maximum efficiency is for this refrigerator so that we can find what it, what its actual operating efficiency is. So we can recall that for the Carno system or that max efficiency K is going to be equal to T sub C divided by T sub H minus T sub C. And so that's the maximum. So our K value is going to be 0.6 or 60% of that. So this K term um can also be referred to as the coefficient of performance, sometimes written as cop. And so we can plug in our known values. We recall that we need to be working in Calvin for this and Calvin to Celsius is just Celsius plus 273. And so five °C becomes write this down here. A me five °C becomes 278 Kelvin and then 25 °C 298 Kelvin for th minus TC. Again, 278 Kelvin, you plug that in and we get our coefficient of performance of 8.34. We could also recall that K is equal to QC divided by WN for refrigerators. And so that's the heat extracted is equal to this coefficient performance. When we're rearranging it, that heat is equal to this coefficient of performance multiplied by the work in. And so we have 8.34 multiplied by our work. Now we're given power in watts and so watts are jewels per second. And so in one second, we have a work of 250 jewels. And so that gives us 2085 jewels. And in time space, if we want to look at a rate per second, that gives us 2085 watts, because in the end, they are asking us how long it would take. So now we need to determine the heat required to be removed from the ethanol to cool it from 25 °C to negative 10 °C. So that heat transfer equation you can recall is, and I'll just do Q sub E for ethanol is equal to MC delta T, we weren't given mass but we were given density and volume. And so we can recall that mass is equal to density multiplied by volume. And then see our specific heat capacity and delta T. So we have all of those terms where our density was given to us as 0.789. And that was in grams per milliliter. We need to keep everything in standard units. So that can be rewritten as kilograms per liter multiplied by our volume. 4 L multiplied by our specific heat capacity, 2440 jewels per kilogram Calvin and then multiplied by our delta T. So our final temperature negative 10 degrees Celsius minus 25 °C, it's going to be negative 35 °C. And then recalling that one °C is equal to one degree Kelvin as far as the um measure of the unit, negative 35 °C is the same as negative 35 degrees Kelvin. When we're looking at deltas, you can plug all of that into our calculator and get negative two point 695 times 10 to the five jewels. And then the last step will be to figure out how long it will take. So our time, our time is going to be our total heat that we need to be removed, divided by the rate at which we can remove heat. And so that's two point I'm sorry, six 95 times 10 to the fifth jewels divided by 2085 jewels per second. And that equals 129.2 seconds. Looking at our multiple choice answers that aligns with answer choice C so C is the correct answer for this problem. That's all we have for this one. We'll see you in the next video.

Verified Solution

Key Concepts

Coefficient of Performance (COP)

Phase Change and Latent Heat

Energy Transfer and Power