An ideal refrigerator utilizes a Carnot cycle operating between 0℃ and 25℃. To turn 10 kg of liquid water at 0℃ into 10 kg of ice at 0℃, (a) how much heat is exhausted into the room and (b) how much energy must be supplied to the refrigerator?

Video transcript

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A cooling unit operates with a carno cycle between negative 15 degrees Celsius and 30 degrees Celsius find how much I heat is dissipated into the surrounding environment. And I I energy must be supplied to the cooling unit in order to bring 18 kg of a liquid substance from 30 degrees Celsius to its solidification point at five degrees Celsius. It's latent heat of fusion. LF is 2.34 multiplied by 10 to the power of five joules per kilogram. OK. So that's our angle. We're trying to find two separate answers. We're trying to find an answer for part I and an answer for part I, I, so we're given some multiple choice answers and both the answers. For part I and I, I are all in the same units of jewels. So let's read them off to see what our final answer pair might be. A is 4.94 multiplied by 10 to the power of six and 0.73 multiplied by 10 to the power of six. B is 2.13 multiplied by 10 to the power of five and 1.54 multiplied by 10 to the power of five C is 5. multiplied by 10 to the power of six and 0.14 multiplied by 10 to the power of six. And D is 4.94 multiplied by the 10, multiplied by 10 to the power of five and 1.16 multiplied by to the power of five. OK. So first off, let us recall and use the equation for the thermal efficiency of a heat engine operating between the temperature for the hot reservoir and the temperature for the cold reservoir of a Carno engine. So, Ada represents the Carno cycle is equal to one minus the temperature of the cold reservoir divided by the temperature of the hot reservoir. And let's call this equation one and equation two. So we need to set equation one equal to equation two. But equation two, we need to consider the heat transfer. So a is equal to one minus the heat transfer of the cold reservoir divided by the heat transfer of the hot reservoir. So we need to set equations one and two equal to each other. And then we need to rearrange to solve for qh which is the heat transfer for the hot reservoir. So let's get going on that. So one minus the temperature of the cold reservoir divided by the temperature of the hot reservoir is equal to one minus the heat transfer of the cold reservoir divided by the heat transfer of the hot reservoir. So note that the ones cancel out when you move them over. So the temperature of the cold reservoir divided by the temperature of the hot reservoir is equal to the temperature of the cold. I mean the heat transfer of the cold reservoir divided by the heat transfer of the hot reservoir. OK. So then when we rearrange the sol for qh, all by itself, we get QH as equal to QC multiplied by th divided by TC. And let's call this equation three. OK. So the heat that is released when 18 kg of a mass is frozen is QC is equal to the mass multiplied by the latent heat of fusion. LF. OK. So now we need to plug in our known variables to sol for QC. So let's do that. So the mass is 18 kg and the latent heat of fusion is 2. multiplied by 10 to the power of five Jews per kilogram. So when we plug that into a calculator, we should get that QC is equal to 4.21 multiplied by 10 to the power of six Jews. So now we need to take this value and substitute it back into equations three to sol for Qh. So let's do that. So Qh is equal to 4. multiplied by 10 to the power of six jewels multiplied by the temperature of the hot reservoir divided by the temperature of the cold reservoir. But we need to pause really quick here. We need to quickly convert our TH and TC values from degrees Celsius to degrees CAL to from degrees Celsius to Kelvin. So note that the temperature for the hot reservoir is degrees C and the temperature for the cold reservoir is negative 15 degrees C. So to convert it, all we have to do is just take our degree Celsius and just add 273 to convert to Kelvin. So when we add that in the calculator, we'll get 303 Calvin doing the exact same thing for our cold reservoir, negative 15 plus 2, 73 equals 258 Calvin. So now let's plug it back into our QH equation. So it'd be 303 Calvin divided by Calvin is equal to 4.94 multiplied by 10 to the power of six jewels. And that is how we find part I nice going everyone, we did it. So I equals 4.94 multiplied by 10 to the power of six jewels. So to find out the energy supplied to the cooling unit or the amount of work done to a our work done to freeze 18 kg of water. We will need to use the following relation. So let's recall that, that the heat transfer for the hot reservoir is equal to the work plus the heat transfer of the cold reservoir. So we need to rearrange this equation to solve for the work. So when we do that, we get that work is equal to the heat transfer for the hot reservoir minus the heat transfer of the cold reservoir. So now we could plug in our knownn variables to solve for w the work. So work is equal to the hot, the heat transfer for the hot reservoir, which was 4.94 multiplied by 10 to the power of six Jews minus the heat transfer for the cold reservoir, which we determined it to be 4.21 multiplied by 10 to the power of six jewels, which is equal two. When you plug that into a calculator, 0.73 multiplied by 10 to the power of six jewels. And that is our answer for part I I and we did it. So let's go back to the top and see what our correct multiple choice letter is. So the correct answer is the letter A I is 4. multiplied by 10 to the power of six jewels. And I I is 0.73 multiplied by 10 to the power of six jewels. Thank you so much for watching. Hopefully that helps and I can't wait to see you in the next video. Bye.