A Carnot refrigerator operates between energy reservoirs at 0℃ an...

Question

A Carnot refrigerator operates between energy reservoirs at 0℃ and 250℃. A 2.4-cm-diameter, 50-cm-long copper bar connects the two energy reservoirs. At what rate, in W, must work be done on the refrigerator to remove heat from the cold reservoir at the same rate that it arrives through the copper bar?

Accepted Answer

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. Data center servers are cooled by a cooling system. The cooling system works between a cold room at five °C and a hot room at 50 °C. A 40 centimeter long steel bar with a diameter of 3.3 centimeters connects the rooms. The cooling system extracts heat from the cold room at the same rate that it is transferred through the steel bar. Compute the required power input in watts or the cooling mechanism to achieve this heat extraction rate. The cooling system operates on the basis of a carno cycle and the thermal conductivity of steel is 45 watts per meter multiplied by Calvin. OK. So our end goal is to find the required power input in watts for the cooling mechanism to achieve the heat extraction rate. So we're given some multiple choice answers. They're all in the same units of watts. Let's read them off to see what our final answer might be. A is 0.6 B is 0.4 C is 0.9 and D is 0.7. So first off, let us recall the equation for the heat conduction rate or the flow of heat through a steel bar. Let's call it equation one. And that states that the heat energy of the steel bar, let's call it Q subscript B to keep it simple divided by the change in time is equal to K multiplied by a multiplied by delta capital T where capital T represents the change in temperature or the temperature difference between the hot and cold reservoir divided by L and L is the distance between the hot and cold reservoir. And lower case K is the thermal conductivity of steel and A is the cross sectional area. OK. And let's note really fast here that delta T is equal to the temperature of the hot reservoir minus the temperature of the cold reservoir. And that the cross sectional area A is equal to pi multiplied by radius squared, which is the same as saying pi multiplied by diameter divided by two squared. So we can now rewrite equation one as QB divided by delta lowercase T is equal to K multiplied by pi multiplied by D divided by two squared multiplied by th minus TC divided by L. So we can simplify this equation. And let's call it equation too. We can simplify it to look like so or look like such So it's QB divided by delta, lowercase T is equal to K multiplied by pi multiplied by D squared multiplied by th minus TC, all divided by four L. OK. So now we must recall that for a Carno refrigerator, the coefficient of performance K is, so the equation for K can be written as K is equal to TC divided by T H minus TC. And that A equals QC divided by the work input, which is equal to like is the, it's the same as saying what you get is what you pay as simple as that. OK. So recall that the work done in a cycle is related to the transfer of heat by combining the expressions for the coefficient of performance as and you can write it as QC is equal to the work input is equal to TC divided by TH minus TC. Thus QC divided by delta lowercase T is equal to TC divided by TH minus TC multiplied by the work input divided by delta. Lower case T let's call this equation three. So now we need to set equations three and two equal to each other. So let's do that. So TC divided by th minus TC multiplied by the work input divided by delta lowercase T is equal to K multiplied by pi multiplied by D squared multiplied by TH minus TC divided by four L. So we can simplify a little bit and we can write that the work input divided by delta T delta, lowercase T is equal to th minus TC, divided by TC, multiplied by K multiplied by Pi multiplied by D square multiplied by T H minus TC, divided by four L. At this stage, we could plug in our known variables and note that we have to convert degrees Celsius to degrees Kelvin. So let's do that. So the work input divided by delta, lower case T is equal to. So the temperature of the hot reservoir was 50 degrees C. But we need to convert that to degrees Kelvin. So when we do that, we get 323 Kelvin. So as a quick note, 50 plus equals 300 in Kelvin. And you can follow that method to solve or any conversions from degrees Celsius to Kelvin. OK, minus the temperature of the cold reservoir. Which when you convert that to Kelvin, it's 278 Kelvin divided by 278 Alvin multiplied by watts per meter multiplied by Alvin multiplied by pi multiplied by 3.3 multiplied by to the power of negative 2 m squared. And this is the diameter, but we had to convert from centimeters to meters multiplied by and we're running out of room. So I'll write it down below. So multiplied by Calvin minus Calvin divided by or multiplied and the value of L was 40 centimeters of the length of the rod or steel bar, I should say was 40 centimeters when we need to convert centimeters to meters. So it's gonna be 40 multiplied by 10 to the power of m. So when we plug all of this into a calculator, we should get 0.7 watts, which is our final answer. Fantastic. We did it. So let's go look at our multiple choice answer to see what the correct answer should be. So the correct letter is D 0.7 watts. Thank you so much for watching. Hopefully that helps and I can't wait to see you in the next video. Bye.

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