Determine the work available in a 2.2-kg block of copper at 49...

Question

Determine the work available in a 2.2-kg block of copper at 490 K if the surroundings are at 290 K. Use results of Problem 57.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem, find out how much maximum work can be obtained from pulling down a 5.0 kg iron block that is initially heated up to a temperature of 800 Kelvin and is later brought to ambient conditions with an average temperature of around 295. Kelvin note that the specific heat capacity of iron is 450 joules per kilogram multiplied by Kelvin. Awesome. So it appears that our final answer that we're ultimately trying to solve for is we're trying to figure out what the maximum work done is for this particular problem given the provided conditions. Awesome. So with that in mind, let's read off our multiple choice answers to see what our final answer might be. And let us note that they're all in the same units of jewels. So A is 4.8 multiplied by 10 to the power of P five B is 5.9 multiplied by 10 to the power of Phi C is 6.6 multiplied by 10 to the power of Phi and D is 7.2 multiplied by 10 to the power of five. OK. So first off, let us write down all of our known variables. So let us note that the mass which we're gonna call m of the iron block is equal to 5.0 kg. Awesome. We also know that the initial temperature of the iron block and we're gonna call this T I. We know that T I is equal to 800 Kelvin. We also know that the final temperature or the ambient temperature is given to us as TF and TF is equal to 295 Kelvin. And we also know that the specific heat capacity of iron, which we're gonna call this value C is equal to 450 its units are Jos Joel per kilogram multiplied by Kelvin. Awesome. So now at this point, we must solve for the total energy stored in the iron block. And we need to note and recall that the total energy stored in the iron block is found from the heat flow that is initially raised its temperature above the surroundings temperature. So therefore, we could, at this point, we should be able to recall and use the following equation to help us solve with the total energy which states that Q is equal to M multiplied by C multiplied by delta T. And let us recall and note that for this case, which is important to me, the case of this problem is slightly different. Delta T is equal to T I minus TF, which means that this is equal to 800 Kelvin minus 295 Alvin, which means that delta T is equal to 505 Kelvin when we perform that calculation. Awesome. So with that in mind, moving right along here, we can now plug in our known variables to sol for the numerical value of Q. So we will find that Q is equal to 5.0 kg. And once again, all we, all we're doing is we're plugging in our known variables to solve for Q. So we know that M is 5.0 kg. We know that C is 450 or 100 and 50 jewels per kilogram multiplied by Calvin multiplied by delta T which is 505 Kelvin, which is equal to let me put that into our calculator. 1136, 250 jewels. So it's 1 million, 136,250 jewels. So now we must solve for the enthropy change of the iron block by recalling and using the following equation that states that delta s capitalized. So the change in enthropy means delta S means change in enthropy and I denotes iron. So the change in entropy of the iron block is equal to the integral evaluated at T I to TF of M multiplied by C multiplied by D capital T all divided by capital T which will we perform the inte goal or the integration. To be precise, we will get that delta SI is equal to M multiplied by C multiplied by the natural log of TF divided by T I. So now we can plug in all of our known variables to software Delta Si. So that means that we will find that delta si is equal to 5.0 kg multiplied by 450 jewels per kilogram multiplied by Calvin multiplied by the natural log of TF which is 295 Alvin divided by 800 Calvin. So when we plug that into our calculator or this equation into our calculator, we will get negative 2244 0.68 when we round to two decimal places and its units will be jewels per Kelvin. So now we must solve for the enthropy change of the surroundings by recalling and using the following equation which states that delta Ss or delta SS is the enthropy change of the surroundings. And this is equal to Q divided by the temperature of the surroundings which are called TS Awesome. And this is equal to 1,136,250. So now we're just plugging in our known variables to solve for delta ss divided by 295 Calvin B is the temperature of the surroundings is told that is gonna be 295 Calvin be it's being later brought to the ambient condition. So that means the what the air surrounding the iron block. OK. So moving right along. So when we plug it into our calculator, we will get 3000 851.695 will be round to three decimal places and its units will be jewels per Kelvin. OK. So now at this point, we can solve for the total entropy change by recalling and using the following equation, they're gonna call the total enthropy change delta ST and this is equal to delta SI plus delta ss which when we add these two together, we will get 1607 0.015 when we round to three decimal places and its units will be jewels per Kelvin. Awesome. So that's when we add the value for delta SI plus delta ss Awesome. So now at this point, we need to solve for the energy that becomes unavailable to do useful work, which we can recall and use the following equation to help us solve for this particular value, which we're gonna call this capital E subscript capital L to denote the energy lost. And the energy loss is equal to delta TF multiplied by delta S which is equal to the plug in or known values delta TF in this case is 295 Alvin and delta S in this case, which to be specific, this is delta S ST. So we're gonna be plugging in our value for delta S ST which is 1000 607 0.015 and its units are jewels per Kelvin. So when we plug that into our calculator, we will get or seven four, zero 69.43 jewels. So we're gonna be getting 474,069 0.43 jewels. So now at this point, we can solve for the maximum work available, which is the final answer that we're ultimately trying to solve for. So we're gonna call this value W subscript capital A. So this is the maximum work available and this is equal to Q minus el awesome, which is equal to, to quickly recap 1,136,250 tools minus our value, which we just determined to be 474,069 0.43 jewels. So point or three jewels. So when we plug that into our calculator, we will find that the final answer rounded to one decimal place, written in scientific notation is approximately equal to 6.6 multiplied by 10 to the power of five. And its units will be Jews and that's it we've solved for this problem. Hooray, we did it. So that means we can look at our multiple choice answers to see what our final answer is that matches up with the answer that we just found together. So looking at our multiple choice answers, the correct answer has to be the letter C 6.6 multiplied by 10 to the power of five Jews. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Determine the work available in a 2.2-kg block of copper at 490 K if the surroundings are at 290 K. Use results of Problem 57.

Key Concepts

Thermodynamics

Heat Transfer

Carnot Efficiency

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