A 75 g ice cube at 0℃ is placed on a very large table at 20℃. You...

Question

A 75 g ice cube at 0℃ is placed on a very large table at 20℃. You can assume that the temperature of the table does not change. As the ice cube melts and then comes to thermal equilibrium, what are the entropy changes of (a) the water, (b) the table, and (c) the universe?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the buying 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 150 g piece of solid wax at °C is placed on an 80 °C metal plate, determine the entropy changes of I the molten wax. I I, the metal plate and II I, the universe assuming that the temperature of the metal plate remains constant as the wax melts and subsequently reaches thermal equilibrium. Take the latent heat of fusion of wax to B 2.0 multiplied by 10 to the power of five Jews per kilogram specific heat to be 2100 Jews per kilogram multiplied by Kelvin and its melting temperature to be °C. OK. So our end goal is to find the entropy changes for three different conditions. And the first condition is the molten wax. The second condition is the metal plate and the third condition is the universe. So we're given some multiple choice answers and for parts II I and II I and they're all in the same units of jewels per Kelvin. So let's read them off to see what our final answer set might be. A is 1.4 multiplied by 10 to the power of three and negative 1.8 multiplied by 10 to the power of two and 16. B is 1.4 multiplied by 10 to the power of two and negative 1.2 multiplied by 10 to the power of three and 18 C is 1.3 multiplied by 10 to the power of three and negative 1.8 multiplied by 10 to the power of three and 14. And D is 1.3 multiplied by 10 to the power of two and negative 1.2 multiplied by 10 to the power of two and 14. OK. So first off, let us note that the wax melts at a constant temperature which is an isothermal process and then the temperature increases, which is an isobaric process. So the pressure remains constant. Now, we need to recall and use the equation to find the entropy change in a melting process. And that states that delta S one is equal to Q divided by T one where Q is equal to the mass of the wax multiplied by the la multiplied by LF or LF is the latent heat of fusion of wax. And T one is the temperature of solid wax. So using this information, we can rewrite that delta S one is equal to the mass of wax multiplied by the latent heat of fusion of wax divided by T one. OK. So the anthropology change in an isobaric process is written as delta S two is equal to the mass of wax multiplied by CP multiplied by Ln, which is the natural log of T two divided by T one where CP is the specific heat of molten wax which in this case is 2100 Jews per kilogram multiplied by Kelvin and T two is the temperature of them of the metal plate. OK. So the entropy change of the molten wax is equal to the sum of the entropy change in each phase of the process can be written as delta S M W. So the entropy of the molten wax wax, which I'm gonna denoted as MW is equal to delta S one plus delta S two. OK. So now we can plug in our values for delta S one and delta S two. And so for delta SMW. So delta SMW is equal to MW. And let's be very clear that it's delta S capital M capital W for molten wax. And then lower case M subscript WMW is the mass of the wax multiplied by LF divided by T one plus MW multiplied by CP multiplied by Ln of T two divided by T one. So delta MW or delta SMW, I should say delta S capital M capital W so is equal to MW multiplied by LF divided by T one plus CP multiplied by LNT two divided by T one. So at this stage, we can plug in all of our known variables to solve for the entropy of the molten wax. So let's do that. So delta S MW, so the entropy of the molten wax is equal to 0.15 kg. Usually you have to convert 150 g to kilograms. Multiplied by 2. multiplied by 10 to the power of five joules per kilogram divided by 40 which in this case, rt one values 40 °C, but we need to convert to Kelvin. So 40 plus equals 313 Kelvin. So for simplicity's sake, let's just write 313 Calvin. So we don't get confused in our calculations. 313 Kelvin plus CP which was 2100 jewels per kilogram multiplied by Kelvin multiplied by Allen, of which we have to take our T two value which was °C and convert it to Kelvin. So 80 plus 273 is 353 Kelvin divided by our T one temperature which was 313 Kelvin. So when we plug it into a calculator, our entropy for the molten wax is equal to 0.7. So 133 0. jewels per Calvin. But when we write it in scientific notation, we get one point three multiplied by 10 to the power of two jewels per Kelvin, which is our first answer or part I, OK. So, so for I I note that the heat is transferred from the metal plate at a constant temperature. The entropy change of the plate is the ratio of the heat required to melt the solid wax and increase the temperature of the molten wax to the temperature of the surrounding area. So T two, the temperature of the metal plate, we need to remember that it's important. So considering what we just talked about delta sp, so delta sp is equal to Q divided by T two and Q in this case is the heat flowed out of the plate which is equal to the heat required to set or I should say. So the heat flow that the, so the heat that flows out of the plate, which is equal to the heat required to melt the solid wax and increase the temperature of the molten wax. OK. So we need to know what Q is. So Q is equal to negative MW multiplied by LF plus MW multiplied by CP multiplied by delta T. So it's negative to all of that inside of the parentheses. So delta sp where delta SP is the metal plate. So the entropy of the plate of the metal plate is equal to negative parentheses. MW multiplied by LF plus MW multiplied by CP multiplied by delta T parentheses, divided by delta, sorry M divided by T two. So delta SP is equal to negative parentheses. MW multiplied by LF plus MW multiplied by CP multiplied by T two minus T one divided by T two. OK. So at this stage, we can plug in all of our known variables and solve for the entropy of the metal plate. So let's do that. So the mass of the wax is 0. kg multiplied by 2.0 multiplied by 10 to the power of five jewels per kilogram plus the mass of the wax 10.15 kg multiplied by 1 jewels per kilogram multiplied by Calvin multiplied by 353 Kelvin minus Calvin fantastic, all divided by 353 albums. OK? So when you plug into a calculator, the entropy of the metal plate is equal to negative one point two, multiplied by 10 to the power of two jewels per Kelvin. And note that this is in scientific notation form because when you type into a calculator and if it's not set scientific notation, you'll get an answer that's slightly different. But this is the correct answer since it matches the multiple choice answers that are provided to us in the prom itself. So the entropy of change for the universe. So this is the sol for part II I so to solve for II I, the anthropic change of the universe. Let's note that delta su the anthropic change for the universe is equal to delta of the entropy of the molten wax plus the entropy of the metal plate. So when we plug in our known values, so delta SU is equal to one point 34 multiplied by 10 to the power of two jewels per Calvin plus negative 1.2 multiplied by 10 to the power of two jewels per Calvin. Delta SU the entropy of change of the universe when you plug into a calculator is 14 jewels per Kelvin and we did it. So let's go look at our multiple choice answers to see what the correct answer should be. So the correct answer is the letter D I is 1.3 multiplied by 10 to the power two Jews per Kelvin. I I is negative 1.2 multiplied by 10 to the power two Jews per Kelvin and II I is 14 Jews per Kelvin. Thank you so much for watching. Hopefully that helped and I can wait to see you next video. Bye.

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A 75 g ice cube at 0℃ is placed on a very large table at 20℃. You can assume that the temperature of the table does not change. As the ice cube melts and then comes to thermal equilibrium, what are the entropy changes of (a) the water, (b) the table, and (c) the universe?

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