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course Phy 122
I'm not totally sure if this is how you were wanting us to submit our questions, so I won't bombard you with questions right now. I did have a bit of a question in chapter 15 about entropy of water and how to find it, especially in problem 36. I'm having difficulty figuring out how to use the specific heat of water and mass to find the change in entropy (J/K). I've looked through the book, but it's still eluding me. Thanks for your time.
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The change in entropy is given by
`dS = `dQ / T.
Your book uses Q / T, but since the internal energy of the water is changing, it is more appropriate in my opinion to use `dQ to designate the change.
Using the specific heat, mass and temperature change of the water you easily find that for a kg of water changing from 273 to 373 K, `dQ is about 419 000 Joules.
So if you use the right value of T, the calculation of `dS is very straightforward.
However T changes during this process from 273 K to 373 K (note that as with all thermodynamic calculations we use absolute temperatures). We can't use all those temperatures (actually we can, using integration, but this is not a calculsus-based course). We have to decide on the most appropriate temperature to use.
In the absence of any other obvious choice, we will choose the mean temperature. Halfway between 273 and 373 K is 323 K. Using this for T we get
`dS = 419 000 J / 323 K = 1300 J / K approximately.
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We could obtain a more accurate estimate by dividing the 100 K interval between 273 and 373 K into smaller subintervals. Using subintervals of 10 K, in each of which `dQ is about 41 900 J, the first subinterval will run from 273 K to 283 K, with a midpoint temperature of 278 K, so we obtain
`dS_1 = 41 900 J / (278 K).
Subsequent 10 K intervals have mean temperatures of 288 K, 298 K, 308 K, ..., 368 K, giving us entropy changes
`dS_2 = 41 900 J / (288 K)
`dS_3 = 41 900 J / (298 K)
`dS_4 = 41 900 J / (308 K)
...
`dS_10 = 41 900 J / (368 K)
Calculating these entropy changes and adding them, we get a refined estimate, which differs from the original be a few J / K.
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We could divide the 100 K interval into even smaller intervals, and this would be fairly easy to do using a spreadsheet or computer algebra software. The changes in our estimates would get smaller and smaller.
Or we could just use calculus and integrate dQ / T between 273 K and 373 K. This is not a calculus based course, but the integral can be stated easily. dQ = 4190 J * dT (for this sample of water you just multiply 4190 J by the change in temperature to get the change in internal energy). So our integral is just
integral ( dQ / T) = integral ( 4190 J * dT / T, T form 273 K to 373 K) = 4190 J * ln(373 / 273) = 1306 J, to four significant figures.
We will note that the specific heat of water changes between 0 C and 100 C, so the assumed 4190 J/ (kg C) is itself an approximation whose percent error on this temperature range is in the same ballpark at the percent error we got by using the midpoint temperature in our first estimate.
Had the ratio of absolute temperatures been greater, the error in assuming the midpoint temperature would have been more significant.
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Final note:
Questions need to be self-contained. To make your quesiton self-contained you would have needed to provide a brief synopsis of the information given in the problem. For example, your
"especially in problem 36. I'm having difficulty "
would have satisfied this condition with the addition of just a little information, e.g.,
"especially in problem 36 (1 kg of water from 273 to 373 K). I'm having difficulty "
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