Estimate the overall change of entropy for this system.

  • Thread starter mlostrac
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  • #1
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Homework Statement


A piece of metal at 80°C is placed into 1.2 litres of water at 72°C. This thermally
isolated system reaches a final temperature of 76°C. Estimate the overall change
of entropy for this system.



Homework Equations


(delta)S = Q/T


The Attempt at a Solution



1) Found Q of water (**C = degrees celsius):

Q = mc(delta)T = 1.2kg (1.00 kcal/kg-C) (4C) = 4.8 kcal

2) Found the Average temp of the water:

T = (76 + 72)/2 = 74C = 347 K

3) Change in Entropy, (delta)S = Q/T = 4.8kcal/347K = .0138 kcal/K

Question: Is the Q (heat) of the metal just -4.8kcal?

I assumed so, and found the change in entropy for the metal to be .0136 kcal/K

Therefore giving me an overall entropy for the entire system of:
(delta)S = .0138 - .0136 = .0002 kcal/K or 0.2 cal/K

This number looks to small to me though. Any help would be appreciated!
 
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Answers and Replies

  • #2
799
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It looks fine. The heat is small so the change in entropy should be small.
 
  • #3
134
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I always have to think three times when working with entropy, but I'm almost sure this is the wrong method. (Others, please correct me if I am wrong).

The formula [itex]\Delta S = \Delta Q / T[/itex] is only valid for isothermal processes (constant temperature).

So what you need to do is imagine another thermodynamic process with the same initial and final states, but that uses an isothermal process for the heat flow.

Since entropy is a state variable, its change only depends on the initial and final states regardless of the "path" used to get between them. So the entropy change in the new imaginary process is the same as the real irreversible one.
 
  • #4
799
0


I always have to think three times when working with entropy, but I'm almost sure this is the wrong method. (Others, please correct me if I am wrong).

The formula [itex]\Delta S = \Delta Q / T[/itex] is only valid for isothermal processes (constant temperature).

So what you need to do is imagine another thermodynamic process with the same initial and final states, but that uses an isothermal process for the heat flow.

Since entropy is a state variable, its change only depends on the initial and final states regardless of the "path" used to get between them. So the entropy change in the new imaginary process is the same as the real irreversible one.

Yes, this method is wrong. But it can be approximately true (Oh, how illogical!) in this problem particularly. The initial and final temperatures are so close to each other, so doing as mlostrac did is a good estimation. Estimation can never be completely precise, right?

Just my personal opinions: Though there may be some other method that can precisely solve the problem, I think I like this estimation more. It's simply more simple and more "physics". Scientists always simplify the complex (and complicate the simple at the same time :D).
 
  • #5
134
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Agreed, as long as one is aware of the assumptions made. You are right in emphasizing the small temperature differential in the problem. Thanks!
 
  • #6
Andrew Mason
Science Advisor
Homework Helper
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The accurate calculation requires calculus.

For the water, [itex]dQ = c_wm_wdT[/itex]. For the metal, [itex]dQ = c_mm_mdT[/itex]

One can see from the fact that the heat gained by the water is equal to the heat lost by the metal and that the temperature changes are equal and opposite, so that [itex]c_mm_m = c_wm_w[/itex]

So the change in entropy is:

[tex]\Delta S_{total} = \Delta S_w + \Delta S_m = \int_{T_{iw}}^{T_f} dQ/T + \int_{T_{im}}^{T_f} dQ/T= c_wm_w \int_{349}^{345} dT/T + c_wm_w \int_{349}^{353} dT/T[/tex]

[tex]\Delta S_{total} = c_wm_w (\ln\left(\frac{349}{345}\right) + \ln\left(\frac{349}{353}\right))[/tex]

[tex]\Delta S_{total} = 1 \times 1200 \times (.01153 - .01140) =.158 Cal/g K[/tex]

AM
 

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