Entropy Calc: 1kg Water Heated from 20C to 80C

In summary: I get that as well. The 1074 J/K is the answer for 6-2(b) which is the same problem with the exception that the water is heated from 20C to 80C in a constant pressure process.
  • #1
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Homework Statement


Find the entropy change of the system.

1kg of water is heated reversibly by an electric heating coil from 20C to 80C([tex]c_{p}=4.18 X 10^{3}Jkg^{-1}K^{-1}[/tex]).

Homework Equations



[tex]dU=TdS-PdV[/tex]

The Attempt at a Solution



dV=0
[tex]dS=\frac{dU}{T}[/tex]....

The answer is 1074J/K

I think I'm stuck on how to handle the heat capacity. If someone could just point me in the right direction so I know what to go over and learn to complete this problem that would be very helpful.
 
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  • #2
Since no work is done, dU is the same as dQ.

Use the formula relating heat Q, mass, Cp, and temperature change.
 
  • #3
Redbelly98 said:
Since no work is done, dU is the same as dQ.

Use the formula relating heat Q, mass, Cp, and temperature change.

Ditto.

DU = DQ...

So, what equation do you have for DQ that relates to heat capacity?
 
  • #4
Thanks for the replies, it may take me awhile to get back to this but I do appreciate the help.

Correct me if I'm wrong but it seems you are getting at using,

[tex]du=c_{p}dT=dq [/tex]

[tex]dS\equiv\frac{dQ}{T}=C_{p}m\frac{dT}{T} [/tex]

[tex]\Rightarrow \Delta S=C_{p}mln(\frac{T_{2}}{T_{1}})=4180ln(\frac{353}{293})=779JK^{-1}[/tex]

This was the first thing I tried and is not right. Where did I go wrong?
 
  • #5
If you're doing Carter problem 6-2 (a), the 1074 J/K answer in the back of the text is wrong. I did it the same was as you and got 778.35 J/K.
 
  • #6
Thanks, that is indeed the problem I'm working on. And while I always like to think it's the book that's wrong it rarely turns out to be the case. If somebody(s) else could please confirm/disprove this I would feel better about letting this problem go.
 
  • #7
Yeah...in thermo I often feel I've got a problem right, but for some frustrating reason or other my solution turns out to be quite wrong. I think you're alright on this one though, as before posting earlier I had referred to my text so see that that particular answer was indeed on the list of errata my prof pointed out. We can wait for a second opinion though :smile:
 
  • #8
Physics Fern said:
I had referred to my text so see that that particular answer was indeed on the list of errata my prof pointed out.

Ahh thank you very much, that's certainly good enough for me. I don't even want to say how much time I spent over complicating this trying to get the book's answer...
Is that list online anywhere that you know of? I couldn't find it in the five minutes I spent looking. It may be of help later. :)
 
  • #9
I don't believe Carter has put out a list; I haven't been able to find one online either. Our prof didn't provide us with an official list either--he sometimes just gives us a heads up if one of the text answers may be incorrect. Also, when students bring up questions in our tutorials he works through some of them as examples. We haven't gotten too far through the text, but here are some of the "corrections" that have come up so far:

4-1 (c) 2312.5 Jkmol-1K-1
6-2 (a) 779 JK-1
6-3 0.127 JK-1
7-2 Not too sure about this question...answer is possibly 704 JK-1 (instead of 660 JK-1).
7-11 311 ms-1

There may be others... Students have also pointed out typos in units and a couple other things (e.g. questionable atomic weight in 4-5), but I'm sure you'll realize those when doing the problems.
 
  • #10
I agree with the 779 J/K answer.
 

1. What is entropy?

Entropy is a measure of the disorder or randomness of a system. It is often described as a measure of the amount of energy in a system that is no longer available to do useful work.

2. How is entropy calculated?

The change in entropy of a system is calculated by dividing the heat transfer by the temperature at which the heat transfer occurred. In the case of this specific scenario, the change in entropy can be calculated by dividing the heat added to the water (in joules) by the temperature change (in Kelvin).

3. What is the initial entropy of the water at 20C?

The initial entropy of the water at 20C is not provided in this scenario. Entropy is a measure of change, so the initial entropy would be considered to be zero for this calculation.

4. How does heating water affect its entropy?

Heating water increases its entropy by increasing the disorder of the water molecules. As the temperature of the water increases, the molecules move more rapidly and become more disordered, resulting in an increase in entropy.

5. Is entropy the same as energy?

No, entropy and energy are not the same. While entropy is a measure of the disorder of a system, energy is a measure of the ability to do work. However, there is a relationship between them in thermodynamics, as an increase in entropy often results in a decrease in available energy to do work.

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