Changes
Jump to navigation
Jump to search
Line 814:
Line 814: + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Talk:Counterexamples to Evolution (view source)
Revision as of 18:20, September 29, 2016
, 18:20, September 29, 2016Added a new example, perhaps this will be clearer
[[User:Richardm|Richardm]] ([[User talk:Richardm|talk]]) 07:42, 27 September 2016 (EDT)
[[User:Richardm|Richardm]] ([[User talk:Richardm|talk]]) 07:42, 27 September 2016 (EDT)
The conservapedia article refers to what I would call a "closed system" as an "isolated system", that is one that exchanges niether matter nor energy with its surroundings. I shall use this terminology from now on.
I shall rephrase my argument using another example, perhaps this will be clearer. Suppose we have two systems 1 and 2 which are in thermal contact and thermally isolated with their surroundings. Together they form another system. Each of these systems are open as they can exchange energy with the other, but the combined system is isolated. The two systems look like this:
-----------------------------------
| System 1 | System 2 |
-----------------------------------
The two systems both have the same mass, <math>m</math> and the same specific heat capacity, <math>c</math>. System 1 has an initial temperature <math>T_1</math> and system 2 has an initial temperature <math>T_2</math>. We shall say that system 1 has a greater temperature than system 2 so that <math>T_1 > T_2</math>.
We now leave the systems for a while and eventually they reach thermal equilibrium and so have the same temperature <math>T_f</math>.
Considering that the heat transfer into or out of a system, <math>Q</math>, can be related to the temperature difference, <math>\Delta T</math>, as <math>q=mc \, \Delta T</math> we can relate the two temperatures as:
<math>
mc \, (T_1 - T_f) = mc \, (T_f-T_2)
</math>
since system 1 loses <math>mc \, (T_1 - T_f)</math> energy and system two gain <math>mc \, (T_f-T_2)</math>. Rearranging, we find it no surprise that the final temperature is the average of the two initial temperatures:
<math>
T_f = \frac{T_1+T_2}{2}
</math>
I am arguing that the entropy of an open system '''can''' decrease. I am '''not''' saying that it '''cannot''' increase. So let's consider the entropy of each system, <math>\Delta S_1</math> and <math>\Delta S_2</math> respectively, and of the combined system (the overall change in entropy), <math>\Delta S_t</math>.
The entropy <math>\Delta S_t = \Delta S_1 + \Delta S_2</math> so we find each entropy first individually. Let us consider system 1 first. The entropy change <math>\Delta S_1</math> is:
<math>
\Delta S_1 = \int^{final state}_{initial state} \frac{dq}{T}
</math>
We can use the equation for specific heat capacity to derive a substitution to produce an integral that we can evaluate. This substitution is <math>dQ = mc \, dT</math>, where <math>dQ</math> is the infinitesimal flow of heat, <math>m</math> is the mass, <math>c</math> is the specific heat capacity and <math>dT</math> is the infinitesimal change in temperature. Hence:
<math>
\Delta S_1 = \int^{T_f}_{T_1} \frac{mc}{T} \, dT
</math>
with an upper limit of <math>T_f</math> as this is the final temperature and a lower limit of <math>T_1</math> as this is the initial temperature. Similarly, the change of entropy of system 2, <math>\Delta S_2</math>, can be expressed as:
<math>
\Delta S_2 = \int^{T_f}_{T_2} \frac{mc}{T} \, dT
</math>
Performing the integration one finds:
<math>
\Delta S_1 = mc \, \ln{\frac{T_f}{T_1}} = mc \, (\ln{T_f} - \ln{T_1})
</math>
<math>
\Delta S_2 = mc \, \ln{\frac{T_f}{T_2}} = mc \, (\ln{T_f} - \ln{T_2})
</math>
Since <math>T_1 > T_f > T_2</math>, it is clear that <math>\Delta S_1 < 0</math> and <math>\Delta S_2</math> (remember that <math>\ln{a} > \ln{b}</math> for any real numbers <math>a>b</math>)
Hence we have two open systems, the entropy of one (system 1) has '''decreased''' and the other (system 2) has '''increased'''. Hence you can see that I am '''not''' denying that the entropy of an open system can increase as system 2 does exactly that.
What about the combined system?
<math>
\Delta S_t = \Delta S_1 + \Delta S_2 = mc \, \ln{\frac{T_f}{T_1 T_2}}
</math>
Substituting in for <math>T_f</math> as above we find:
<math>
\Delta S_t = mc \left( \ln{\left( \frac{T_1}{T_2} + \frac{T_2}{T_1} + 2 \right)} - \ln{4} \right)
</math>
Expressing <math>T_2</math> as <math>cT_1</math> for some real constant <math>c</math>, we see:
<math>
\frac{T_1}{T_2} + \frac{T_2}{T_1} + 2 = \frac{1}{c} + c + 2 > 4 \text{ for } c > 0
</math>
This can be seen here [http://math.stackexchange.com/questions/657706/prove-that-a1-a-%E2%89%A5-2-and-a1-a-%E2%89%A42]. Since temperatures are always greater than 0 :
<math>
\frac{T_1}{T_2} + \frac{T_2}{T_1} + 2 > 4
</math>
Hence the overall change in entropy is greater than zero.
So we can conclude that the second law is '''not''' valid for an open system (i.e. we cannot say "for any open system its entropy cannot decrease") as in the example there is a system that entropy decreases. However for the overall, isolated system we see the second law holds true in this case.
Since entropy of an open system can decrease, this argument in the article using the second law is invalid and should be removed. [[User:Richardm|Richardm]] ([[User talk:Richardm|talk]]) 14:20, 29 September 2016 (EDT)
== Note to AlecT ==
== Note to AlecT ==
(username removed)