September 23, 2020, 04:52:50 AM
Forum Rules: Read This Before Posting

### Topic: Sign convention when proving 2nd law with Clausius principle  (Read 3271 times)

0 Members and 1 Guest are viewing this topic.

#### idest

• Regular Member
•   • Posts: 11
• Mole Snacks: +0/-0 ##### Sign convention when proving 2nd law with Clausius principle
« on: April 06, 2016, 11:28:16 AM »
Hello. Recently I'm reviewing physical chemistry text, and I want to get some help.

I have a question about sign convention of work and heat in a proof of the equivalence between Clausius principle and the 2nd law. My lecture note brings up a model where two heat engines are combined like this: Heat engine 1 releases $-Q_{1c}$ and do work $-W_{1}$ when it's given $Q_{1h} >0$ . However, Heat engine 2 is working reversely with usual manner, nearly as refrigerator. It steals $Q_{2c}>0$ from a cold source, and give away $-Q_{2h}$ when it's given $W_{2}$. Then apply the 1st law to each heat engine and a combined one.

$$Q_{1h}+Q_{1c}+W_{1}=0$$

$$Q_{2h}+Q_{2c}+W_{2}=0$$

$$Q_{h}+Q_{c}+W=0$$,

where $Q_{h}$ is heat provided with the combined engine, and $-Q_{c}$ is heat released to the sink, and $-W$ is work done by the combined one.

The note says that heat and work of combined system can be expressed as

$$Q_{h}=Q_{1h}-Q_{2h}$$

$$Q_{c}=Q_{2c}-Q_{1c}$$

$$W=W_{2}-W_{1}$$

Now I'm confused by these relations. $-Q_{2h}$ is a released heat from the engine 2, and it seems like getting out from the whole system. Then why not $Q_{h}=Q_{1h}+Q_{2h}$ in the whole system? In the same reason, I thought $Q_{c}=Q_{1c}+Q_{2c}$, but it was also a different result.

I really want to know where I'm wrong.

To love and win is the best thing. To love and lose, the next best.

- William Makepeace Thackeray

#### mjc123

• Chemist
• Sr. Member
• • Posts: 1775
• Mole Snacks: +252/-11 ##### Re: Sign convention when proving 2nd law with Clausius principle
« Reply #1 on: April 06, 2016, 12:28:58 PM »
Quote
Q1h+Q1c+W1=0

Q2h+Q2c+W2=0
These expressions are wrong. Can you see why?

#### idest

• Regular Member
•   • Posts: 11
• Mole Snacks: +0/-0 ##### Re: Sign convention when proving 2nd law with Clausius principle
« Reply #2 on: April 06, 2016, 09:20:46 PM »
Thanks for your reply. Still, I have no idea of error of the expressions.

A heat engine is a cycle, which means that it recovers its initial state when a process is over ( $\Delta U=0$ ). Then work performed by the engine is equal to heat inserted(>0) and released(<0). That makes $-W_{x}=Q_{xh}+Q_{xc}$.

Is there something that I missed?
To love and win is the best thing. To love and lose, the next best.

- William Makepeace Thackeray

#### mjc123

• Chemist
• Sr. Member
• • Posts: 1775
• Mole Snacks: +252/-11 ##### Re: Sign convention when proving 2nd law with Clausius principle
« Reply #3 on: April 07, 2016, 04:31:34 AM »
If an amount of heat Qxc is released, that is equivalent to saying an amount -Qxc is inserted. Therefore -Wx = Qxh - Qxc.
The sum of the inputs is zero; an output counts as a negative input. So in your diagram Q1h - Q1c - W1 = 0

#### idest

• Regular Member
•   • Posts: 11
• Mole Snacks: +0/-0 ##### Re: Sign convention when proving 2nd law with Clausius principle
« Reply #4 on: April 07, 2016, 05:17:45 AM »
I get it now.

I think there is inconsistency of convention between diagram and equations then...

I checked that your revision is compatible with Qh=Q1h-Q2h, Qc=Q2c-Q1c, W=W2-W1
Quote
If an amount of heat Qxc is released, that is equivalent to saying an amount -Qxc is inserted. Therefore -Wx = Qxh - Qxc.
The sum of the inputs is zero; an output counts as a negative input. So in your diagram Q1h - Q1c - W1 = 0

Thank you.
« Last Edit: April 07, 2016, 08:50:07 AM by idest »
To love and win is the best thing. To love and lose, the next best.

- William Makepeace Thackeray

#### mjc123

• Chemist
• Sr. Member
• • Posts: 1775
• Mole Snacks: +252/-11 ##### Re: Sign convention when proving 2nd law with Clausius principle
« Reply #5 on: April 07, 2016, 06:22:24 AM »
You can look at it in one of two ways. Either you can say there is a heat input Qi and a heat output Qo, both being positive quantities. Then the heat change is input - output = Qi - Qo.
Or you can say there is a heat input Qi and a "negative input" -Qo , and the heat change is the sum of inputs = Qi - Qo.
What you mustn't do is talk of a "heat output of -Qo" and subtract this from the input, giving Qi - (-Qo) = Qi + Qo. This is effectively taking account of the negative sign twice, and cancelling it out. Your diagram may possibly lead to confusion here, as you have e.g. an inward arrow with a value of Q1h and an outward arrow with a value of -Q1c. Now you should either add all the values together, taking account of their signs, or treat all values as positive, and add the nunbers on the inward arrows and subtract the numbers on the outward arrows. But if you try to do both (e.g. subtract -Q1c) you will get it wrong.
I think the preferable way to work is to treat everything as input, and outputs as negative inputs. Thus in thermodynamic convention dQ is the change in heat content of the system, and is positive for heat put into the system, and negative for heat given out by the system. dW is the work done on the system by the surroundings; work done by the system is -dW. (For chemists. As another thread suggests, physicists seem to have the reverse convention for dW, but ignore that for now.) So ΔQ for engine 1 is Q1h - Q1c, and ΔU = 0 = ΔQ + ΔW = Q1h - Q1c - W1.