# $\sum{\frac{Q}{T}}$ is zero

I was going through a proof in my book and couldn't understand the last part of the proof. I mean I understand

$$\frac{Q_1^{\prime}}{T_1}-\frac{Q_2^{\prime}}{T_2}\leq 0$$

But after that how did author arrived at end ?

What i am saying is

$$\frac{Q_1}{T_1}-\frac{Q_2}{T_2}\leq 0$$ And hence for total cycle it should be $$\frac{Q_1}{T_1}-\frac{Q_2}{T_2}+\frac{Q_3}{T_3}-\frac{Q_4}{T_4}........\leq 0$$﻿ ﻿not $$\frac{Q_1}{T_1}+\frac{Q_2}{T_2}+\frac{Q_3}{T_3}+\frac{Q_4}{T_4}........\leq 0$$﻿ ﻿

Or

$$\frac{Q_1^{\prime}}{T_1}-\frac{Q_2^{\prime}}{T_2}\leq 0$$ Let's take it as $a-b\leq 0$ and later it says $$\sum{\frac{Q}{T}}\leq 0$$ Which means $a+b\leq 0$

• Which part don't you understand? The mathematics? – Chet Miller Dec 5 '17 at 13:02
• Sir at first book says $$\frac{Q_1^{\prime}}{T_1}-\frac{Q_2^{\prime}}{T_2}\leq 0$$ Let's take it as $a-b\leq 0$ and later it says $$\sum{\frac{Q}{T}}\leq 0$$ Which means $a+b\leq 0$ – user177597 Dec 6 '17 at 0:21
• @ChesterMiller Sir ! $$\frac{Q_1}{T_1}-\frac{Q_2}{T_2}\leq 0$$ And hence for total cycle it should be $$\frac{Q_1}{T_1}-\frac{Q_2}{T_2}+\frac{Q_3}{T_3}-\frac{Q_4}{T_4}........\leq 0$$﻿ ﻿not $$\frac{Q_1}{T_1}+\frac{Q_2}{T_2}+\frac{Q_3}{T_3}+\frac{Q_4}{T_4}........\leq 0$$﻿ ﻿ – user177597 Dec 6 '17 at 0:37
• I see what the problem is. They are "fooling with" the mathematics. In the equation $\sum{\frac{Q}{T}}\leq 0$, all the Q's are supposed to represent heat transferred from the surroundings to the system. If, in one of the steps for an actual process, heat is transferred from the system to the surroundings, the corresponding Q will be negative. They handle this by adding a minus sign to the equation and treating the corresponding Q as positive. Crazy, huh. But it has been done like this for hundreds of year. – Chet Miller Dec 6 '17 at 1:59
• @ChesterMiller I got it sir ! Is it all about "sign convention" foolery ? – user177597 Dec 6 '17 at 2:17

What the author is saying is let $Q_2$ be the heat put in the system and $Q_1$ the heat taken out of the system. Then

$$\frac{Q_1}{T_1}-\frac{Q_2}{T_2}\leq0\ .$$

Now, one could have more heat taken out of the system, say $\{Q_a\}$, $a=1,\ldots,p$ and more put in the system $\{Q_b\}$, $b=p+1,\ldots,n$ (what I mean by more is in more than one step). Here we take $Q_a>0$ and $Q_b<0$. Then what is meant by the formula is that the sum of all these must be zero or negative. In other words

$$\underbrace{\frac{Q_1}{T_1}+\frac{Q_2}{T_2}+\cdots+\frac{Q_p}{T_p}}_{\text{Heat taken out}}+\underbrace{\frac{Q_{p+1}}{T_{p+1}}+\cdots+\frac{Q_{n-1}}{T_{n-1}}+\frac{Q_{n}}{T_{n}}}_{\text{Heat put in}}\leq0\ .$$

One can then write this in a more compact way as

$$\sum_{i=1}^{n}\frac{Q_i}{T_i}\leq0\ ,$$

or, even in a more compact form where the summation indices are implicit,

$$\sum\frac{Q}{T}\leq0\ .$$

The compact notation is useful if the idea is clear, but the more explicit version is better to getting the hang of what is going on. Hope it helps!

• Sir ! $$\frac{Q_1}{T_1}-\frac{Q_2}{T_2}\leq 0$$ And hence for total cycle it should be $$\frac{Q_1}{T_1}-\frac{Q_2}{T_2}+\frac{Q_3}{T_3}-\frac{Q_4}{T_4}........\leq 0$$﻿ ﻿not $$\frac{Q_1}{T_1}+\frac{Q_2}{T_2}+\frac{Q_3}{T_3}+\frac{Q_4}{T_4}........\leq 0$$﻿ – user177597 Dec 6 '17 at 0:43
• @AnswerSeekingPenguin Hello sir! I understand what you mean, but as I stated, I assume $\{Q_b\}$ are negative quantitites, i.e. $Q_b<0$. Thus, performing $Q_a+Q_b$ is the same as doing $Q_a-|Q_b|$. – L. Werneck Dec 6 '17 at 15:08
• Thanks sir I got it ! It is about sign convention . + comes when we takes absolute value . – user177597 Dec 7 '17 at 2:57