Temperature and entropy in relation to energy

Question

In our textbook's formulation of the 1st law of thermodynamics, the amount of heat $Q$ supplied to the system is given equal to $TdS$, where $T$ is the absolute temperature, and $S$ is entropy per unit of mass. I don't understand why. There was no explanation given.

I have two questions:

1) Why is $Q=TdS$?

2) Why is entropy given in units per mass?

3) Under which assumptions this is correct (since there was no assumptions given in our textbook either)

The law is: $TdS=d\epsilon +pdV$, $\epsilon$ is internal energy, $p$ is pressure and $V$ is volume per unit of mass. Sorry to not have included it right away

Answer 1

Answers:

1. The relationship comes from the definition of a differential change in entropy

$$ds=\frac{\delta q_{rev}}{T}$$

Where $\delta q$ is a reversible transfer of heat at temperature $T$. So $q$ in your equation only applies to reversible heat transfer.

2. Entropy can either be expressed per unit mass, which is called its specific entropy $s$ or as total entropy $S$ which is simply the specific entropy times the mass of the system, $ms$. Note specific entropy should be lower case not upper case.

3. Unless otherwise stated, a lower case indicates entropy per unit mass and upper case total entropy. The same applies to other extensive properties, such as internal energy $U$, $u$.

Given the above, if your first equation is intended to be differential change in specific entropy (entropy per unit mass) it should technically read

$$\delta q_{rev}=Tds$$

Where $q=Q/m$ and $s=S/m$

If its a differential change in total entropy, it should read

$$\delta Q_{rev}=TdS$$

If the reversible heat transfer occurs at constant temperature, then you can integrate the last two equations and get:

$$Q=T\Delta S=T(S_{2}-S_{1})$$

or

$$q=T\Delta s=T(s_{2}-s_{1})$$

Hope this helps.