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I teach undergraduate thermodynamics and I was quite ashamed that I couldn't explain to a student, the following. I thought I'd bring it to physics.SE in hope of providing my student a good explanation.

The question was concerning the energy equation for an open system:

$$\underbrace{\frac{\mathrm{d} E}{\mathrm{d}t}}_\text{Rate of change of total energy in the system} = \underbrace{\delta \dot{Q}}_\text{Rate of heat transfer} - \underbrace{\delta \dot{W}}_\text{Work extracted/input} + \underbrace{\dot{m} \left(h_1 + \frac{V_1^2}{2} + g z_1 \right)}_\text{energy of inlet stream} - \underbrace{\dot{m} \left(h_1 + \frac{V_1^2}{2} + g z_1 \right)}_\text{energy of outlet stream}$$

The young lady asked me that for steady state operations, the rate of change of total energy, $dE/dt$ is zero. So why are $\delta \dot{Q}$ and $\delta \dot{W}$ are not zero as they are also rates of heat exchange and work generation/input. It is just obvious to me but I don't know how to explain this to a 17 year old.

I'd appreciate it if someone could help me out on this.

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It would help if you explain what the terms are. I don't know about everyone else, but it is not at all obvious to me what any of the symbols in the third and fourth terms on the right hand side mean. Are they supposed to exactly cancel, as you've written? Anyway, the steady state solution $\frac{dE}{dt}=0$ just means that the average amount of energy in the system is not changing with time. So presumably any amount of heat that is transferred to the system is completely converted to work by whatever process within the system that you're modelling, meaning $\delta \dot{Q} = -\delta\dot{W}$. – Mark Mitchison Feb 26 at 22:25
@MarkMitchison Well this was my exact explanation! It apparently isn't obvious to 17-18 year old freshmen/sophomores! :( – drN Feb 26 at 22:43
Well that is the answer! The best you can do is explain it clearly, she will work it out for herself eventually. Why don't you try asking her why she thinks that the rates of heat and work flow should be zero? Another useful thing is to think of an example that would be familiar to her which demonstrates this principle. I would try and suggest something, but you never actually explained what on Earth we are talking about here! Failing that, try to concoct your own example. The best demonstration of a physical principle is usually provided by Nature herself :) – Mark Mitchison Feb 26 at 22:51
@MarkMitchison Yes, I will try and turn to nature. My explanation (which should have been included) was to show that the dE/dt term is an accumulation rate which is zero for steady state. – drN Feb 26 at 22:53
Sorry but I have no idea what that means! What is the physical system, what is coming in the 'inlet stream', what is $\dot{m}$, $V_1$ etc. etc.?? I appreciate you might not be bothered to include these things, but if you have time it would be helpful to future users of this site with a similar question if you were to edit this information into your post. Also, other people might have some better suggestions than me once they know what you are talking about. It really is not clear to anyone except maybe a specialist in whatever field of physics we are supposed to be talking about here. – Mark Mitchison Feb 26 at 22:58
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1 Answer

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The key to explaining this lies in two pieces that you have correctly written in the equation but have to make the student appreciate:

1) The difference between a $d$ and a $\delta$

$d$ means a small differential change, while $\delta $ just means a small quantity($\delta$ is used to denote a change in variational calculus, but that is different). Therefore in your equation $\delta Q/dt$ is just a small quantity per unit time (or per small change in time).

2) Steady state is when a system's properties don't change with time: Heat is not a systems property it is a transfer of energy. This is precisely why one never says $dQ$. Same for work, or matter. Now matter is a funny one it can be a change and a transfer. So technically if one is doing an atom balance one says \begin{align*} dN=\delta N \end{align*} or if your system leaks a particular molecules while also produces it by a chemical reaction one would write, \begin{align*} dN_i = \delta N_{i, leak}+\delta N_{i, reaction} \end{align*} Also energy transfers are always path dependent since they are not a system's property but depend on the process path. This is why $\delta $ also reminds us of that path dependence.

Essentially in steady state no changes to system properties occur but the quantity of heat or work is not zero it is still what it is.

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This is a good point that $\delta$ represents an inexact differential and that $\mathrm{d}$ is an exact differential. The fact that $\delta \dot{Q}$ and $\delta \dot{W}$ are path dependent but $\mathrm{d}\dot{E}$ is state dependent is certainly the clue to this answer. – drN Feb 27 at 9:56
I will also point out that $\delta \dot{Q}$ is not wrong but can be confusing for a student. It gives the impression that it is the change in a time derivative or at the very least a derivative while its neither. Usually in batch systems one tends to write $dE=\delta Q+\delta W$ (ignore the time ) or in steady-flow case one tends to highlight the by writing out $\delta Q/dt$ – Sankaran Feb 27 at 13:16

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