Thermodynamic sign convention for heat (i.e. in heat engines)

In summary, the first law of thermodynamics states that the change in a system's internal energy is equal to the heat transferred to the system plus the work done on the system. There are two equivalent formulations of this law, one with a positive sign for work done by the system and one with a negative sign. However, there is often confusion surrounding the conventions for heat transfer, with some sources using positive signs for both Qh and Qc, while others use a negative sign for Qc. It is good practice to reserve symbols for single use and to use Qnet for any net heat transfer, with Qh and Qc following their respective conventions.
  • #1
etotheipi
Just to clarify, I'm aware of the two equivalent expressions of the first law ##\Delta U = Q + W## and ##\Delta U = Q - W## when applied to a certain system, though my question is primarily about ##Q## - for which, as far as I am aware, the convention is almost universally that ##Q > 0## if heat is transferred to the system.

Firstly, suppose we have a heat engine operating between ##T_h## and ##T_c##. On a diagram, and in calculations, we'd usually write that ##Q_h## is transferred from the source to the engine and ##Q_c## from the engine to the sink. We might then define other quantities like ##\eta = \frac{Q_h - Q_c}{Q_h}##, except clearly this doesn't obey the sign convention. Because if we stick to it, then ##Q_c## is negative and we should really write ##\eta = \frac{Q_h + Q_c}{Q_h} = \frac{Q_h - |Q_c|}{Q_h}##. Also, on heat engine diagrams, we generally label the arrow from the engine with the sink with ##Q_c##. Though should we be writing ##Q_c## or ##-Q_c##?

And on a sort of related note, often we will want to know the heat given out to the surroundings (e.g. maybe for calculating an enthalpy change of a reaction). Is it good practice to "reverse" the system and surroundings in this case? That is, we might first define the system to be the reactants, which have heat ##Q## transferred to them. Then, we let the previous surroundings be the new system, and the reactants the new surroundings, and now we may say ##-Q## is the heat transferred to the surroundings. Is the "reversing" part necessary? I ask because it seems somewhat dubious to refer to the heat transferred to the surroundings, since our sign convention is defined for the system!

Thank you!
 
Science news on Phys.org
  • #2
In ##\Delta U=Q+W##, ##Q## is the heat that goes in the system and ##W## is the work done on the system by what's outside the system. It says "the change in the system's internal energy is equal to what goes into the system as heat and as work."

In ##\Delta U=Q-W##, ##Q## is the heat that goes in the system and ##W## is the work done by the system on what's outside the system. It says "the energy tat stays in the system's is equal to what goes into the system as heat less what comes out as work."

Both formulations assume that you have already specified what the system is. The formulations are system-independent. If you compress the gas in a syringe adiabatically, you do positive work on the gas and the gas does negative work on you. This means that ##W## in the first formulation is positive and that ##W## in the second formulation is negative. It's a simple idea, but most people are confused when they think about "work done" without a clear idea of who does work on whom.
 
  • Like
Likes etotheipi
  • #3
kuruman said:
Both formulations assume that you have already specified what the system is. The formulations are system-independent. If you compress the gas in a syringe adiabatically, you do positive work on the gas and the gas does negative work on you. This means that ##W## in the first formulation is positive and that ##W## in the second formulation is negative. It's a simple idea, but most people are confused when they think about "work done" without a clear idea of who does work on whom.

Thanks for your reply; I am comfortable with the ideas and conventions regarding the work done.

I'm just a little confused about the conventions for heat. I'm trying to follow along with Peskin and Schroeder but notice that they use ##Q_h## and ##Q_c##, and don't bother with any negative signs. And I did some reading around and also found that the formulations they use are quite common!
 
  • #4
etotheipi said:
Thanks for your reply; I am comfortable with the ideas and conventions regarding the work done.

I'm just a little confused about the conventions for heat. I'm trying to follow along with Peskin and Schroeder but notice that they use ##Q_h## and ##Q_c##, and don't bother with any negative signs. And I did some reading around and also found that the formulations they use are quite common!
I don't have Peskin and Schroeder but I am pretty sure that Qh is the heat that goes into the system and Qc is the heat goes out of the system. In this convention, both heats are positive numbers. Their relation to ##Q## in the context of the first law is ##Q=Q_h## and ##Q=-Q_c##.
 
  • Like
  • Informative
Likes hutchphd and etotheipi
  • #5
kuruman said:
I don't have Peskin and Schroeder but I am pretty sure that Qh is the heat that goes into the system and Qc is the heat goes out of the system. In this convention, both heats are positive numbers. Their relation to ##Q## in the context of the first law is ##Q=Q_h## and ##Q=-Q_c##.

Ah OK, so do we generally reserve ##Q## and ##W## to be the total/net work done conforming to whatever we choose as our sign convention, but we can also speak of any other old heat transfer (i.e. ##Q = Q_h - Q_c##) and these don't necessarily need to follow the same sign convention?
 
  • #6
etotheipi said:
Ah OK, so do we generally reserve ##Q## and ##W## to be the total/net work done conforming to whatever we choose as our sign convention, but we can also speak of any other old heat transfer (i.e. ##Q = Q_h - Q_c##) and these don't necessarily need to follow the same sign convention?
Good habits eliminate unnecessary confusion. It is always good practice to reserve symbols for single use. I would use symbol ##Q## solely within the context of the first law. That is what Peskin & Schroeder seem to be doing. For any old net heat transfer, I would write ##Q_{net}=Q_h-Q_c.## This assumes that ##Q_h## and ##Q_c## are positive as defined in post #4. Then ##Q_{net}## is positive in the case of a heat engine and negative in the case of a refrigerator as the system. The converse is true if the system is the environment.
 
  • Like
  • Love
Likes hutchphd and etotheipi
  • #7
kuruman said:
Good habits eliminate unnecessary confusion. It is always good practice to reserve symbols for single use. I would use symbol ##Q## solely within the context of the first law. That is what Peskin & Schroeder seem to be doing. For any old net heat transfer, I would write ##Q_{net}=Q_h-Q_c.## This assumes that ##Q_h## and ##Q_c## are positive as defined in post #4. Then ##Q_{net}## is positive in the case of a heat engine and negative in the case of a refrigerator as the system. The converse is true if the system is the environment.

Awesome, that sounds like a plan. Having another read over, that seems right.

I wonder, just out of interest, what the convention is for anything to do with the surroundings. A fairly standard school Chemistry experiment might consist of mixing two reagents and measuring the rise in temperature of the surroundings (the solvent). We then might want to calculate ##\Delta H##.

For the system, at constant ##V## and ##P##, ##\Delta H = Q##. However, when we measure our temperature change of the solvent and work out ##C_{v} \Delta T##, we're actually calculating ##-Q##.

I'm not sure if this is a moot point (I could just be getting paranoid over things that don't really matter!), but do we then always write the heat to the surroundings as ##-Q##. Or would we swap the places of the system and surroundings and call it something like ##Q_{surr}##, which would be numerically equal to ##-Q##?
 
  • #8
Any symbol is shorthand notation for a string of words and sentences in the English (or any other) language. As long as (a) you are aware of what the symbols stand for in plain language and (b) you make certain to explain that unambiguously to the reader, you'll be OK.
 
  • Like
Likes etotheipi
  • #9
etotheipi said:
I'm just a little confused about the conventions for heat. I'm trying to follow along with Peskin and Schroeder but notice that they use QhQhQ_h and QcQcQ_c, and don't bother with any negative signs.
There should be an explanatory statement or two near the beginning of the chapter. It is quite common practice for authors to refer to the ##Q##'s used in the analysis of heat engines as absolute values. Only in that way are they consistent with things like the 1st Law and relations like ##Q=cm\,\Delta T##.
 
  • Like
Likes etotheipi
  • #10
Mister T said:
It is quite common practice for authors to refer to the ##Q##'s used in the analysis of heat engines as absolute values. Only in that way are they consistent with things like the 1st Law and relations like ##Q=cm\,\Delta T##.
I always thought that ##Q## in ##Q=cm\,\Delta T## is defined in the context of the first law as the heat that goes in the system. It can be positive (heat is added) or negative (heat is removed) matching a respective rise or drop of the system's temperature.
 
  • #11
kuruman said:
I always thought that ##Q## in ##Q=cm\,\Delta T## is defined in the context of the first law as the heat that goes in the system.

Right. In the chapter on heat engines it means ##\mid Q \mid##.
 
  • #12
etotheipi said:
We might then define other quantities like η=Qh−QcQhη=Qh−QcQh\eta = \frac{Q_h - Q_c}{Q_h}, except clearly this doesn't obey the sign convention.
Maybe this helps.
For an exothermic reaction, or anything giving off heat the Q is less than zero.
For enothermic the Q>0, or absorbing heat.

Here Qin is endothermic so >0
Qout is exothermic so <0
You have two Q's interacting with the system.
We could write Qtotal = Q1 + Q2, as a general statement.
Substituting Qin for Q1, and Qout for Q2, and knowing their sign, we have
Qtotal = Qin + ( -Qout )
or
Qtotal = Qin-Qout.

This is the same thing as on a number line moving to the right has a is positive sign and moving to the left has a negative sign.
Moving to in any direction we speak of the magnitude and the direction.
So three units to the right and 2 to the left becomes 3 + ( -2).
and not 3 - ( -2 ).
 
  • Like
Likes etotheipi

1. What is the thermodynamic sign convention for heat?

The thermodynamic sign convention for heat states that heat transfer from a hotter object to a colder object is considered positive, while heat transfer from a colder object to a hotter object is considered negative.

2. Why is the thermodynamic sign convention for heat important?

The thermodynamic sign convention for heat is important because it helps us to keep track of the direction of heat flow in thermodynamic processes. It also allows us to accurately calculate the efficiency of heat engines.

3. How does the thermodynamic sign convention for heat relate to the first law of thermodynamics?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted. The thermodynamic sign convention for heat is a way of quantifying the transfer of energy in the form of heat in thermodynamic processes.

4. Does the thermodynamic sign convention for heat apply to all thermodynamic processes?

No, the thermodynamic sign convention for heat only applies to heat engines, which are systems that convert heat into work. For other thermodynamic processes, such as refrigeration or heat pumps, the convention is reversed.

5. How can the thermodynamic sign convention for heat be used to determine the efficiency of a heat engine?

The efficiency of a heat engine can be calculated by dividing the work output by the heat input. The heat input is considered positive according to the thermodynamic sign convention for heat, while the work output is considered negative. Therefore, the efficiency can be expressed as a negative value, representing the amount of heat that is converted into work.

Similar threads

Replies
3
Views
553
Replies
1
Views
764
Replies
18
Views
2K
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
765
  • Thermodynamics
Replies
15
Views
1K
Replies
5
Views
1K
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
4K
  • Introductory Physics Homework Help
Replies
1
Views
774
Back
Top