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Good evening. I've happened to be sitting down today and just couldn't wrap my head around this question which seems rather simple at first. From reading about the first law and sign conventions as it deals with work done by a system and it's surroundings, when the system does work it has a positive sign convention and when the surroundings does work it has a negative sign convention.

The first Law of thermodynamics for a closed system reads $Q-W=U_2 - U_1$.

So let's assume the system undergoes some random process of which $Q=0$, $K.E.=0$ and $P.E.=0$ (for the system itself), and W is positive (indicating that the system is doing work on its surroundings). The First Law Equation then becomes $-W=U_2 - U_1$ (whereas, by convention, $U_2 - U_1$ would be taken as an increase in the internal energy of the system). We know that this is normally associated with an increase in temperature because in most cases, temperature increases are linked to increases in molecular interactions, which are internal to the system.

My question here (based on every accepted convention and equation) is why is there a decrease in the internal energy of a system if it is doing work on the surroundings? To provide a little more clarity here, suppose you were pushing a heavy load up a ramp. You(the system) would be doing work on the surrounding object(the heavy load). But at the same time, the temperature within your body is continuing to increase for as long as you exert a force for the total length of application. So therefore, rather than having a decrease in your bodily internal energy, that internal energy should be increasing, correct? In other words the work done by the system is generally increasing the internal energy of the system, rather than the opposite. Can anyone explain to me where the flaw is in my understanding of this, because I feel like I'm missing something.

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    $\begingroup$ I think the basic mistake here is equating body temperature to internal energy. $\endgroup$ – paisanco Aug 1 '15 at 19:51
  • $\begingroup$ When you do some work, like pushing a heavy load up a ramp, you are using your muscles. They are expanding and contracting, using up chemical energy stored in your body. This stored energy thus decreases and that is all that the First Law states, though this example is really crude. It happens to be that in gases, this internal energy is related to the temperature, which is again the average molecular kinetic energy, and hence internal energy change means temperature change in general for gases but not everywhere else. $\endgroup$ – Aritra Das Aug 1 '15 at 20:05
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  1. With no heat input (Q=0) and KE and PE of the system =0 it can't do work on the surrounding.. note: rise in temperature, rise in internal energy is in fact rise in KE of gas molecules in a P-V-T system.

  2. Don't complicate things, the 1st law is simply conservation of energy.. if you do work then you lose energy; as simple as that. The example you gave is a bit misleading, there the body temperature increases because of increased metabolism and cellular respiration, that energy comes from breaking of carbohydrates and thus the energy stored as chemical bonds now partly goes in pushing the object and is partly released as heat.. you can think of it like Total internal energy in form of bonds in glucose ==> excess heat + work. So here in fact Q is negative as body is giving out heat, W is positive, and internal energy decreases as some energy from the burger you had a couple of days ago is now gone..!! Don't confuse internal energy and temperature, they sometimes go hand in hand in simple systems but are completely different things. Look for the source of all energies, which form of energy is converted to what other form, and remember energy can neither be created nor destroyed (in a weaker sense, STR says something more serious) as that's all you need for the 1st law of thermodynamics.

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Your problem has to do with your example system rather then sign convention. To understand all of the energy dynamics of the human body is rather complex and increase in body temperature doesn't imply increase in internal energy. The increase in temperature is due to reactions taking place in your muscle but if were to add up to energy due to Kinetic fluxuations(temperature) and chemical bond energy in the body it would have decreased after pushing an object.

Physically you notice the loss in energy as being more tired after lifting many objects. After lifting the objects your body temperature has returned to the temperature from before but you have less energy(less carbohydrate bonds that your muscles can break for energy) to continue doing more work.

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I've dealt with this same issue in the past. To some extent, the issue you are describing is a nomenclature issue. In other words, some thermo text books say work leaving a system is positive, and some thermo text books say that work leaving a system is negative (I'm writing about high school physics and college undergrad text books - I don't know about graduate level thermo texts).

I have solved this problem the following way, and I teach my physics students (I'm a high school physics teacher) to work their 1st law problems this way to avoid confusion:

The first law is typically defined in text books as a process where heat is entering the system, work is leaving the system, and the system internal energy is changing as a result. The equation that describes this process is $\Delta U = Q - W$, where $Q$ and $W$ are positive. By implication, this means that heat entering the system is positive and work leaving the system is positive.

Unfortunately, there are other processes where work is being done on the system (work enters) and heat may or may not be entering or leaving the system. To explicitly use the equation noted above, one would have to keep the subtraction in the equation AND put a negative sign on $W$, which gets somewhat confusing, particularly when one realizes that some texts maintain that work entering a system is positive.

To avoid confusion, I teach students to draw a thermo diagram for each and every problem statement, whereby they also show the proper entry and exit for work and heat. After this is done, treat both heat and work as energy, and construct the equation that describes the given process, under the condition that any energy entering the process increases the system's internal energy and any energy leaving the process decreases the system's internal energy.

Example: For a heat pump, work is done on the system to pump low temperature heat (which enters the system) to a high temperature environment (heat leaves the system). All variables are positive, and the equation for this particular system is:

$\Delta U = W + Q_{low} - Q_{high}$ , meaning that internal energy change is determined by work and low temperature heat entering the system and high temperature heat leaving the system.

This approach avoids confusion on the answer to a large extent, but if you are taking a test from a particular text, you will still often have to worry about how the author defined work leaving a system if the question needs to match the nomenclature of your particular text (which it usually does).

In conclusion, as another poster stated, the 1st law of thermo is just an energy balance. Due to this, it is easily possible (and in my opinion desirable) to formulate the first law with an equation that describes your particular system and that doesn't blindly follow a single equation in your text book. By doing so, a lot of confusion can be avoided, both for you and for the people who are interpreting your work.

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  • $\begingroup$ Yes, I realize that I need more practice with LaTex formatting. Suggestions from those experienced in LaTex would be appreciated. $\endgroup$ – David White Aug 2 '15 at 6:25

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