Sign of work in the first law of thermodynamics

Question

The first law of thermodynamics can be expressed mathematically as

$dU = \bar{d}Q + \bar{d}W$ or as $\Delta U = \Delta Q + \Delta W$.

So lets suppose we had some system and suppose $80J$ of heat flow into the system, and the system does $30J$ of work. Would the change in internal energy of the system be expressed as

$\Delta U = 80J -30J = 50J$

or as

$\Delta U = 80J +30J = 110J$

I think it should be the second one, however my lecture notes indicate that the first expression for $\Delta U$ is correct. Why is that so?

A quick lookup on Wikipedia indicates that there is some sort of sign convention that one chooses, could that play a part in this? https://en.wikipedia.org/wiki/First_law_of_thermodynamics#Description

Answer 1

You are correct in that there is some sign convention. However, when calculating the change in internal energy, both must yield the same result, as the underlying physics must not change.

In the form you stated the first law, work done to the system, e.g. you compress a gas or something the like, is positive, work done by the system, e.g. expanding a gas and driving a piston, is negative. So the $30\,J$ in your example are negative and the first result is correct. It is also quite logical. If the second answer were correct, you could increase the internal energy of a system by making it perform work, which does not make sense. The work performed by the system leaves it, thus reducing the internal energy.

If you give the work done to and by the system the opposite signs than mentioned above, then the formula is $$\Delta U = \Delta Q -\Delta W$$

In other words, the first law of thermodynamics states that it does not matter in which form you add energy to the system - mechanical or thermal - they both increase the internal energy. Or: thermal energy is just another form of energy.

Answer 2

So lets suppose we had some system and suppose $80J$ of heat flow into the system, and the system does $30J$ of work.

Heat is energy transferred between a system and its surroundings by virtue of a temperature difference only and any other means for changing the energy of a system is called work.
Just think of the first law of thermodynamics as a restatement of the law of conservation of energy about the flow of energy into and out of a system.
You have $80\, \rm J$ of energy entering the system as heat and $30\,\rm J$ of energy leaving the system because of the work that the system has done.
The system has a net gain in energy of $80-30= 50 \,\rm J$ which is called the change in internal energy of the system.

Physicists generally use the equation
$\Delta U = Q-W$
where $\Delta U$ is the change in internal energy of the system, $Q$ is the heat input to the system and $W$ is the work done by the system.
For your example $\Delta U = 80 -(+30) = +50 \,\rm J$.

Chemists generally use the equation
$\Delta U = Q+W$
where $\Delta U$ is the change in internal energy of the system, $Q$ is the heat input to the system and $W$ is the work done on the system.
For your example $\Delta U = 80 +(-30) = +50 \,\rm J$.

Answer 3

In undergraduate texts, the 1st Law of Thermodynamics can be confusing. This law is actually another way to express conservation of energy, but the equation that is listed in texts typically describes only one process. This often leads to confusion regarding the sign associated with work, and in fact, some texts state that work entering a system (work done by the environment on the system) is positive while some texts state that work exiting a system (work done on the environment by the system) is positive.

From a practical standpoint, after having worked on thermodynamic problems in industry for a number of years, it is apparent that the following applies regarding the 1st Law of Thermodynamics:

1) work is equivalent to energy; 2) heat is equivalent to energy; 3) if there is a net accumulation of energy in a system, the internal energy of that system increases; 4) if there is a net reduction of energy in a system, the internal energy of that system decreases.

Based on the above items, any energy entering a thermodynamic system adds to the system's internal energy, and any energy leaving a thermodynamic system subtracts from that system's internal energy. In addition, in order to keep all of the signs straight, it is best to draw the equivalent of a free body diagram for each thermodynamic system that you are working with, properly assign the energy flows to that system, and set up an energy balance for that particular system.

For the stated problem, where 80J of heat enter the system, and 30J of work leave the system, the attached drawing shows the process that is involved, and indicates that the change in internal energy is 50J.