# 3.3 First law of thermodynamics  (Page 2/6)

 Page 2 / 6

Often the first law is used in its differential form, which is

$d{E}_{\text{int}}=dQ-dW.$

Here $d{E}_{\text{int}}$ is an infinitesimal change in internal energy when an infinitesimal amount of heat dQ is exchanged with the system and an infinitesimal amount of work dW is done by (positive in sign) or on (negative in sign) the system.

## Changes of state and the first law

During a thermodynamic process, a system moves from state A to state B , it is supplied with 400 J of heat and does 100 J of work. (a) For this transition, what is the system’s change in internal energy? (b) If the system then moves from state B back to state A , what is its change in internal energy? (c) If in moving from A to B along a different path, ${{W}^{\prime }}_{AB}=400\phantom{\rule{0.2em}{0ex}}\text{J}$ of work is done on the system, how much heat does it absorb?

## Strategy

The first law of thermodynamics relates the internal energy change, work done by the system, and the heat transferred to the system in a simple equation. The internal energy is a function of state and is therefore fixed at any given point regardless of how the system reaches the state.

## Solution

1. From the first law, the change in the system’s internal energy is
$\text{Δ}{E}_{\text{int}AB}={Q}_{AB}-{W}_{AB}=400\phantom{\rule{0.2em}{0ex}}\text{J}-100\phantom{\rule{0.2em}{0ex}}\text{J}=300\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$
2. Consider a closed path that passes through the states A and B . Internal energy is a state function, so $\text{Δ}{E}_{\text{int}}$ is zero for a closed path. Thus
$\text{Δ}{E}_{\text{int}}=\text{Δ}{E}_{\text{int}AB}+\text{Δ}{E}_{\text{int}BA}=0,$

and
$\text{Δ}{E}_{\text{int}AB}=\text{−}\text{Δ}{E}_{\text{int}BA}.$

This yields
$\text{Δ}{E}_{\text{int}BA}=-300\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$
3. The change in internal energy is the same for any path, so
$\begin{array}{cc}\hfill \text{Δ}{E}_{\text{int}AB}& =\text{Δ}{{E}^{\prime }}_{\text{int}AB}={{Q}^{\prime }}_{AB}–{{W}^{\prime }}_{AB};\hfill \\ \hfill 300\phantom{\rule{0.2em}{0ex}}\text{J}& ={{Q}^{\prime }}_{AB}–\left(-400\phantom{\rule{0.2em}{0ex}}\text{J}\right),\hfill \end{array}$

and the heat exchanged is
${{Q}^{\prime }}_{AB}=-100\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$

The negative sign indicates that the system loses heat in this transition.

## Significance

When a closed cycle is considered for the first law of thermodynamics, the change in internal energy around the whole path is equal to zero. If friction were to play a role in this example, less work would result from this heat added. [link] takes into consideration what happens if friction plays a role.

Notice that in [link] , we did not assume that the transitions were quasi-static. This is because the first law is not subject to such a restriction. It describes transitions between equilibrium states but is not concerned with the intermediate states. The system does not have to pass through only equilibrium states. For example, if a gas in a steel container at a well-defined temperature and pressure is made to explode by means of a spark, some of the gas may condense, different gas molecules may combine to form new compounds, and there may be all sorts of turbulence in the container—but eventually, the system will settle down to a new equilibrium state. This system is clearly not in equilibrium during its transition; however, its behavior is still governed by the first law because the process starts and ends with the system in equilibrium states.

## Polishing a fitting

A machinist polishes a 0.50-kg copper fitting with a piece of emery cloth for 2.0 min. He moves the cloth across the fitting at a constant speed of 1.0 m/s by applying a force of 20 N, tangent to the surface of the fitting. (a) What is the total work done on the fitting by the machinist? (b) What is the increase in the internal energy of the fitting? Assume that the change in the internal energy of the cloth is negligible and that no heat is exchanged between the fitting and its environment. (c) What is the increase in the temperature of the fitting?

## Strategy

The machinist’s force over a distance that can be calculated from the speed and time given is the work done on the system. The work, in turn, increases the internal energy of the system. This energy can be interpreted as the heat that raises the temperature of the system via its heat capacity. Be careful with the sign of each quantity.

## Solution

1. The power created by a force on an object or the rate at which the machinist does frictional work on the fitting is $\stackrel{\to }{\text{F}}·\stackrel{\to }{\text{v}}=\text{−}Fv$ . Thus, in an elapsed time $\text{Δ}t$ (2.0 min), the work done on the fitting is
$\begin{array}{cc}\hfill W& =\text{−}Fv\text{Δ}t=\text{−}\left(20\phantom{\rule{0.2em}{0ex}}\text{N}\right)\left(0.1\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\left(1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}\phantom{\rule{0.2em}{0ex}}\text{s}\right)\hfill \\ & =-2.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}\hfill \end{array}$
2. By assumption, no heat is exchanged between the fitting and its environment, so the first law gives for the change in the internal energy of the fitting:
$\text{Δ}{E}_{\text{int}}=\text{−}W=2.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$
3. Since $\text{Δ}{E}_{\text{int}}$ is path independent, the effect of the $2.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{J}$ of work is the same as if it were supplied at atmospheric pressure by a transfer of heat. Thus,
$2.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{J}=mc\text{Δ}T=\left(0.50\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(3.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}\phantom{\rule{0.2em}{0ex}}\text{J/kg}·\text{°}\text{C}\right)\text{Δ}T,$

and the increase in the temperature of the fitting is
$\text{Δ}T=12\phantom{\rule{0.2em}{0ex}}\text{°}\text{C},$

where we have used the value for the specific heat of copper, $c=3.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}\phantom{\rule{0.2em}{0ex}}\text{J/kg}·\text{°}\text{C}$ .

## Significance

If heat were released, the change in internal energy would be less and cause less of a temperature change than what was calculated in the problem.

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