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According to the Kinetic Theory of Matter, temperature is nothing but a measure of the kinetic energy of matter. My textbook says that the change in internal energy of a system is the heat gained plus the work done on it. The heat change, apparantly, is measured using the change in temperature in a calorimeter, using the ideas of specific heat capacity.

Let's say a piston is kept inside a calorimeter, and it undergoes a reaction in which volume is increased. Work is done on the molecules of water in the calorimeter, and radiation is emitted in the course of the reaction. Both the radiation and the work increase the KE of the water, and therefore contribute to the temperature change. So what part of this change is due to work, and what is due to "heat"? How exactly are they differentiated?

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    $\begingroup$ This is a very deep question, where heat can be considered "low quality" energy and work is "high quality" energy, since all work can be transformed into heat but not all heat can be transformed into work (unless you have an infinite heat bath at absolute zero temperature). $\endgroup$ – user288447 Oct 13 '14 at 10:30
  • $\begingroup$ Other questions about the difference between heat and work: physics.stackexchange.com/q/135539/58382, physics.stackexchange.com/q/23582/58382, physics.stackexchange.com/q/8522/58382 $\endgroup$ – glS Oct 13 '14 at 12:00
  • $\begingroup$ Keep in mind that the first law of thermo also applies to systems undergoing a phase change. If your system contains a boiling liquid, the temperature can remain constant even while heat is being added or subtracted from that system, and even while work is being done on or by that system. $\endgroup$ – David White Feb 27 at 18:49
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Work is force acting over a distance. That means there can only be work done on the gas if the piston moves, i.e. if the gas changes volume. In that case, $W = \int F \text{d} x = \int P\, \text{d}V$, where P is the pressure difference between the inside and outside of the calorimeter. Note, however, that in your example the gas is doing work on the piston, not the other way around, so the work done on the gas is negative, and the average KE per particle would decrease as a result of that part. Any transfer of energy into or out of the system not associated with a change of volume may be referred to as heat (positive if transferred in, negative if transferred out).

Your question suggests you may be confusing radiation with heat. In a good calorimeter on the time scale of the experiment there is negligible loss from the system due to radiation. In a system that is not so well isolated the radiation usually represents heat loss (i.e. negative heat in), but can also represent negative work if, for example, the radiation is strong enough to push out the piston. In general, though, heat may be lost from the system via conduction through the walls, or by other means if the system is not well-sealed (most importantly convection and evaporation). Depending how you do your accounting and what kind of reaction is going on, heat can also represent transfer of thermal energy out of or into your system as chemical energy (into = exothermic reaction, out of = endothermic).

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First law of thermodynamics states that the change in internal energy, $dU$, of a system is equal to $\delta Q + \delta W$. The only way to separate the contributions from the two is to arrange the experiment so that one of the quantities is zero.

If the change is adiabatic so that there is no heat exchange, $\delta Q = 0$ and hence $dU = \delta W$. On the other hand, if the change is isochoric, that is there is no change in volume, $\delta W = 0$ and hence $dU = \delta Q$.

Except for these special circumstances, the only information we can get is what proportion of $dU$ is because of heat supplied and work done.

You may want to note that although $\delta Q$ and $\delta W$ depend on the path taken by the system on the $P-V$ diagram, their sum is independent of a path. Thus the internal energy, $U$, is a state function.

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