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There is a cylinder-piston system with some gas in it, and let's say this system is a closed system. Say that the system is submerged in a large beaker of water whose temperature is $25\ \mathrm{^\circ C}$. A spark is used to trigger a reaction in the system, and after the reaction is completed, the temperature of the water bath increased to $28\ \mathrm{^\circ C}$, and the piston went inside the cylinder.

By instinct, I think that heat is lost by the system since the surrounding(water)'s temperature increased, and work has been done to the system. However, when I consider the thermal equilibrium, the system's temperature must be $28\ \mathrm{^\circ C}$ as well. In this case, is heat added to the system or not? Furthermore, in the equation $\Delta E = q+w$, what are the signs (plus or minus) of the three terms?

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  • $\begingroup$ You have a system and a water bath, both at 25°C before and 28°C after. I would definitively say q is negative, wouldnt you. And you have done no work on the outside, on the contrary, the outside has pushed in the cylinder, so w is positive. Whats your guess on the sign of deltaE? $\endgroup$ – Karl Sep 7 at 8:22
  • $\begingroup$ @Karl Maybe negative, but we don't know for sure unless the actual numbers are goven, right? $\endgroup$ – John. P Sep 7 at 8:34
  • $\begingroup$ Thats it, exactly. $\endgroup$ – Karl Sep 8 at 17:12
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Chemical energy of system has been converted to thermal energy of both the system and the surrounding water.

Additionally, water did work on system, as the system shrinked down.

$$\Delta E_\mathrm{system}=-C_\mathrm{water} \cdot \Delta T_\mathrm{water}-p \cdot \Delta V_\mathrm{system}$$

For $H=E + p \cdot V$, $\Delta H=\Delta E + p \cdot \Delta V + V \cdot \Delta p$

As p is constant,

$\Delta H=\Delta E + p \cdot \Delta V $

$\Delta H_\mathrm{system}=-C_\mathrm{water} \cdot \Delta T_\mathrm{water}$

where $H$ means enthalpy, energy at the constant pressure.

If we considered other thermodynamic quantities like Gibbs energy or entropy, we would need to examine the underwent reaction and system composition. But if the system energy is considered, its change is related just to exchanged thermal energy and work.

$q$ in the equation $\Delta E = q+w$ means heat transfers to the system, so it is negative. $w$ is work done on system and is positive. $\Delta E$ is very probably negative, as work value would be much less than transferred heat.

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    $\begingroup$ In my judgment, this answer is not 100 % correct and, in addition, does not discuss an important feature of the reaction. See my answer. $\endgroup$ – Chet Miller Sep 7 at 15:25
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    $\begingroup$ I always leave space for other answers. $\endgroup$ – Poutnik Sep 7 at 15:28
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The first law of thermodynamics tells us that, for this process at constant pressure, $$\Delta E_\text{system}=q_\text{system}-p\,\Delta V\tag{1}$$and $$MC_\mathrm s\,\Delta T_\mathrm w=-q_\text{system}\tag{2}$$where $\Delta V$ is the change in volume of the system. Eqn. 1 and 2, in terms of system enthalpy, can also be combined and written as $$\Delta H_\text{system}=q_\text{system}=-MC_\mathrm s\,\Delta T_\mathrm w\tag{3}$$So there is no $p\,\Delta V$ in this equation.

What is happening in this system is that (1) the reaction is exothermic so that heat is transferred from the system to the water and the water temperature rises and (2) the total number of moles of gas decreases in the reaction so that the volume of the system decreases and the piston moves inward.

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  • $\begingroup$ So, what about the signs $\Delta E$, $p$, and $w$ of OP's equation in this instance? $\endgroup$ – Mathew Mahindaratne Sep 7 at 18:04
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    $\begingroup$ @Mathew Mahindaratne The op is using the convention that w represents work done on the system. $\endgroup$ – Chet Miller Sep 7 at 18:35
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I would like to address one of the OP's questions about heat transfer directly because the existing answers did not (yet):

[OP:] I think that heat is lost by the system since the surrounding(water)'s temperature increased, and work has been done to the system. However, when I consider the thermal equilibrium, the system's temperature must be 28 ∘C as well. In this case, is heat added to the system or not?

The short answer is: no heat is added to the system; the system gives off heat.

Heat is a transfer of energy from one place to another. So you can't "add heat" somewhere without "losing heat" somewhere else. In a heat transfer across a boundary, $q$ is positive for one side of the boundary and negative (but of equal magnitude) for the other.

The surrounding (in this case the water bath) is often chosen to be "boring" (no chemical reactions, for sure), while the system can have all kinds of stuff going on (a spark, a chemical reaction, large fluctuations in temperature). This way, you can measure the heat transfer via temperature changes in the water bath.

In the case at hand, both system and surrounding show an increase in temperature after things have calmed down. That tells us that some other form of energy (potential energy or specifically "chemical" energy in this case) has been transformed into thermal energy. This took place in the system (the surrounding is too "boring" for that).

To get back to the question I wanted to address:

... is heat added to the system or not?

The thermal energy of the system increases because of the exothermic chemical reaction. Some of that thermal energy is lost to the surrounding through heat transfer, warming up the surrounding. You could imaging that the temperature of the system first went up a lot (if the reaction is rapid), and then cooled down to 28 $^\circ$C, heating up the surrounding from 25$^\circ$C to 28 $^\circ$C. Or you could skip the details and just compare the initial and final state and not worry what happened in between, in the spirit of thermodynamics. The conclusion is that heat is transferred from the system to the surrounding. Nevertheless, the temperature of the system increased because of the exothermic reaction, which generated more thermal energy than the system gave off. In other words, no heat is added to the system; the system gives off heat.

In this explanation, I have set aside the work done on the system. The OP is vague about whether the process runs at constant pressure or not, and we don't know the magnitude of the work compared to the heat transfer; for my answer, you should assume that the magnitude of the work is negligible compared to that of the heat transfer, otherwise some of the statements might become inaccurate (such as "the exothermic reaction, which generated more thermal energy than the system gave off").

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The problem is that in your last wondering, i.e. "wouldn't the system be at a higher temperature because it's 28 C?" is that you're confusing the system and its environment. The system here is the cylinder full of explosive chemical, the environment is the surrounding water. Note that technically, though, there is no reason that you can't call the surrounding water a "system" and the cylinder an "environment" - it sounds weird to our ears but it works out in theory, it's just that usually an "environment" is much larger/more complex than we can describe in detail in real cases. The real trick is in being consistent in your identification of such and such throughout.

In the thermodynamic equation

$$\Delta E = Q + W$$

the $E$ referred to here is the energy of the system, its internal energy (sometimes also denoted by $U$). Hence, $Q$ is the heat input to the system, and likewise, $W$ is the work input: to see this, consider the case where one is zero and the other nonzero, to leave only one to consider. Now, if $\Delta E$ is positive, then the nonzero one must be positive. Moreover, since this represents an increase in the system energy, that implies that both $Q$ and $W$ measure inputs, and their sign comes to be thus.

The energy change in the environment, e.g. $\Delta E_\mathrm{env}$, is, then, just the negative of this (alternatively, just exchange the roles of system and environment as mentioned before), in order to make it balance.

Hence, in your particular case, since the system both deposited heat into the environment and did work on it, for the system, $Q$ and $W$ are both negative: negative heat/work "received by" the system is equivalent to such done by the system. But in the case of the environment-as-system, its $Q_\mathrm{env}$ and $W_\mathrm{env}$ will then be positive, since it is receiving both from the cylinder.

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