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Consider a room with air and an electric heater inside. I want to determine what will be the temperature of the air in the room after 1 hour of starting the heater (whose power rating is known). The walls have cracks from which some air can escape. The air remains at a constant pressure.

For this analysis

  1. I can choose the air as the system. So heater, walls, and everything else in the universe constitute surroundings. When the heater is turned on, and heat transfer takes place to the air, it expands and some of it leaves through the cracks. Since air is the system we will need to keep track of it as it leaves the room. Since only energy crosses the boundaries but no mass, this is a closed system.

  2. I can take the volume formed by the inner surfaces of the walls as the system. Both mass and energy cross the boundaries of this system so it is an open system.

In this problem, it will be better to take air as the system, i.e. it will be better to analyze this problem by considering a closed system. This is because the change in enthalpy of the air will be equal to the heat transferred. In analyzing this problem using an open system I will have to take care of the rate at which air leaves the cracks.

As seen in this problem, even though it is difficult to analyze the situation by considering an open system, it is at least possible.

So my question is, does this apply to every analysis? Every problem can be solved by considering an open or a closed system, what matters is which kind of system makes your analysis easier, is it like that?

For instance, can I analyze biological systems by considering them as closed systems?

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  • $\begingroup$ If air can leave through cracks, this cannot be considered as a closed system. Also, there are sources of energy in the setup - as, e.g., the heater, so it is not even something that could be expected to be in thermodynamic equilibrium. Related: physics.stackexchange.com/a/645441/247642 $\endgroup$
    – Roger V.
    Commented Jun 27, 2022 at 7:18
  • $\begingroup$ Even with the closed system case, you will need to specify the amount of air leaving through the cracks, and account for the changes in this air. Its final state will certainly not be the same as that of the air remaining in the room. $\endgroup$
    – Chet Miller
    Commented Jun 27, 2022 at 10:03
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    $\begingroup$ This problem is solved incorrectly for the case of the house with cracks. This is because not all the air leaving through the cracks is heated to 70 F. Some of it leaves at temperatures close to 50 F. $\endgroup$
    – Chet Miller
    Commented Jun 27, 2022 at 13:41
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    $\begingroup$ The case of leaking cracks is an interesting problem in the application of the open system version of the 1st law of thermodynamics. Any interest in solving it correctly? $\endgroup$
    – Chet Miller
    Commented Jun 27, 2022 at 22:37
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    $\begingroup$ This is a good start. Rather than including the heater in the control volume, choose the room minus the heater as the control volume. Then the focus will be on the air, and you can work with the heat Q from the heater rather than the electrical work W. Also, assume that the heat addition is gradual, so that the 1st law can be expressed in differential form: dU=d(nu)=dQ+hdn, where n is the number of moles of air in the room (such that dn is negative). Here h=u+Pv=u+RT $\endgroup$
    – Chet Miller
    Commented Jun 28, 2022 at 14:56

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