Worked examples — System vs surroundings; open, closed, isolated

You have met three boundaries: open, closed, isolated. Now we drive them through every case they can throw at you: matter crossing or not, heat crossing or not, work switched on or off, degenerate zeros, limiting values, a word problem, and an exam twist. We build every symbol before using it.

Look at the first figure: the orange arrows are matter crossing, the magenta arrows are heat/work energy crossing. Reading left to right you can watch both switches turn off one at a time — open (both on), closed (matter off), isolated (all off).

Before any numbers, three words earn their meaning:

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The scenario matrix

Every situation this topic hands you lands in one of these cells. The examples that follow are tagged to the cell they cover.

Cell	Matter crosses?	Heat crosses?	Work ($\Delta V$)?	Kind of case	Example
A	❌	✅	rigid, $w=0$	closed + fixed volume	Ex 1
B	❌	✅	movable, $w\ne 0$	closed + expansion	Ex 2
C	❌	✅	movable, compression ($\Delta V<0$)	sign flip on $w$	Ex 3
D	❌	❌ (adiabatic)	movable	[[Adiabatic Processes	adiabatic]] closed
E	❌	❌	rigid	isolated (all switches OFF)	Ex 5
F	✅	✅	—	open system, real world	Ex 6
G	✅	✅	—	open, steady state ($\Delta U=0$ limiting case)	Ex 7
H	boundary choice / degenerate	—	—	zero-input & classification twist	Ex 8

The dimensions we sweep: matter ON/OFF, heat ON/OFF, work +/0/−, the degenerate zero ($\Delta V=0$ or $\Delta T=0$), the limiting case (steady state $\Delta U=0$), a word problem (Ex 6), and an exam twist (Ex 8).

Constants we reuse: for one mole of an ideal diatomic gas the molar heat capacity at constant volume is $C_V\approx 20.8{\text{J⋅mol}}^{-1}{\text{K}}^{-1}$, and $C_P=C_V+R\approx 29.1{\text{J⋅mol}}^{-1}{\text{K}}^{-1}$ with $R=8.314{\text{J⋅mol}}^{-1}{\text{K}}^{-1}$.

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Cell A — closed, rigid: all heat becomes internal energy

Forecast: guess now — will the work be positive, negative, or zero? Where does the heat energy go?

1. Classify. Sealed ⇒ matter switch OFF ⇒ closed. Steel conducts ⇒ heat switch ON. Why this step? The switches tell us which terms in $\Delta U=q+w$ survive.
2. Kill the work term. Rigid ⇒ $\Delta V=0$ ⇒ $w=-P_{\text{ext}}\Delta V=0$. Why this step? "Rigid" is a mechanical fact, separate from "sealed". No volume change means no push-through-distance, so no work.
3. Reduce the First Law. $\Delta U=q+0=q$. Why this step? With $w=0$, every joule of heat is stored, not spent.
4. Compute using $C_V$. At constant volume, $\Delta U=n{C}_V\Delta T$. $\Delta U=3\times 20.8\times (350-300)=3120\text{J}$ Why this step? $C_V$ is defined as energy-per-degree at fixed volume — exactly our situation (Heat Capacity at Constant Volume).
5. So $q=3120\text{J}$, $w=0$, $\Delta U=3120\text{J}$.

Verify: Units: $\text{mol}\times \frac{\text{J}}{\text{mol⋅K}}\times \text{K}=\text{J}$ ✓. Heat entered ($\Delta T>0$) so $q>0$ ✓, consistent with $\Delta U>0$.

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Cell B — closed, movable: gas pushes out, work is negative

Forecast: the gas pushes the piston outward — does the gas gain or lose energy from that push?

1. Classify. Piston seals gas in ⇒ closed. Piston moves ⇒ work switch ON. Heat given ⇒ heat switch ON. Why this step? Naming the switches tells us that both $q$ and $w$ are live terms here — unlike Ex 1 where the rigid wall zeroed $w$.
2. Volume change. $\Delta V=V_f-V_i=(5.0-2.0)\times 10^{-3}=3.0\times 10^{-3}{\text{m}}^3>0$. Why this step? The sign of $\Delta V$ decides the sign of the work.
3. Work on the system. $w=-P_{\text{ext}}\Delta V=-(1.00\times 10^5)(3.0\times 10^{-3})=-300\text{J}$. Why this step? Expansion pushes the surroundings, so energy leaves the system as work ⇒ $w<0$.
4. First Law. $\Delta U=q+w=600+(-300)=+300\text{J}$. Why this step? $\Delta U=q+w$ is the only bridge from the two crossing quantities to the stored energy; with both terms now known we simply add them.

Verify: $\text{Pa}\times {\text{m}}^3=\frac{\text{N}}{{\text{m}}^2}{\text{m}}^3=\text{N⋅m}=\text{J}$ ✓. Heat in beat work out, so internal energy still rose ✓.

Look at the second figure: the orange arrow shows the piston sliding outward, and the magenta labels mark $\Delta V>0$ leading to $w=-P\Delta V<0$. Compare the wider "final" cylinder to the narrow "initial" one — that widening is the expansion that costs the gas energy.

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Cell C — closed, movable: compression flips the sign

Forecast: we push in on the gas — should its energy go up or down from the push?

1. Volume change. $\Delta V=2.0\times 10^{-3}-5.0\times 10^{-3}=-3.0\times 10^{-3}{\text{m}}^3<0$. Why this step? Compression means $\Delta V$ is negative — the mirror of Ex 2.
2. Work. $w=-P_{\text{ext}}\Delta V=-(1.00\times 10^5)(-3.0\times 10^{-3})=+300\text{J}$. Why this step? Two minus signs give $+$: energy is delivered into the gas ⇒ $w>0$.
3. First Law. $\Delta U=q+w=-250+300=+50\text{J}$. Why this step? $\Delta U=q+w$ combines the two channels; here heat left ($q<0$) but work entered ($w>0$), and only the First Law tells us the net effect on stored energy.

Verify: Sign check — pushing in adds energy ($w>0$), but heat leaked ($q<0$); the small net $+50\text{J}$ says the push won narrowly ✓. This is the sign-flipped partner of Ex 2, covering the $\Delta V<0$ corner.

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Cell D — closed, adiabatic: no heat, work only

Forecast: if you squeeze a gas that can't dump heat, does it get hotter or colder?

1. Classify. Sealed ⇒ closed; insulated ⇒ heat switch OFF ⇒ $q=0$. This is an adiabatic process. Why this step? Setting the heat switch OFF is what makes $q=0$; recognizing it now is what lets us drop the heat term in the very next line.
2. First Law with $q=0$. $\Delta U=0+w=+450\text{J}$. Why this step? Adiabatic kills the heat term, so all the energy change is the work done on the gas.
3. Back out the temperature rise. Internal energy of an ideal gas depends only on $T$: $\Delta U=n{C}_V\Delta T$, so $\Delta T=\frac{\Delta U}{n{C}_V}=\frac{450}{1\times 20.8}=21.6\text{K}.$ Why this step? $C_V$ converts stored energy into a temperature change even though the volume itself changed — because for an ideal gas $U$ is a function of $T$ alone.

Verify: $\Delta U>0$ and $\Delta T>0$ — compressing an insulated gas heats it ✓ (this is why a bike pump warms). Units: $\text{J}/(\text{mol}\cdot \frac{\text{J}}{\text{mol⋅K}})=\text{K}$ ✓.

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Cell E — isolated: all switches OFF (equilibration)

Forecast: the masses are unequal — will $T_f$ be exactly halfway ($50{}^{\circ }\text{C}$), above, or below?

1. Classify. No mass leaves, no heat leaves, rigid ⇒ isolated. Hence $q_{\text{system}}=0$ and $\Delta U_{\text{total}}=0$.
2. Energy bookkeeping inside. Heat lost by hot = heat gained by cold (Calorimetry): $m_{h}c(T_f-80)+m_{c}c(T_f-20)=0.$ Why this step? Inside an isolated system energy is merely redistributed; $c$ cancels since both are water.
3. Solve. $200(T_f-80)+300(T_f-20)=0\Rightarrow 500T_f=200(80)+300(20)=22000$, so $T_f=\frac{22000}{500}=44{}^{\circ }\text{C}.$ Why this step? The final temperature is the mass-weighted average, not the plain average — the larger cold mass pulls it below $50{}^{\circ }\text{C}$.

Verify: Heat from hot $=200(4.18)(80-44)=30,096\text{J}$; heat to cold $=300(4.18)(44-20)=30,096\text{J}$. Equal ✓, so $\Delta U_{\text{total}}=0$ ✓.

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Cell F — open system, real-world word problem

Forecast: which switches are ON, and does energy leave only as heat?

1. Classify. Steam escapes ⇒ matter switch ON; burner heats and vapour carries energy out ⇒ heat switch ON. Both ON ⇒ open system.
2. Track matter. $\Delta m=-12\text{g}$ of water leaves the system boundary (the kettle interior). Why this step? In open systems mass itself carries energy across, so $\Delta m\ne 0$ matters.
3. Why the closed formula fails. Escaping steam carries enthalpy out with it. Enthalpy, symbol $H$, is defined as $H=U+PV$ — the internal energy of a chunk of matter plus the $PV$ "room-making" energy it needed to occupy its space (Enthalpy and Constant Pressure). Every gram of steam takes its share of $H$ away, an extra energy channel beyond $q$ and $w$. So the plain $\Delta U=q+w$ (which assumes fixed mass) must be extended with a mass-flow term — that is exactly why chemistry prefers closed systems.
4. Quantify the carried energy. Latent heat of vaporization $L\approx 2260{\text{J⋅g}}^{-1}$, so energy leaving as vapour $=12\times 2260=27120\text{J}\approx 2.71\times 10^4\text{J}$.

Verify: Units $\text{g}\times {\text{J⋅g}}^{-1}=\text{J}$ ✓. This $27\text{kJ}$ is separate from any $q$ or $w$ — confirming an open system needs its own accounting.

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Cell G — open, steady state: the limiting $\Delta U=0$ case

Forecast: mass flows in AND out — is $\Delta U$ of the heater zero, positive, or negative?

1. Classify. Water in, water out ⇒ open. But the heater's own contents don't accumulate energy ⇒ steady state, the limiting case where $\Delta U_{\text{heater}}=0$. Why this step? This kills the trap "open ⇒ $\Delta U\ne 0$": here input rate = output rate.
2. Name the flow rate. Let $\dot{m}$ = mass-flow rate, the mass of water crossing the boundary each second; here $\dot{m}=0.10{\text{kg⋅s}}^{-1}$. Energy balance per second = heat needed to warm the mass passing through in that second: $P=\dot{m}c\Delta T.$ Why this step? With $\Delta U=0$, all supplied heat goes straight into the outgoing stream; per-second accounting turns a heat into a power (energy per second).
3. Compute. $P=0.10\times 4180\times (45-15)=0.10\times 4180\times 30=12540\text{W}\approx 12.5\text{kW}$. Why this step? Plugging the named numbers into $P=\dot{m}c\Delta T$ turns the balance equation into the single quantity asked for — the power the heater must deliver.

Verify: Units $\frac{\text{kg}}{\text{s}}\cdot \frac{\text{J}}{\text{kg⋅K}}\cdot \text{K}=\frac{\text{J}}{\text{s}}=\text{W}$ ✓. Positive power to warm water ✓; and $\Delta U_{\text{heater}}=0$ despite mass flowing — the steady-state limiting corner.

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Cell H — degenerate & exam-twist classification

Forecast: does the system type of the very same ampoule change when its surroundings change?

1. (a) Degenerate zero-input case. Sealed ⇒ closed. Room and ampoule same $T$ ⇒ no heat flows: $q=0$. Rigid glass ⇒ $w=0$. Hence $\Delta U=0+0=0$. Why this step? Being closed does not force any energy transfer; with no driving temperature difference and no volume change, every term is zero. Closed $\ne$ "something must happen."
2. (b) Same object, hot bath. Still sealed ⇒ still closed. Now a temperature difference exists ⇒ $q>0$; glass rigid ⇒ $w=0$; so $\Delta U=q>0$. Why this step? The object's type is set by its boundary's matter switch, but the values of $q$ depend on the surroundings — two different questions.
3. (c) Re-drawn boundary. Ampoule + oil, insulated: no matter out, no heat out, rigid overall ⇒ isolated. Inside, heat flows from oil to gas until equal $T$, but $\Delta U_{\text{total}}=0$. Why this step? System type is a choice of boundary, not a property of the stuff. Move the boundary, change the classification.

Verify: (a) $\Delta U=0+0=0$ ✓ (degenerate corner). (c) energy conserved across the enlarged boundary, $\Delta U_{\text{total}}=0$ ✓. This example alone sweeps the zero-input degenerate case and the boundary-choice exam twist.

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Recall Quick self-test

Ex 1 answer $\Delta U$ (J) ::: $3120$ Ex 2 work $w$ (J) ::: $-300$ Ex 3 work $w$ (J), and $\Delta U$ (J) ::: $w=+300$, $\Delta U=+50$ Ex 4 temperature rise $\Delta T$ (K) ::: $21.6$ Ex 5 final temperature (°C) ::: $44$ Ex 6 energy carried by steam (J) ::: $27120$ Ex 7 heater power (kW) ::: $12.5$ Ex 8a value of $\Delta U$ ::: $0$

See also: State Functions vs Path Functions (why $\Delta U$ cares only about start/end while $q,w$ depend on route), and Entropy and Second Law (isolated systems only).