Exercises — System vs surroundings; open, closed, isolated

(Recall: ✅ = can cross, ❌ = cannot cross.) (a) Open. The cup is open to air: steam (matter) leaves, and heat leaves to the room. Matter ✅, heat ✅. (b) Closed. The balloon is sealed → no gas escapes (matter ❌). Heat enters from the sun (heat ✅), and because the balloon is flexible it expands, doing work (work ✅). Energy crosses, matter does not → closed. (c) Isolated. Insulated → no heat; sealed → no matter; rigid → no work. Nothing crosses → isolated. (d) Closed. Sealed → matter ❌. Steel conducts → heat ✅. Rigid → volume fixed → no PV work. Energy still crosses (as heat), so it is closed, not isolated.

(a) Sealed → matter ❌; steel conducts → heat ✅; rigid → $\Delta V=0$. Energy crosses, matter does not → closed. (b) Rigid means $\Delta V=0$, so $w=-P_{\text{ext}}\Delta V=0.$ With no work, the First Law collapses to $\Delta U=q$. At fixed volume, $q=q_V=n{C}_V\Delta T=3\times 20.8\times (450-300)=3\times 20.8\times 150=9360\text{J}.$ Therefore $\boxed{w=0,q=9360\text{J},\Delta U=9360\text{J}}$. Every joule of heat became internal energy because no work escaped.

The gas does work on the surroundings, so work done on the system is negative: $w=-2100\text{J}.$ First Law: $\Delta U=q+w\Rightarrow q=\Delta U-w=9360-(-2100)=11460\text{J}.$ $\boxed{q=11460\text{J}}$. More heat is needed here than in 2.1 because part of the heat pays for the expansion work.

This is an open system: mass flows in and out (matter ✅) and heat is supplied (heat ✅). In steady state the temperature inside the heater is constant, so $\Delta U_{\text{device}}=0$ — but energy still streams through as enthalpy carried by the moving matter. The power needed is $\dot{q}=\dot{m}c\Delta T=0.50\times 4.18\times (60-20)=0.50\times 4.18\times 40=83.6\text{kW}.$ $\boxed{\dot{q}=83.6\text{kW}}$. Key open-system idea: matter crossing the boundary carries energy with it, so we track an enthalpy flow $\dot{m}c\Delta T$, not a change in the device's own $\Delta U$.

Take both water masses together as the system. Insulated → no heat leaves → for this system $q=0$; rigid liquid container → $w\approx 0$; so $\Delta U_{\text{total}}=0$. Energy lost by the hot mass equals energy gained by the cold mass: $m_{\text{hot}}c(T_f-70)+m_{\text{cold}}c(T_f-10)=0.$ $c$ cancels. Using $m_{\text{hot}}=2000\text{g}$, $m_{\text{cold}}=1000\text{g}$: $2000(T_f-70)+1000(T_f-10)=0.$ $2000T_f-140000+1000T_f-10000=0\Rightarrow 3000T_f=150000\Rightarrow T_f=50{}^{\circ }\text{C}.$ $\boxed{T_f=50{}^{\circ }\text{C}}$ — closer to the hot side, because there is twice as much hot water. See the balance below.

No — $T_f=50{}^{\circ }\text{C}$ still. The two halves are separate systems that exchange heat through the diathermal wall but no matter (they don't mix). The combined system is still isolated (outer wall insulating), so total internal energy is conserved and the same energy balance holds. Insight: whether matter mixes or only heat flows, energy conservation fixes the same $T_f$. Only the entropy generated differs — mixing (3.1) is more irreversible.

(a) Sealed → matter ❌; adiabatic → heat ❌; movable → work ✅. Matter and heat blocked but work allowed → closed (an adiabatic closed system). It is not isolated, because work can still cross. (b) Adiabatic ⇒ $q=0$. Work is done on the gas ⇒ $w=+500\text{J}$. Then $\Delta U=q+w=0+500=500\text{J}.$ (c) $\Delta U>0$ and for an ideal gas $\Delta U=n{C}_V\Delta T$, so $\Delta T>0$ → temperature rises. All the compression work went straight into internal energy because none could leak out as heat. (This is exactly the logic behind Adiabatic Processes.)

Isothermal ⇒ $\Delta T=0$ ⇒ for an ideal gas $\Delta U=n{C}_V\Delta T=0$. First Law: $0=q+w=q+500\Rightarrow q=-500\text{J}.$ $\boxed{q=-500\text{J}}$: the gas must expel exactly $500\text{J}$ of heat to the surroundings. The same $500\text{J}$ of work entered, but here it leaves as heat instead of raising $T$. The boundary decides the fate of the energy: adiabatic wall → stays as $\Delta U$; diathermal + isothermal → exits as heat.

(a) Insulated → no heat to the outside; sealed bomb → no matter escapes; rigid → no PV work. Toward the outside world it is isolated: $\Delta U_{\text{cal+contents}}=0$. (See Calorimetry.) (b) Split the isolated whole into two internal parts: the reaction and the calorimeter hardware. Their energies must sum to zero: $q_{\text{reaction}}+q_{\text{calorimeter}}=0.$ Now fix the sign convention for the hardware sub-system: when heat enters the calorimeter hardware its temperature rises, so by the same rule as our main system ($q>0$ = heat in), $q_{\text{calorimeter}}>0$ here and equals $q_{\text{calorimeter}}=C_{\text{cal}}\Delta T$ with $\Delta T=299.47-298.00>0$. Substituting, $q_{\text{reaction}}=-q_{\text{calorimeter}}=-C_{\text{cal}}\Delta T.$ The minus sign is now earned: the reaction loses exactly what the hardware gains. (c) $\Delta T=299.47-298.00=1.47\text{K}$, so $q_{\text{reaction}}=-8.50\times 1.47=-12.495\text{kJ}.$ $\boxed{q_{\text{reaction}}\approx -12.50\text{kJ}}$ — negative, meaning the reaction releases energy (exothermic). Because the bomb is rigid, this measured quantity is $\Delta U$ of the reaction, not $\Delta H$ (that distinction is Enthalpy and Constant Pressure).

$q_{\text{reaction}}=-C_{\text{cal}}\Delta T=0$. Two readings:

1. Thermoneutral reaction: the reaction genuinely released/absorbed no net energy ($\Delta U_{\text{rxn}}\approx 0$). Rare but real for some balanced bond-energy cases.
2. Simultaneous exo + endo internal processes that cancel (e.g. bond-forming heat consumed by a phase change inside), so the net $q$ is zero even though energy moved around internally. Either way, the isolated condition still holds — total internal energy is conserved; a flat thermometer just means no net energy landed in the calorimeter's heat-capacity store.

Isolated ⇒ energy conserved ⇒ (equal $m$, equal $c$) final $T$ is the average: $T_f=\frac{400+200}{2}=300\text{K}.$ Why hot→cold? Consider transferring a small positive amount of heat, $\delta q>0$, out of the hot block and into the cold block — this is the candidate "hot→cold" direction, and we take $\delta q$ positive by definition of that direction. The hot block loses $\delta q$ at temperature $T_H$, the cold block gains $\delta q$ at $T_C$, so the total entropy change is $\delta S=-\frac{\delta q}{T_H}+\frac{\delta q}{T_C}=\delta q\left(\frac{1}{T_C}-\frac{1}{T_H}\right).$ Since $\delta q>0$ and $T_C<T_H$ makes $\frac{1}{T_C}>\frac{1}{T_H}$, the bracket is positive, so $\delta S>0$. The Entropy and Second Law permits this. The reverse (cold→hot) would flip the sign of $\delta q$ and give $\delta S<0$ — forbidden in an isolated system. Energy conservation fixes where it ends ($300\text{K}$); the Second Law fixes which way it gets there.

An open system exchanges both matter and energy. A closed system exchanges energy but not matter. An isolated system exchanges neither matter nor energy. A rigid wall forbids PV (boundary) work but says nothing about heat. A bomb calorimeter is rigid, so its measured reaction heat equals ==$\Delta U$== (not $\Delta H$).