3.P — Worked Problems: Thermodynamics

Fourteen problems across Part 3, climbing from standard to genuinely awkward. Every solution shows the arithmetic. Attempt each before opening it.

Problem 1 — The mixture that ends up as ice

A 40 g ice cube at −15 °C is dropped into 200 g of water at 25 °C in an insulated cup. Find the final temperature and the final contents. Take c_{\text{ice}} = 2100, c_{\text{water}} = 4186, L_f = 3.34\times10^5 J/kg.

Solution

Never assume the answer. Compute the energy available and the energy demanded, then see which runs out first.

Energy the warm water can give before it reaches 0 °C:

Q_{\text{available}} = (0.200)(4186)(25) = 20{,}930\ \text{J}

Energy the ice demands to reach 0 °C and then melt fully:

Q_{\text{warm ice}} = (0.040)(2100)(15) = 1260\ \text{J}

Q_{\text{melt}} = (0.040)(334000) = 13{,}360\ \text{J}

Q_{\text{demanded}} = 14{,}620\ \text{J}

Available (20,930) exceeds demanded (14,620), so all the ice melts and the final state is entirely liquid above 0 °C. Now solve for that temperature.

Heat given up by the original water, cooling from 25 °C to T:

Q_{\text{out}} = (0.200)(4186)(25 - T) = 837.2(25-T)

Heat taken by the ice: warming to 0, melting, then warming as water from 0 to T:

Q_{\text{in}} = 1260 + 13360 + (0.040)(4186)T = 14620 + 167.4\,T

Set them equal:

837.2(25-T) = 14620 + 167.4T

20930 - 837.2T = 14620 + 167.4T

6310 = 1004.6\,T

T = 6.28\ ^\circ\text{C}

What to notice. Change the ice to 80 g and the demanded energy becomes 29,240 J, more than the 20,930 available, so not all of it melts. The final temperature would then be exactly 0 °C with ice and water coexisting, and the unknown becomes how much melted rather than what temperature was reached. Always check which case you are in before writing an equation.

Problem 2 — Rms speed and a light gas

Find v_{\text{rms}} for helium (molar mass 4.0 g/mol) at 300 K, and the temperature at which oxygen (32 g/mol) would have the same v_{\text{rms}}.

Solution

Mass of one helium atom:

m = \frac{0.004}{6.022\times10^{23}} = 6.64\times10^{-27}\ \text{kg}

v_{\text{rms}} = \sqrt{\frac{3k_BT}{m}} = \sqrt{\frac{3(1.381\times10^{-23})(300)}{6.64\times10^{-27}}} = \sqrt{\frac{1.243\times10^{-20}}{6.64\times10^{-27}}}

v_{\text{rms}} = \sqrt{1.872\times10^{6}} = 1368\ \text{m/s}

For oxygen to match, 3k_BT/m must be equal, so T \propto m. Oxygen is 8 times heavier:

T_{\text{O}_2} = 8\times300 = 2400\ \text{K}

What to notice. 1368 m/s at room temperature is four times the speed of sound. Earth's escape velocity is 11.2 km/s, only about 8 times this, and the Maxwell–Boltzmann tail reaches that far often enough over billions of years that Earth has lost essentially all its primordial helium. The helium in party balloons and MRI magnets is dug out of the ground, produced by alpha decay of uranium and thorium and trapped under rock, and it is a genuinely finite resource — once released it leaves the planet.

Problem 3 — A cycle with three legs

Two moles of a monatomic ideal gas start at 300 K and 2.0×10⁵ Pa. The gas is (a) heated at constant pressure until its volume doubles, (b) cooled at constant volume back to 300 K, (c) compressed isothermally back to the start. Find W, Q and \Delta U for each leg and the cycle efficiency.

Solution

Starting volume:

V_1 = \frac{nRT}{P} = \frac{(2)(8.314)(300)}{2.0\times10^{5}} = 0.02494\ \text{m}^3

Leg (a), isobaric, volume doubles to 0.04988 m³. At constant pressure V \propto T, so the temperature doubles to 600 K.

W_a = P\Delta V = (2.0\times10^{5})(0.02494) = 4988\ \text{J}

\Delta U_a = nC_V\Delta T = (2)(1.5\times8.314)(300) = 7483\ \text{J}

Q_a = \Delta U_a + W_a = 12{,}471\ \text{J}\ \text{(in)}

Leg (b), isochoric, back to 300 K.

W_b = 0

\Delta U_b = (2)(1.5\times8.314)(-300) = -7483\ \text{J}

Q_b = -7483\ \text{J}\ \text{(out)}

Leg (c), isothermal at 300 K, from 0.04988 back to 0.02494 m³.

W_c = nRT\ln\frac{V_1}{V_2} = (2)(8.314)(300)\ln(0.5) = 4988\times(-0.6931) = -3457\ \text{J}

\Delta U_c = 0, \qquad Q_c = -3457\ \text{J}\ \text{(out)}

Cycle totals.

\Delta U = 7483 - 7483 + 0 = 0\ \checkmark

W_{\text{net}} = 4988 + 0 - 3457 = 1531\ \text{J}

Q_{\text{net}} = 12471 - 7483 - 3457 = 1531\ \text{J}\ \checkmark

Efficiency uses only the heat that went in, which is leg (a) alone:

\eta = \frac{1531}{12471} = 12.3\ \%

Compare with a Carnot engine between 600 K and 300 K: \eta = 1 - 300/600 = 50 %.

What to notice. This cycle achieves a quarter of the Carnot efficiency, and the reason is that most of its heat transfer happens across large temperature gaps. Leg (a) heats gas from 300 K to 600 K, which means putting heat into gas that is at every temperature between, from a source that must be at 600 K — and every joule crossing a gap generates entropy, which Chapter 3.5 showed is work destroyed.

Problem 4 — Adiabatic compression in a diesel

A diesel engine draws in air at 300 K and 1.0 atm and compresses it adiabatically with a compression ratio of 20. Find the final temperature and pressure. Then find the work done per mole. Take \gamma = 1.40, C_V = \frac{5}{2}R.

Solution

Temperature, from TV^{\gamma-1} constant:

T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} = 300\times20^{0.4}

\ln 20 = 2.9957, \quad 0.4\times2.9957 = 1.1983, \quad e^{1.1983} = 3.314

T_2 = 300\times3.314 = 994\ \text{K}

Pressure, from PV^{\gamma} constant:

P_2 = P_1\left(\frac{V_1}{V_2}\right)^{\gamma} = 1.0\times20^{1.4}

1.4\times2.9957 = 4.194, \quad e^{4.194} = 66.3

P_2 = 66.3\ \text{atm}

Work. Adiabatic means Q = 0, so W = -\Delta U:

\Delta U = nC_V\Delta T = (1)(2.5\times8.314)(994-300) = 20.785\times694 = 14{,}425\ \text{J}

W = -14{,}425\ \text{J per mole}

Negative, meaning 14.4 kJ of work was done on each mole of air.

What to notice. Check the pressure ratio against the naive expectation. Compressing 20-fold isothermally would give 20 atm; adiabatically it gives 66 atm, because the gas also got hot. The ratio 66.3/20 = 3.31 is exactly the temperature ratio, as PV = nRT demands. This is why diesel engines must be built far more heavily than petrol engines: they contain three times the pressure.

Problem 5 — Two Carnot engines in series

A Carnot engine works between 800 K and an intermediate temperature T_m; its entire waste heat feeds a second Carnot engine working between T_m and 300 K. Show that the combined efficiency does not depend on T_m, and find it.

Solution

Let the first engine take in Q_H at 800 K. Its efficiency is \eta_1 = 1 - T_m/800, so its work is:

W_1 = Q_H\left(1 - \frac{T_m}{800}\right)

and it rejects:

Q_m = Q_H\frac{T_m}{800}

The second engine takes that in and produces:

W_2 = Q_m\left(1 - \frac{300}{T_m}\right) = Q_H\frac{T_m}{800}\left(1 - \frac{300}{T_m}\right) = Q_H\left(\frac{T_m}{800} - \frac{300}{800}\right)

Total work:

W = W_1 + W_2 = Q_H\left(1 - \frac{T_m}{800} + \frac{T_m}{800} - \frac{300}{800}\right) = Q_H\left(1 - \frac{300}{800}\right)

The T_m terms cancel exactly, and:

\eta = 1 - \frac{300}{800} = 62.5\ \%

which is precisely the efficiency of a single Carnot engine between 800 K and 300 K.

What to notice. This had to happen, and the reason is the argument from Chapter 3.4: all reversible engines between the same two reservoirs have the same efficiency. The pair, viewed as one box, takes heat at 800 K and rejects it at 300 K reversibly, so it must match. If cascading engines could beat a single one, you could cascade forever and exceed the Carnot limit. The result is also why real combined-cycle power plants work: a gas turbine's exhaust at 900 K drives a steam cycle, and the pair reaches about 60 % where each alone reaches 40 %. They are not beating Carnot — they are stopping the exhaust heat from being thrown away across a gap.

Problem 6 — Entropy of an irreversible mixing

Two kilograms of water at 80 °C are mixed with 3 kg at 20 °C in an insulated container. Find the final temperature and the entropy change of the universe.

Solution

Final temperature, from heat lost equals heat gained:

2(4186)(80-T) = 3(4186)(T-20)

The 4186 cancels:

2(80-T) = 3(T-20) \quad\Longrightarrow\quad 160 - 2T = 3T - 60 \quad\Longrightarrow\quad 220 = 5T

T = 44\ ^\circ\text{C} = 317\ \text{K}

Entropy. Mixing is irreversible, so use a reversible path: imagine cooling the hot water slowly through a sequence of reservoirs, and warming the cold water the same way. Each gives \Delta S = mc\ln(T_f/T_i) with temperatures in kelvin.

Hot water, 353 K → 317 K:

\Delta S_1 = (2)(4186)\ln\frac{317}{353} = 8372\times\ln(0.8980) = 8372\times(-0.10758) = -900.7\ \text{J/K}

Cold water, 293 K → 317 K:

\Delta S_2 = (3)(4186)\ln\frac{317}{293} = 12558\times\ln(1.08191) = 12558\times0.07872 = 988.6\ \text{J/K}

\Delta S_{\text{universe}} = -900.7 + 988.6 = +87.9\ \text{J/K}

Positive, as required.

What to notice. The lost work is T_0\Delta S = 293\times87.9 = 25.8 kJ, taking the ambient as 20 °C. That is what you could have extracted by running an engine between the two bodies as they equilibrated instead of simply pouring them together. Every time hot and cold are mixed directly, that much work is thrown away — which is why industrial plants use heat exchangers to transfer heat across the smallest temperature gap they can afford, rather than mixing streams.

Problem 7 — Which is bigger, the melt or the heat?

A 2 kg block of ice at 0 °C is melted and the resulting water is then heated to 100 °C. Compare the entropy change of each stage.

Solution

Melting, at constant 273 K:

\Delta S_1 = \frac{mL_f}{T} = \frac{(2)(334000)}{273} = \frac{668000}{273} = 2447\ \text{J/K}

Heating, 273 K → 373 K:

\Delta S_2 = mc\ln\frac{373}{273} = (2)(4186)\ln(1.3663) = 8372\times0.31220 = 2614\ \text{J/K}

Almost equal — 2447 against 2614.

What to notice. Melting a block of ice changes its entropy by about as much as heating the resulting water across the entire liquid range. The melt happens at one temperature and takes 668 kJ; the heating takes 837 kJ spread over 100 degrees. The reason they come out similar is the 1/T weighting: the melt's heat all enters at the lowest temperature in the problem, where each joule buys the most entropy. Now compare with boiling: \Delta S = 2\times2260000/373 = 12{,}118 J/K, nearly five times either. The gas phase is where the arrangements really multiply.

Problem 8 — Free expansion versus isothermal expansion

One mole of ideal gas at 300 K doubles its volume, once by free expansion into vacuum and once by slow isothermal expansion. Compare Q, W, \Delta U, \Delta S_{\text{gas}} and \Delta S_{\text{universe}} for the two.

Solution

Free expansion. No external pressure, so W = 0. Insulated, so Q = 0. Therefore \Delta U = 0 and T is unchanged.

For entropy, use a reversible path between the same two states:

\Delta S_{\text{gas}} = nR\ln 2 = 8.314\times0.6931 = 5.76\ \text{J/K}

Nothing else changed, so \Delta S_{\text{universe}} = +5.76 J/K.

Isothermal expansion. \Delta U = 0 (same temperature), so Q = W:

W = nRT\ln 2 = (1)(8.314)(300)(0.6931) = 1729\ \text{J}

Q = 1729\ \text{J}

\Delta S_{\text{gas}} = \frac{Q}{T} = \frac{1729}{300} = 5.76\ \text{J/K}

The same, as it must be — same initial and final states, and S is a state function. But the reservoir lost 1729 J at 300 K:

\Delta S_{\text{reservoir}} = -\frac{1729}{300} = -5.76\ \text{J/K}

\Delta S_{\text{universe}} = 5.76 - 5.76 = 0

What to notice. The gas ends in exactly the same state both times, and its own entropy change is identical. What differs is the universe's total: zero for the reversible route, +5.76 J/K for the free one. And the difference in usefulness is stark — the isothermal route delivered 1729 J of work, the free expansion delivered none. The 1729 J is precisely T\Delta S_{\text{universe}} = 300\times5.76. Entropy generated is work forfeited, exactly.

Problem 9 — Van der Waals versus ideal

One mole of CO₂ occupies 0.500 L at 300 K. Find the pressure from the ideal gas law and from van der Waals, and say which correction dominates. Take a = 0.364 Pa m⁶/mol², b = 4.27\times10^{-5} m³/mol.

Solution

V = 5.00\times10^{-4} m³.

Ideal:

P = \frac{RT}{V} = \frac{(8.314)(300)}{5.00\times10^{-4}} = \frac{2494}{5.00\times10^{-4}} = 4.99\times10^{6}\ \text{Pa} = 49.2\ \text{atm}

Van der Waals:

P = \frac{RT}{V-b} - \frac{a}{V^2}

V - b = 5.00\times10^{-4} - 4.27\times10^{-5} = 4.573\times10^{-4}\ \text{m}^3

\frac{RT}{V-b} = \frac{2494}{4.573\times10^{-4}} = 5.454\times10^{6}\ \text{Pa}

\frac{a}{V^2} = \frac{0.364}{(5.00\times10^{-4})^2} = \frac{0.364}{2.50\times10^{-7}} = 1.456\times10^{6}\ \text{Pa}

P = 5.454\times10^{6} - 1.456\times10^{6} = 4.00\times10^{6}\ \text{Pa} = 39.5\ \text{atm}

The ideal law overestimates by 25 %.

What to notice. The two corrections pull opposite ways. The excluded volume b raised the pressure by 0.46\times10^6 Pa, because the gas has less room than you thought. The attraction a lowered it by 1.46\times10^6 Pa, three times more, because molecules pull each other back from the walls. At this density the attraction wins, which is the general rule at moderate compression — and it is why real gases are easier to compress than the ideal law predicts, right up until the molecules are nearly touching and the b term takes over and makes them suddenly much harder.

Problem 10 — What temperature does a spacecraft reach?

A spherical satellite with no internal heating orbits at Earth's distance from the Sun, where the intensity is 1361 W/m². Its surface absorbs a fraction \alpha_s of sunlight and has infrared emissivity \varepsilon. Find its temperature (a) for a black surface with \alpha_s = \varepsilon = 1, and (b) for a surface with \alpha_s = 0.20, \varepsilon = 0.85.

Solution

A sphere of radius R intercepts sunlight on its cross-section \pi R^2 and radiates from its whole surface 4\pi R^2.

\alpha_s S \pi R^2 = \varepsilon\sigma T^4\, 4\pi R^2 \quad\Longrightarrow\quad T = \left(\frac{\alpha_s S}{4\varepsilon\sigma}\right)^{1/4}

(a) \alpha_s = \varepsilon = 1:

T = \left(\frac{1361}{4\times5.67\times10^{-8}}\right)^{1/4} = \left(\frac{1361}{2.268\times10^{-7}}\right)^{1/4} = (6.001\times10^{9})^{1/4}

\sqrt{6.001\times10^9} = 7.747\times10^4, \qquad \sqrt{7.747\times10^4} = 278.3

T = 278\ \text{K} = 5\ ^\circ\text{C}

(b) \alpha_s = 0.20, \varepsilon = 0.85:

T = 278.3\times\left(\frac{0.20}{0.85}\right)^{1/4} = 278.3\times(0.2353)^{1/4}

\sqrt{0.2353} = 0.4851, \qquad \sqrt{0.4851} = 0.6965

T = 278.3\times0.6965 = 194\ \text{K} = -79\ ^\circ\text{C}

What to notice. Only the ratio \alpha_s/\varepsilon matters, and it is raised to the one-quarter power, so a ratio of 4.25 in your favour buys a factor of only 1.44 in temperature. This ratio is the entire discipline of spacecraft thermal control. White paint and second-surface mirrors have low \alpha_s and high \varepsilon, so they run cold; polished metal has moderate \alpha_s and very low \varepsilon, so it runs hot. The gold foil wrapped around spacecraft is doing exactly this arithmetic. Note also that a black sphere at Earth's distance sits at 5 °C, not at Earth's −18 °C, because Earth reflects 30 % away — the albedo term.

Problem 11 — The lagged pipe

A steam pipe of outer radius 5.0 cm carries steam at 150 °C and is lagged with 4.0 cm of insulation with k = 0.05 W m⁻¹ K⁻¹. The outside of the lagging is at 30 °C. Find the heat loss per metre of pipe.

Solution

The area is not constant — heat spreads outwards through ever-larger cylindrical shells — so the slab formula does not apply and the integral must be done.

Through a thin shell at radius r and thickness dr, over a length L, the area is 2\pi rL:

\frac{Q}{t} = -k(2\pi rL)\frac{dT}{dr}

In the steady state the same power crosses every shell, so Q/t is a constant, call it \dot{Q}. Separate and integrate from the inner radius r_1 to the outer r_2:

\dot{Q}\int_{r_1}^{r_2}\frac{dr}{r} = -2\pi kL\int_{T_1}^{T_2}dT

\dot{Q}\ln\frac{r_2}{r_1} = 2\pi kL(T_1 - T_2)

\boxed{\dot{Q} = \frac{2\pi kL(T_1-T_2)}{\ln(r_2/r_1)}}

Per metre, with r_1 = 0.050, r_2 = 0.090:

\ln\frac{0.090}{0.050} = \ln 1.80 = 0.5878

\dot{Q} = \frac{2\pi(0.05)(1)(120)}{0.5878} = \frac{37.70}{0.5878} = 64.1\ \text{W per metre}

What to notice. The logarithm is why doubling the lagging thickness does not halve the loss. Going from 4 cm to 8 cm makes r_2 = 0.13 and \ln(2.6) = 0.9555, giving 39.5 W/m — a 38 % reduction for 100 % more insulation. This also produces a genuinely surprising effect on thin wires: adding a little insulation can increase heat loss, because the added outer surface area gains more convective loss than the added thickness blocks by conduction. The radius where the two effects balance is called the critical radius, r_c = k/h, and for a thin electrical cable it can be larger than the cable itself.

Problem 12 — Cooling with a time constant

A cup holding 250 g of coffee at 85 °C cools in a 20 °C room. After 5.0 minutes it is at 70 °C. How long until it reaches 50 °C? Assume Newton's law of cooling.

Solution

Excess temperature \theta = T - 20 decays exponentially:

\theta(t) = \theta_0 e^{-t/\tau}

\theta_0 = 65, \qquad \theta(5) = 50

\frac{50}{65} = e^{-5/\tau} \quad\Longrightarrow\quad \ln(0.7692) = -\frac{5}{\tau} \quad\Longrightarrow\quad -0.26236 = -\frac{5}{\tau}

\tau = \frac{5}{0.26236} = 19.06\ \text{min}

For T = 50 °C, \theta = 30:

\frac{30}{65} = e^{-t/19.06} \quad\Longrightarrow\quad \ln(0.46154) = -\frac{t}{19.06} \quad\Longrightarrow\quad -0.77319 = -\frac{t}{19.06}

t = 14.7\ \text{minutes}

What to notice. It took 5 minutes to lose the first 15 degrees and takes another 9.7 to lose the next 20. The rate falls with the excess, so the last few degrees take effectively forever — the coffee approaches room temperature asymptotically and never formally arrives. This is also why the milk-first strategy wins: dropping the coffee to 75 °C immediately means the whole subsequent decay starts from a smaller \theta_0 and therefore proceeds more slowly in absolute terms.

Problem 13 — The heat pump that beats the heater

A house loses heat at 8.0 kW when it is 21 °C inside and 2 °C outside. Compare the electricity needed to hold that temperature with (a) a resistance heater and (b) a heat pump running at 45 % of its Carnot COP.

Solution

(a) Resistance heater. All electrical energy becomes heat, one for one:

P_{\text{electric}} = 8.0\ \text{kW}

(b) Heat pump. Ideal COP:

\text{COP}_{\text{ideal}} = \frac{T_H}{T_H - T_C} = \frac{294}{294-275} = \frac{294}{19} = 15.5

At 45 % of ideal:

\text{COP}_{\text{real}} = 0.45\times15.5 = 6.96

P_{\text{electric}} = \frac{8000}{6.96} = 1149\ \text{W} = 1.15\ \text{kW}

A factor of seven less electricity for the same warmth.

Now repeat for a cold night at −8 °C (265 K), with the house losing more heat, say 12 kW:

\text{COP}_{\text{ideal}} = \frac{294}{294-265} = \frac{294}{29} = 10.1, \qquad \text{COP}_{\text{real}} = 4.56

P_{\text{electric}} = \frac{12000}{4.56} = 2632\ \text{W}

What to notice. The heat pump is still four and a half times better than resistance heating on the cold night, but the load has risen and the COP has fallen simultaneously — the two move against you together, which is why heat pump systems are sized for the cold extreme rather than the average. The physics behind the whole advantage is that a resistance heater creates heat from work at an efficiency that can never exceed 1, while a heat pump moves heat that already exists, and moving is cheaper than creating.

Problem 14 — Entropy of the universe from a cup of tea

A 300 g cup of tea at 90 °C cools to room temperature at 20 °C. Find the entropy change of the tea, of the room, and of the universe. Then find how much work could in principle have been extracted, and compare it with the heat released.

Solution

Heat released by the tea:

Q = mc\Delta T = (0.300)(4186)(70) = 87{,}906\ \text{J}

Tea's entropy change, cooling from 363 K to 293 K:

\Delta S_{\text{tea}} = mc\ln\frac{293}{363} = (0.300)(4186)\ln(0.80716) = 1255.8\times(-0.21423) = -269.0\ \text{J/K}

Room's entropy change. The room is huge, so it stays at 293 K and absorbs all the heat at that one temperature:

\Delta S_{\text{room}} = \frac{87906}{293} = +300.0\ \text{J/K}

\Delta S_{\text{universe}} = -269.0 + 300.0 = +31.0\ \text{J/K}

Work available. The maximum work is what a sequence of Carnot engines could extract, running between the tea at its current temperature and the room, as the tea cools. That equals the heat released minus the part that must be dumped:

W_{\max} = Q - T_0|\Delta S_{\text{tea}}| = 87906 - (293)(269.0) = 87906 - 78817 = 9089\ \text{J}

What to notice. Out of 88 kJ of heat, only 9 kJ was ever extractable as work — about 10 %. The rest was locked away by the second law from the moment the tea was poured, because most of the cooling happens at temperatures close to the room's, where the Carnot efficiency is nearly zero. And by simply letting it cool on the table, all 9 kJ was destroyed: T_0\Delta S_{\text{universe}} = 293\times31.0 = 9083 J, matching to rounding. Every irreversible process in the world does this quietly and continuously, and the running total is the arrow of time.