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University Physics II · Optional extension · Institutional Extension · Temperature, Heat, and Kinetic Theory · E2.10

Thermal Physics Problem Studio

Eight problems, from a heater's energy bill to the molecular speeds inside a compressed cylinder. Each names the model it may use — all-melted or slush, ideal gas or real — and holds you to three significant figures.

Open the lab ↓Practice & worked examples ↓
01

Build the model

Connect the measurement to the mechanism.

Calorimetry, phase changes, transfer rates, gas states, and kinetic-theory estimates with explicit model and assumption checks. Treat this course-map statement as a claim to test rather than an invitation to import a familiar equation. In Institutional Extension · Temperature, Heat, and Kinetic Theory, begin from microscopic-macroscopic connection, then state the system, observable, assumptions, and evidence before calculating.

Simple definition
Calorimetry, phase changes, transfer rates, gas states, and kinetic-theory estimates with explicit model and assumption checks.
Example
A strong response uses energy bar charts and states where the model stops being reliable.
01

The subsection's claim

Calorimetry, phase changes, transfer rates, gas states, and kinetic-theory estimates with explicit model and assumption checks.

02

How to work with it

Start from microscopic-macroscopic connection. Then declaring the system, the energy-transfer mechanism, and the gas model before substituting into thermal equations. Select an equation only after its variables and assumptions match the stated system.

03

What evidence would decide

What value of absolute zero is supported by constant-volume pressure-temperature data? Useful evidence includes pressure-temperature pairs, a linear fit, the extrapolated intercept, residuals, uncertainty, and comparison with the accepted value.

04

Keep the boundary visible

This GioPhysics course map is an adaptable learning sequence, not academic credit, accreditation, or a universal university syllabus. Departments may redistribute weeks, laboratory hours, optics, or the modern-physics survey to match local requirements. This GioPhysics course map is an adaptable learning sequence, not academic credit, accreditation, or a universal university syllabus. Departments may redistribute weeks, laboratory hours, optics, or the modern-physics survey to match local requirements. Thermal physics appears as an unnumbered institutional extension: some universities assess it within Physics II, while others teach it in a separate course, so include the thermal extensions only where the local syllabus requires them. A result should be checked against units, signs, limiting cases, and the conditions under which its model was derived.

02

Change one variable at a time

Make the relationship visible.

Interactive model
0.12 kg
26 °C

The sliders open on problem 3: 0.12 kg of ice, water at 26 °C, and the crossing lands just past the melt mark at 0.60 °C. Pull the water down to 25 °C and the crossing slides back onto the flat plateau — the answer stops being a temperature and becomes 0.00 °C with about a gram of ice left. That is the melt test, drawn.

Interactive physics modelTemperature against the energy handed from the water to the ice. 0.12 kg of ice at −8.50 °C goes into 0.400 kg of water at 26 °C: the falling line is the water, the rising line is the ice, which warms to 0 °C, holds flat through 40.1 kJ of melting, then warms again as liquid. The lines meet at equilibrium, 0.60 °C, with 0.0 g of ice still solid.T / °C260−8.5watericeall melted at 42.2 kJ0.60 °Cenergy handed to the ice / kJ

WATER CAN GIVE43.5 kJ

ICE DEMANDS42.2 kJ

FINAL TEMPERATURE0.60 °C

ICE STILL SOLID0.0 g

Live interpretationWATER CAN GIVE: 43.5 kJ. ICE DEMANDS: 42.2 kJ. FINAL TEMPERATURE: 0.60 °C. ICE STILL SOLID: 0.0 g

03

Catch the common trap

Explain before calculating.

What should a Thermal physics problem studio solution make visible?

Choose an answer to test the model.

04

Practice & worked examples

Reason from the model, then test the result.

01 · EasyAn immersion heater rated 1.50 kW warms 0.850 kg of water in a vacuum flask from 18.0 °C to 92.0 °C. Take cwater = 4186 J kg⁻¹ K⁻¹. Find the energy the water absorbs and the time the heater would need if every joule reached the water. The measured time is 189 s; find the fraction of the electrical energy that actually reached the water.
  1. Heat for a temperature change with no phase change: Q = mcΔT. The rise is ΔT = 92.0 − 18.0 = 74.0 K, and a Celsius interval equals a kelvin interval, so no conversion is needed.
  2. Q = (0.850)(4186)(74.0) = 263 299 J = 2.63 × 10⁵ J.
  3. Ideal time at full transfer: t = Q/P = 263 299/1500 = 175.53 s, so 176 s.
  4. Electrical energy actually supplied in 189 s: E = Pt = (1500)(189) = 283 500 J.
  5. Fraction delivered to the water: 263 299/283 500 = 0.9287, or 92.9%.
  6. Model check: mcΔT holds c constant, and water's c drifts about 0.5% across this range. The missing 7.1% is the flask wall, the heater body, and leakage to the room — thermal mass and losses the audit ignored.

AnswerQ = 2.63 × 10⁵ J, ideal time 176 s, and 92.9% of the electrical energy reached the water.

02 · EasyA 0.235 kg metal cylinder is heated in a steam jacket to 98.5 °C, then transferred quickly into 0.180 kg of water at 21.0 °C inside a copper calorimeter cup of mass 0.0950 kg. The mixture settles at 29.0 °C. Take cwater = 4186 J kg⁻¹ K⁻¹ and cCu = 386 J kg⁻¹ K⁻¹. Find the specific heat of the metal, and say which way an unaccounted heat leak to the room would shift it.
  1. Take the cup and contents as the system: heat lost by the cylinder equals heat gained by the water and by the cup, since all three end at one temperature.
  2. Combined water-plus-cup heat capacity: (0.180)(4186) + (0.0950)(386) = 753.48 + 36.67 = 790.15 J K⁻¹.
  3. Both rise by ΔT = 29.0 − 21.0 = 8.0 K, so they absorb Q = (790.15)(8.0) = 6321.2 J.
  4. The cylinder falls by 98.5 − 29.0 = 69.5 K, so cmetal = Q/(mΔT) = 6321.2/[(0.235)(69.5)] = 6321.2/16.3325 = 387.0 J kg⁻¹ K⁻¹.
  5. Model check: no heat crosses the cup wall. A leak to a cooler room would drain part of that 6321 J before the thermometer settled, lowering the equilibrium reading, shrinking Q, and biasing cmetal low.
  6. The standard defence is to start the water as far below room temperature as it will finish above, so the inward leak in the first half cancels the outward leak in the second.

Answercmetal = 387 J kg⁻¹ K⁻¹, consistent with copper or a brass; an unaccounted heat leak biases the result low.

03 · MediumA 0.120 kg block of ice at −8.50 °C is dropped into 0.400 kg of water at 26.0 °C in an insulated vessel of negligible heat capacity. Take cice = 2100 J kg⁻¹ K⁻¹, Lf = 3.34 × 10⁵ J kg⁻¹, cwater = 4186 J kg⁻¹ K⁻¹. Find the final temperature and the final state of the contents. Do not assume in advance that all the ice melts.
  1. Test the melt before solving. Energy the water can release on the way down to 0 °C: (0.400)(4186)(26.0) = 43 534 J.
  2. Energy the ice needs to become 0 °C water: warming, (0.120)(2100)(8.50) = 2142 J, plus melting, (0.120)(3.34 × 10⁵) = 40 080 J — 42 222 J in total.
  3. 43 534 J exceeds 42 222 J, so every gram melts and the mixture ends above 0 °C. The surplus is 43 534 − 42 222 = 1312 J.
  4. That surplus warms the whole 0.520 kg as liquid: ΔT = 1312.4/[(0.520)(4186)] = 1312.4/2176.7 = 0.603 K, so Tf = 0.603 °C.
  5. Model check on the margin: start the water at 25.0 °C instead and it supplies only 41 860 J, melting (41 860 − 2142)/(3.34 × 10⁵) = 0.1189 kg. Equilibrium is then pinned at 0.00 °C with 1.08 g of ice still solid.
  6. A 1 °C change in one input therefore changes the form of the answer, not just its value — which is why the melt test comes first and the mixing formula second.

AnswerAll the ice melts; Tf = 0.603 °C, leaving 0.520 kg of liquid water and no ice.

04 · MediumA wall panel of area 12.0 m² is built from 0.150 m of concrete (k = 0.800 W m⁻¹ K⁻¹) bonded to 0.0600 m of rigid foam (k = 0.0350 W m⁻¹ K⁻¹). In steady state the inner concrete surface is at 21.0 °C and the outer foam surface at −4.00 °C. Find the conduction rate through the panel, the temperature at the concrete–foam interface, and the share of the total thermal resistance carried by the foam.
  1. In steady state the same H crosses both layers, so their resistances add. Per unit area R = L/k: concrete gives 0.150/0.800 = 0.1875 m² K W⁻¹, foam gives 0.0600/0.0350 = 1.7143 m² K W⁻¹.
  2. Total R = 1.9018 m² K W⁻¹, and H = AΔT/R = (12.0)(25.0)/1.9018 = 157.75 W, which rounds to 158 W.
  3. Interface temperature from the concrete layer alone: its resistance is 0.1875/12.0 = 0.015625 K W⁻¹, so the drop across it is (157.75)(0.015625) = 2.46 K and Tinterface = 21.0 − 2.46 = 18.5 °C.
  4. Check the other layer: the foam must carry 18.54 − (−4.00) = 22.54 K, and (157.75)(1.7143/12.0) = 22.54 K. Consistent.
  5. Resistance share: 1.7143/1.9018 = 0.901, so the foam supplies 90.1% of the resistance from 28.6% of the thickness. Nearly all the concrete does is hold the foam up.
  6. Assumption check: surface temperatures were given, so the still-air films on each face are excluded; adding roughly 0.13 and 0.04 m² K W⁻¹ would raise R by about 9% and cut H by about 8%. Steady state also means nothing is being stored in the wall.

AnswerH = 158 W, the interface sits at 18.5 °C, and the foam carries 90.1% of the thermal resistance.

05 · MediumA bare steam pipe of outer diameter 0.0890 m and length 3.20 m has a surface at 78.0 °C and runs through a room whose walls sit at 19.0 °C. The surface emissivity is 0.900 and σ = 5.670 × 10⁻⁸ W m⁻² K⁻⁴. Find the net radiative power leaving the pipe, ignoring the ends. Then find it again with the surface at 95.0 °C, and say why the increase is so much larger than the temperature increase.
  1. Lateral area: A = πdL = π(0.0890)(3.20) = 0.8947 m².
  2. Radiation needs absolute temperatures: Ts = 351.15 K, Tsurr = 292.15 K. Fourth powers: Ts⁴ = 1.5204 × 10¹⁰ K⁴ and Tsurr⁴ = 7.2849 × 10⁹ K⁴.
  3. Net exchange with the surroundings: Pnet = εσA(Ts⁴ − Tsurr⁴) = (0.900)(5.670 × 10⁻⁸)(0.8947)(7.9196 × 10⁹) = 361.6 W.
  4. At 95.0 °C, Ts = 368.15 K and Ts⁴ = 1.8370 × 10¹⁰ K⁴, so Pnet = (4.5658 × 10⁻⁸)(1.1085 × 10¹⁰) = 506.1 W.
  5. The surface temperature rose 4.84% in kelvin but the net rate rose 40.0%, because the emitted term climbs as T⁴ while the absorbed term stays fixed by the walls. The difference of fourth powers, not the ratio of temperatures, sets the answer.
  6. Model check: this is radiation only. A pipe this hot in still air also sheds a comparable amount by natural convection, so 362 W is a floor on the total loss, not the loss. The grey-body treatment additionally sets absorptivity equal to emissivity at 0.900.

AnswerPnet = 362 W at 78.0 °C and 506 W at 95.0 °C — a 40.0% rise from a 4.84% rise in absolute temperature.

06 · MediumA rigid steel cylinder of internal volume 0.0450 m³ holds nitrogen (M = 28.0 g mol⁻¹) at an absolute pressure of 15.5 MPa and 17.0 °C. Take R = 8.314 J mol⁻¹ K⁻¹. Treating the gas as ideal, find the amount of substance, the mass of nitrogen, and the density. The cylinder is then left in the sun until the gas reaches 52.0 °C; find the new pressure.
  1. Convert first: T₁ = 17.0 + 273.15 = 290.15 K and T₂ = 52.0 + 273.15 = 325.15 K. The pressure is already absolute.
  2. n = pV/(RT) = (15.5 × 10⁶)(0.0450)/[(8.314)(290.15)] = 697 500/2412.3 = 289.1 mol.
  3. m = nM = (289.142)(0.0280) = 8.096 kg, and ρ = m/V = 8.096/0.0450 = 179.9 kg m⁻³.
  4. The cylinder is rigid, so V and n are fixed and p/T is constant: p₂ = p₁(T₂/T₁) = 15.5(325.15/290.15) = 17.4 MPa. Substituting Celsius would have given 15.5(52.0/17.0) = 47.4 MPa, which is nonsense.
  5. Model check: the molar volume is V/n = 1.556 × 10⁻⁴ m³ mol⁻¹, only about four times nitrogen's excluded volume per mole (b ≈ 3.9 × 10⁻⁵ m³ mol⁻¹), so the point-particle assumption is strained. Real nitrogen at this state is roughly 2% less dense than 180 kg m⁻³.
  6. The ratio p₂/p₁ = T₂/T₁ survives that better than the absolute figures do, because the same real-gas correction sits on both sides and largely cancels across a 35 K change.

Answern = 289 mol, m = 8.10 kg, ρ = 180 kg m⁻³, and p₂ = 17.4 MPa after the warming.

07 · HardContinue with the nitrogen of the previous problem: 289.1 mol in 0.0450 m³ at 15.5 MPa and 290.15 K, with M = 28.0 g mol⁻¹, ρ = 179.9 kg m⁻³ and k = 1.381 × 10⁻²³ J K⁻¹. Find the average translational kinetic energy per molecule, the rms molecular speed, and the total translational kinetic energy of the gas. Compare that speed with helium's (M = 4.00 g mol⁻¹) at the same temperature, then evaluate ⅓ρvrms² and say what the agreement does and does not prove.
  1. Average translational kinetic energy depends on temperature alone: Kavg = (3/2)kT = 1.5(1.381 × 10⁻²³)(290.15) = 6.01 × 10⁻²¹ J per molecule.
  2. Setting that equal to ½mv² and working per mole: vrms = √(3RT/M) = √[3(8.314)(290.15)/0.0280] = √(2.5846 × 10⁵) = 508 m s⁻¹. Helium at the same T shares the identical Kavg but runs √(28.0/4.00) = 2.65 times faster, at 1.35 × 10³ m s⁻¹.
  3. Total translational kinetic energy: Ktrans = (3/2)nRT = (3/2)pV = 1.5(15.5 × 10⁶)(0.0450) = 1.05 × 10⁶ J.
  4. Kinetic-theory pressure: ⅓ρvrms² = ⅓(179.91)(2.5846 × 10⁵) = 1.550 × 10⁷ Pa, exactly the stated 15.5 MPa.
  5. That agreement proves nothing new. ρ came from pV = nRT and vrms came from the same T, so ⅓ρvrms² = nRT/V is an identity inside the model rather than an independent test of it.
  6. The real test uses a measured density. At about 176 kg m⁻³ the same formula returns ⅓(176)(2.5846 × 10⁵) = 15.2 MPa, 2.2% under the gauge — and that shortfall is what the point-particle, no-interaction assumptions left out.

AnswerKavg = 6.01 × 10⁻²¹ J, vrms = 508 m s⁻¹ (helium 1.35 × 10³ m s⁻¹), Ktrans = 1.05 MJ; ⅓ρvrms² returns 15.5 MPa as an identity, not a check.

08 · HardA 0.320 kg aluminium pan (c = 900 J kg⁻¹ K⁻¹) holds 0.750 kg of water at 20.0 °C on a hob delivering a steady 1.80 kW into the pan base. Take cwater = 4186 J kg⁻¹ K⁻¹ and Lv = 2.256 × 10⁶ J kg⁻¹ at 1 atm. Find the time to reach 100 °C and the further time to boil away 0.250 kg. The stainless-steel base is 4.50 mm thick and 0.220 m across (k = 15.0 W m⁻¹ K⁻¹); find the temperature difference across it while the water boils.
  1. Sensible-heat stage, pan and water rising together: Q₁ = [(0.750)(4186) + (0.320)(900)](80.0) = (3139.5 + 288)(80.0) = 2.742 × 10⁵ J, so t₁ = 274 200/1800 = 152 s.
  2. Phase-change stage at a fixed 100 °C: Q₂ = mLv = (0.250)(2.256 × 10⁶) = 5.640 × 10⁵ J, so t₂ = 564 000/1800 = 313 s.
  3. Boiling off a third of the water takes 313/152 = 2.06 times as long as heating all of it through 80 K. Latent heat, not specific heat, sets the kitchen timer.
  4. Conduction through the base while boiling: A = π(0.110)² = 3.801 × 10⁻² m², and H = kAΔT/L rearranges to ΔT = HL/(kA).
  5. ΔT = (1800)(0.00450)/[(15.0)(3.801 × 10⁻²)] = 8.10/0.5702 = 14.2 K, so the outer face of the base runs about 14 K above the inner face — near 114 °C with boiling water against the inside.
  6. Model checks: 1.80 kW is what enters the pan, not what the hob draws; c and Lv are held constant; and Lv is pressure-dependent — at 1500 m elevation water boils near 95 °C, which shortens the first stage and slightly lengthens the second.

Answert₁ = 152 s to reach 100 °C, a further t₂ = 313 s to boil off 0.250 kg, and ΔT = 14.2 K across the pan base.