Heat Transfer Practice Problems with Solutions (20 Worked Examples)

This page collects 20 original heat transfer practice problems with complete worked solutions. They cover the three modes of heat transfer (conduction, convection and radiation), combined modes and energy balances, and conduction with internal heat generation. Each solution is typed step by step with the governing equation, the numbers substituted and a check where one is available. For lab-style examples, see the radial heat conduction experiment and the forced convection experiment.

Heat transfer formula cheat sheet

Mode	Rate equation	Notes
Conduction (plane wall)	q″ = k(T₁ − T₂)/L	Fourier’s law; k in W/m·K
Conduction (cylinder)	q′ = 2πk(T₁ − T₂)/ln(r₂/r₁)	Per metre of length
Convection	q″ = h(Ts − T∞)	Newton’s law of cooling
Radiation (to large surroundings)	q″ = εσ(Ts4 − Tsur4)	σ = 5.67×10⁻⁸ W/m²·K⁴; absolute temperatures
Thermal resistance	R″ = L/k, 1/h, 1/hr	Resistances in series add
Lumped capacitance	(T − T∞)/(Ti − T∞) = exp(−t/τ)	Valid when Bi = hLc/k < 0.1
Heat generation (wall, one side insulated)	Tmax − Ts = q̇L²/2k	Maximum at the insulated face
Heat generation (solid cylinder)	T(0) − Ts = q̇ro²/4k	Maximum on the axis

Conduction and thermal resistance

Conduction problems apply Fourier’s law layer by layer. Treat each layer or surface film as a thermal resistance, add them in series, and then work back through the network to find temperatures.

Problem 1: Thermal conductivity from a measured heat flux

Problem. The heat flux through an 80 mm thick slab of insulating board is measured as 25 W/m². The two faces are held at 35 °C and 5 °C. Find the thermal conductivity of the board.

Solution.

1. Apply Fourier’s law for one-dimensional steady conduction through a plane wall: q″ = k(T₁ − T₂)/L.
2. Solve for k: k = q″L/(T₁ − T₂) = (25)(0.08)/(35 − 5) = 2.00/30.
3. k = 0.0667 W/m·K, which is in the range of rigid fibre insulation boards.

Answer. k ≈ 0.067 W/m·K

Problem 2: Heat loss through a composite wall

Problem. A wall consists of 100 mm of brick (k = 0.72 W/m·K), 50 mm of foam insulation (k = 0.04 W/m·K) and 15 mm of plaster (k = 0.5 W/m·K), in that order from inside to outside. The inside surface is at 22 °C and the outside surface is at −8 °C. Find the heat flux and the temperature at each interface.

Solution.

1. Each layer is a thermal resistance per unit area, R″ = L/k, and the layers add in series.
2. Brick: 0.1/0.72 = 0.1389 m²·K/W. Foam: 0.05/0.04 = 1.2500 m²·K/W. Plaster: 0.015/0.5 = 0.0300 m²·K/W.
3. Total R″ = 1.4189 m²·K/W, so q″ = (22 − (−8))/1.4189 = 21.14 W/m².
4. Brick–foam interface: T = 22 − 21.14(0.1389) = 19.06 °C.
5. Foam–plaster interface: T = 19.06 − 21.14(1.2500) = −7.37 °C.
6. Check: −7.37 − 21.14(0.0300) = −8.00 °C, which matches the outside surface temperature.

Answer. q″ ≈ 21.1 W/m²; interface temperatures ≈ 19.1 °C and −7.4 °C. Almost all of the temperature drop is across the foam.

Problem 3: Overall heat transfer coefficient and building heat loss

Problem. The wall of Problem 2 now separates room air at 21 °C from outdoor air at −5 °C. The inside convection coefficient is 8 W/m²·K and the outside coefficient is 25 W/m²·K. Find the overall heat transfer coefficient U and the heat loss through 12 m² of wall.

Solution.

1. Add the convection resistances in series with the conduction layers: R″total = 1/h_i + ΣL/k + 1/h_o.
2. 1/h_i = 0.1250; ΣL/k = 1.4189; 1/h_o = 0.0400. The total is 1.5839 m²·K/W.
3. U = 1/R″total = 0.631 W/m²·K.
4. Q = UA(T_in − T_out) = (0.631)(12)(21 − (−5)) = 197 W.

Answer. U ≈ 0.63 W/m²·K and Q ≈ 197 W

Problem 4: Insulated steam pipe

Problem. A steam pipe has an outer radius of 30 mm and carries steam so that the pipe surface is at 150 °C. It is wrapped with insulation of conductivity 0.06 W/m·K out to a radius of 50 mm, and the outer surface of the insulation is at 30 °C. Find the heat loss per metre of pipe.

Solution.

1. For steady radial conduction in a hollow cylinder, q′ = 2πk(T₁ − T₂)/ln(r₂/r₁).
2. ln(r₂/r₁) = ln(0.05/0.03) = 0.5108.
3. q′ = 2π(0.06)(150 − 30)/0.5108 = 88.6 W/m.
4. Equivalent thermal resistance per metre: R′ = ln(r₂/r₁)/(2πk) = 1.355 m·K/W, and 120/1.355 gives the same 88.6 W/m.

Answer. q′ ≈ 89 W/m

Problem 5: Critical radius of insulation on a thin wire

Problem. A bare wire of radius 4 mm carries current and its surface is held at 80 °C in air at 20 °C with h = 10 W/m²·K. A plastic coating with k = 0.17 W/m·K is proposed. Find the critical radius, and compare the heat loss per metre for the bare wire, for a coating out to the critical radius and for a coating out to 50 mm. At what outer radius does the coating finally reduce the loss below the bare-wire value?

Solution.

1. For a cylinder the critical radius is r_cr = k/h = 0.17/10 = 0.017 m = 17 mm.
2. The wire radius (4 mm) is less than r_cr, so adding coating first increases the heat loss.
3. Bare wire: q′ = h(2πr₀)(T_w − T_∞) = 15.08 W/m.
4. Coating to r = 17 mm: q′ = (T_w − T_∞)/[ln(r/r₀)/(2πk) + 1/(h·2πr)] = 26.19 W/m.
5. Coating to r = 50 mm: q′ = 22.36 W/m, still above the bare-wire value.
6. Setting q′ equal to the bare-wire loss and solving numerically shows that the coating only becomes a net insulator beyond r ≈ 263 mm.

Answer. r_cr = 17 mm. Heat loss: bare 15.1 W/m, at r_cr 26.2 W/m (the maximum), at 50 mm 22.4 W/m. A thin coating on a small wire helps cool it; it only insulates once the outer radius exceeds about 263 mm.

Convection heat transfer

Convection problems need a heat transfer coefficient. Either it is given, it comes from an energy balance on the object, or it is calculated from a Nusselt number correlation after checking the Reynolds number.

Problem 6: Newton’s law of cooling for a hot plate

Problem. A flat plate 0.5 m by 0.5 m has a surface temperature of 90 °C and is cooled on one face by air at 20 °C with h = 25 W/m²·K. Find the convective heat transfer rate and the heat flux.

Solution.

1. Newton’s law of cooling: q″ = h(T_s − T_∞).
2. q″ = 25(90 − 20) = 1,750 W/m².
3. Area A = 0.5 × 0.5 = 0.25 m², so q = q″A = 1,750 × 0.25 = 437.5 W.

Answer. q″ = 1,750 W/m² and q = 438 W

Problem 7: Free convection coefficient from a cooling plate

Problem. A thin aluminum plate, 0.3 m × 0.3 m, hangs vertically in still air at 20 °C. It has a mass of 1.2 kg and a specific heat of 900 J/kg·K. When its temperature is 70 °C it is observed to cool at 0.058 K/s. Neglect radiation and find the average free convection coefficient.

Solution.

1. The plate is thin and highly conductive, so treat it as one lump at temperature T. The energy balance is m c_p dT/dt = −h(2A)(T − T_∞), with both faces exposed.
2. Surface area of both faces: 2A = 2(0.3)(0.3) = 0.18 m².
3. h = m c_p|dT/dt|/[2A(T − T_∞)] = (1.2)(900)(0.058)/[(0.18)(50)] = 62.64/9.00 = 6.96 W/m²·K.
4. This is a typical natural convection value for air. Including radiation would lower the convection coefficient inferred from the same data.

Answer. h ≈ 7.0 W/m²·K

Problem 8: Average versus local coefficient on a plate

Problem. For turbulent flow over a flat plate, the local Nusselt number is correlated by Nu_x = 0.0296 Re_x^0.8 Pr^1/3. Find the ratio of the average coefficient over length L to the local coefficient at x = L, and compute the average coefficient if h_L = 48 W/m²·K.

Solution.

1. Since Re_x = ux/ν, the local coefficient varies as h_x = C x^−0.2, where C collects all the constants.
2. The average over 0 to L is h̄_L = (1/L)∫₀^L C x^−0.2 dx = (C/L)(L^0.8/0.8) = C L^−0.2/0.8.
3. Dividing by the local value at x = L, h_L = C L^−0.2, gives h̄_L/h_L = 1/0.8 = 1.25.
4. h̄_L = 1.25(48) = 60 W/m²·K.

Answer. h̄_L/h_L = 1.25, so h̄_L = 60 W/m²·K. For a laminar boundary layer, where h ∝ x^−1/2, the ratio is 2.

Problem 9: Heated cylinder in cross-flow

Problem. Air at 20 °C flows at 5 m/s across a long cylinder of diameter 25 mm whose surface is held at 90 °C. Use ν = 15.89×10⁻⁶ m²/s, k = 0.0263 W/m·K and Pr = 0.707. With Hilpert’s correlation Nu_D = C Re_D^m Pr^1/3 (C = 0.193, m = 0.618 for 4,000 < Re_D < 40,000), find h and the heat loss per metre.

Solution.

1. Re_D = VD/ν = (5)(0.025)/(15.89×10⁻⁶) = 7,867, which is inside the range of the correlation.
2. Nu_D = 0.193(7,867)^0.618(0.707)^1/3 = 44.0.
3. h = Nu_D k/D = (44.0)(0.0263)/0.025 = 46.2 W/m²·K.
4. q′ = h(πD)(T_s − T_∞) = (46.2)(π × 0.025)(90 − 20) = 254 W/m.

Answer. h ≈ 46 W/m²·K and q′ ≈ 254 W/m

Problem 10: Maximum power of an air-cooled chip

Problem. A square chip, 15 mm on a side, is cooled on its top face by air at 25 °C flowing parallel to it at 8 m/s. The chip temperature may not exceed 75 °C. Assuming laminar flow and negligible radiation, find the maximum allowable chip power. Use ν = 15.89×10⁻⁶ m²/s, k = 0.0263 W/m·K and Pr = 0.707.

Solution.

1. Re_L = VL/ν = (8)(0.015)/(15.89×10⁻⁶) = 7,552, far below the transition value of 5×10⁵, so the laminar average correlation applies.
2. Nu_L = 0.664 Re_L^1/2 Pr^1/3 = 0.664(7,552)^1/2(0.707)^1/3 = 51.4.
3. h̄ = Nu_L k/L = (51.4)(0.0263)/0.015 = 90.1 W/m²·K.
4. q = h̄A(T_s − T_∞) = (90.1)(0.015)²(75 − 25) = 1.01 W.

Answer. Maximum chip power ≈ 1.01 W. A turbulence promoter at the leading edge would raise h̄ and allow more power.

Thermal radiation

Radiation depends on the fourth power of absolute temperature, so convert to kelvin before you start. For a small surface in large surroundings the surroundings temperature is all you need.

Problem 11: Surface temperature of a space probe

Problem. A spherical probe of diameter 0.4 m dissipates 120 W of electronics inside it. The surface emissivity is 0.85, and the probe receives no radiation from the sun or other bodies. What is the equilibrium surface temperature?

Solution.

1. At steady state all the electrical power leaves by emission: P = εσA_sT_s⁴, where the surroundings (deep space) are effectively at 0 K.
2. Surface area: A_s = πD² = π(0.4)² = 0.5027 m².
3. T_s⁴ = P/(εσA_s) = 120/[(0.85)(5.67×10⁻⁸)(0.5027)] = 4.953×10⁹ K⁴.
4. T_s = 265.3 K = −7.9 °C.

Answer. T_s ≈ 265 K (−8 °C)

Problem 12: Net radiation to large surroundings and the radiation coefficient

Problem. A surface of area 0.5 m² and emissivity 0.8 is at 420 K inside a large room whose walls are at 300 K. Find the net radiation heat transfer rate and the equivalent radiation heat transfer coefficient h_r.

Solution.

1. For a small surface in large surroundings, q_rad = εσA(T_s⁴ − T_sur⁴).
2. T_s⁴ = 3.112×10¹⁰ K⁴; T_sur⁴ = 8.100×10⁹ K⁴.
3. q_rad = (0.8)(5.67×10⁻⁸)(0.5)(2.302×10¹⁰) = 522 W.
4. Linearize: h_r = εσ(T_s + T_sur)(T_s² + T_sur²) = 8.70 W/m²·K.
5. Check: h_rA(T_s − T_sur) = (8.70)(0.5)(120) = 522 W.

Answer. q_rad ≈ 522 W and h_r ≈ 8.7 W/m²·K

Problem 13: Convection and radiation from a hot panel

Problem. A panel at 80 °C stands in a room where the air and walls are at 20 °C. The panel has an emissivity of 0.9 and a convection coefficient of 10 W/m²·K. Find the convective and radiative heat fluxes and the fraction of the total that is radiation.

Solution.

1. Both modes act in parallel from the same surface: q″ = q″_conv + q″_rad.
2. Convection: q″_conv = h(T_s − T_∞) = 10(80 − 20) = 600 W/m².
3. Radiation (absolute temperatures 353.15 K and 293.15 K): q″_rad = εσ(T_s⁴ − T_sur⁴) = (0.9)(5.67×10⁻⁸)(8.169×10⁹) = 417 W/m².
4. Total = 1017 W/m², and radiation is 41% of it.

Answer. q″_conv ≈ 600 W/m², q″_rad ≈ 417 W/m²; radiation is about 41% of the total, so it cannot be neglected for a high-emissivity surface in air.

Problem 14: Radiation between parallel plates and the effect of a shield

Problem. Two large parallel plates are at 500 K (ε = 0.8) and 300 K (ε = 0.4). Find the net radiation flux between them. Then a thin shield with ε = 0.1 on both sides is placed between them; find the new flux.

Solution.

1. For two large parallel gray surfaces: q″ = σ(T₁⁴ − T₂⁴)/(1/ε₁ + 1/ε₂ − 1).
2. σ(T₁⁴ − T₂⁴) = (5.67×10⁻⁸)(6.250×10¹⁰ − 8.100×10⁹) = 3,084 W/m².
3. Denominator: 1/0.8 + 1/0.4 − 1 = 2.75. So q″ = 1,122 W/m².
4. With the shield there are two gaps in series. The resistance is (1/0.8 + 1/0.1 − 1) + (1/0.1 + 1/0.4 − 1) = 10.25 + 11.50 = 21.75.
5. q″_shield = 3,084/21.75 = 142 W/m².

Answer. Without the shield q″ ≈ 1,122 W/m²; with it q″ ≈ 142 W/m², a reduction of 87%.

Energy balances and combined modes

These problems combine modes. Write the energy balance first, then replace each term by its rate equation. When the balance is nonlinear, solve by iteration.

Problem 15: Surface temperature of a gearbox housing

Problem. A gearbox housing, a cube 0.3 m on a side with five exposed faces, receives 30 hp from the engine and delivers 95% of it to the output shaft. The remaining power is lost as heat through the housing to air and surroundings at 300 K. The convection coefficient is 12 W/m²·K and the surface emissivity is 0.8. Find the steady surface temperature.

Solution.

1. Input power: 30 hp × 745.7 W/hp = 22,371 W. Heat loss = (1 − η)P = 0.05 × 22,371 = 1,119 W.
2. Exposed area A = 5(0.30)² = 0.45 m².
3. Steady-state energy balance: loss = hA(T_s − T_∞) + εσA(T_s⁴ − T_sur⁴), with the surroundings at the air temperature.
4. This is nonlinear in T_s, so solve by iteration. At T_s = 420.1 K the right side equals 1,119 W.
5. Check the split at that temperature: convection 648 W and radiation 470 W.

Answer. T_s ≈ 420 K (147 °C)

Problem 16: Chip junction temperature with a thermal resistance network

Problem. A processor dissipates 6.0 W. The thermal resistances are 3.5 K/W from junction to case, 1.2 K/W across the thermal paste from case to heat sink, and 6.0 K/W from the heat sink to the ambient air at 35 °C. Find the junction temperature, and the heat sink resistance needed to hold the junction at 85 °C.

Solution.

1. The resistances are in series, so T_j = T_a + P(R_jc + R_cs + R_sa).
2. R_total = 3.5 + 1.2 + 6.0 = 10.7 K/W.
3. T_j = 35 + (6.0)(10.7) = 99.2 °C.
4. For T_j = 85 °C: R_total = (85 − 35)/6.0 = 8.33 K/W, so R_sa = 8.33 − 3.5 − 1.2 = 3.63 K/W.

Answer. T_j ≈ 99 °C, which is above the 85 °C limit. A heat sink with R_sa ≤ 3.6 K/W is needed.

Problem 17: Temperature rise of water in an electric heater

Problem. An electric heater delivers 3,000 W to water flowing through it at 0.02 kg/s. The water enters at 15 °C and the specific heat is 4180 J/kg·K. All the electrical power goes into the water. Find the outlet temperature.

Solution.

1. Apply the steady-flow energy balance: q = ṁc_p(T_out − T_in).
2. ΔT = q/(ṁc_p) = 3,000/[(0.02)(4180)] = 35.9 K.
3. T_out = 15 + 35.9 = 50.9 °C.

Answer. T_out ≈ 50.9 °C

Problem 18: Cooling time of a copper sphere by lumped capacitance

Problem. A copper sphere of diameter 20 mm (ρ = 8933 kg/m³, c_p = 385 J/kg·K, k = 401 W/m·K) is heated to 300 °C and then quenched in air at 25 °C with h = 60 W/m²·K. Check whether a lumped model is valid and find the time to cool to 100 °C.

Solution.

1. Characteristic length L_c = V/A_s = D/6 = 3.33 mm.
2. Biot number Bi = hL_c/k = (60)(0.00333)/401 = 5.0×10⁻⁴, much less than 0.1, so the lumped capacitance method is valid.
3. Time constant τ = ρVc_p/(hA_s) = ρ c_p L_c/h = (8933)(385)(0.00333)/60 = 191 s.
4. (T − T_∞)/(T_i − T_∞) = exp(−t/τ), so t = −τ ln[(100 − 25)/(300 − 25)] = −191 ln(0.2727) = 248 s.

Answer. Bi ≈ 5.0×10⁻⁴ (lumped model valid); t ≈ 248 s (4.1 min)

Conduction with internal heat generation

When heat is generated inside the body, the temperature profile is parabolic. The maximum temperature sits where the heat has farthest to travel: the insulated face of a wall or the axis of a cylinder.

Problem 19: Plane wall with uniform heat generation

Problem. A wall of thickness L = 30 mm generates heat uniformly at 2.0×10⁶ W/m³. One face is perfectly insulated and the other face is held at 100 °C. The conductivity is 20 W/m·K. Find the maximum temperature and the heat flux at the cooled face.

Solution.

1. With the insulated face at x = 0 and the cooled face at x = L, the steady solution is T(x) = T_s + (q̇/2k)(L² − x²).
2. The maximum is at the insulated face: T_max − T_s = q̇L²/(2k) = (2.0×10⁶)(0.03)²/[2(20)] = 45.0 K, so T_max = 145.0 °C.
3. All generated heat leaves through the cooled face: q″ = q̇L = (2.0×10⁶)(0.03) = 60,000 W/m².

Answer. T_max ≈ 145 °C at the insulated face; q″ = 60 kW/m² at the cooled face

Problem 20: Current-carrying wire

Problem. A stainless steel wire of diameter 3 mm carries 250 A. The electrical resistivity is 7.0×10⁻⁷ Ω·m and the thermal conductivity is 15 W/m·K. The wire surface is held at 90 °C by a coolant. Find the volumetric heat generation and the centerline temperature.

Solution.

1. Cross-section area A_c = πD²/4 = 7.069×10⁻⁶ m².
2. Generation per unit volume: q̇ = I²ρ_e/A_c² = (250)²(7.0×10⁻⁷)/(7.069×10⁻⁶)² = 8.756×10⁸ W/m³.
3. For a solid cylinder with uniform generation, T(0) − T_s = q̇r_o²/(4k) = (8.756×10⁸)(0.0015)²/[4(15)] = 32.8 K.
4. T(0) = 90 + 32.8 = 122.8 °C.

Answer. q̇ ≈ 8.76×10⁸ W/m³ and T(0) ≈ 123 °C

Heat transfer FAQ

What are the three modes of heat transfer?

Conduction is heat flow through a solid or stationary fluid driven by a temperature gradient. Convection is heat transfer between a surface and a moving fluid. Radiation is energy emitted as electromagnetic waves, and it needs no medium.

What is Fourier’s law of heat conduction?

Fourier’s law states that the heat flux is proportional to the temperature gradient: q″ = −k dT/dx. For a plane wall at steady state it becomes q″ = k(T₁ − T₂)/L.

How do you find the heat transfer coefficient h?

Use a Nusselt number correlation for the geometry and flow, h = Nu·k/L, after checking the Reynolds number. For a lumped object you can also back it out of a measured cooling rate with an energy balance.

Why must radiation problems use kelvin?

The radiation equation depends on the fourth power of absolute temperature, so using °C gives a wrong answer. Convert with T(K) = T(°C) + 273.15 before using εσT⁴.

When can I use the lumped capacitance method?

When the Biot number Bi = hL_c/k is below about 0.1, the temperature inside the body is nearly uniform and the object can be treated as one lump.

What is the critical radius of insulation?

For a cylinder, r_cr = k/h. If the bare radius is below this value, adding insulation raises the heat loss until the outer radius passes r_cr.

Keep learning

• Heat transfer basics
• Radial heat conduction lab experiment
• Forced convection heat transfer and the stirring-rate experiment
• Design of a heat sink
• Battery thermal management 101
• Thermodynamics, entropy and the second law