Heat Transfer Calculator
Calculate heat transfer rate (W) and thermal resistance for conduction, convection, and radiation. Enter temperature, area, and material properties.
Heat Transfer Calculator
Calculator
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Formula: Q = kAΔT/L (conduction) | Q = hAΔT (convection) | Q = εσA(T₁⁴-T₂⁴) (radiation)
Additional inputs: Surface Emissivity (0 to 1).
Worked example — Q = 160W | R_th = 0.1875 °C/W | R-value = 3.75
Formula
Q = kAΔT/L (conduction) | Q = hAΔT (convection) | Q = εσA(T₁⁴-T₂⁴) (radiation)
Heat conduction rate depends on thermal conductivity, area, temperature difference, and thickness (Fourier's Law). Convection depends on the heat transfer coefficient. Radiation follows the Stefan-Boltzmann law with the fourth power of absolute temperature.
Worked Examples
Example 1: Wall Insulation Heat Loss
Problem:Calculate heat loss through a 150mm fiberglass-insulated wall (k = 0.04 W/m·K), area = 20m², with 20°C inside and -10°C outside.
Solution:k = 0.04 W/(m·K), A = 20 m², ΔT = 30°C, L = 0.15m Q = kA(ΔT)/L = 0.04 × 20 × 30 / 0.15 = 160W Thermal Resistance = L/(kA) = 0.15/(0.04×20) = 0.1875 °C/W R-value = L/k = 0.15/0.04 = 3.75 m²·K/W Heat Flux = 160/20 = 8 W/m²
Result:Q = 160W | R_th = 0.1875 °C/W | R-value = 3.75
Example 2: Cooling Electronics with Forced Convection
Problem:A 50W processor has a heatsink with 0.02m² surface area. Air is forced over it with h = 150 W/(m²·K). What is the temperature rise?
Solution:Q = hA(ΔT), so ΔT = Q/(hA) ΔT = 50 / (150 × 0.02) = 16.7°C If ambient = 25°C, heatsink temp = 41.7°C Thermal Resistance = 1/(hA) = 1/(150×0.02) = 0.333 °C/W
Result:ΔT = 16.7°C | T_surface = 41.7°C | R_th = 0.333 °C/W
Frequently Asked Questions
What are the three modes of heat transfer?
Heat transfers through three fundamental mechanisms: (1) Conduction — heat flows through a solid material or between materials in direct contact, driven by a temperature gradient. Rate depends on thermal conductivity, area, temperature difference, and material thickness. (2) Convection — heat transfer between a surface and a moving fluid (liquid or gas). Can be natural (driven by buoyancy) or forced (driven by fans/pumps). Rate depends on the convection coefficient, area, and temperature difference. (3) Radiation — heat transfer via electromagnetic waves (infrared). Does not require a medium and can occur through vacuum. Rate depends on emissivity, temperature, and area.
What is Fourier's Law of conduction?
Fourier's Law states that the rate of heat conduction through a material is proportional to the negative temperature gradient and the cross-sectional area: Q = -kA(dT/dx). For a flat wall with uniform properties: Q = kA(T_hot - T_cold)/L, where Q is heat transfer rate (Watts), k is thermal conductivity (W/m·K), A is cross-sectional area (m²), ΔT is temperature difference (K or °C), and L is thickness (m). Materials with high k (metals) conduct heat well; materials with low k (insulators) resist heat flow. This law is the basis for insulation design and heat exchanger calculations.
What is thermal resistance?
Thermal resistance (R_th) is the opposition to heat flow, analogous to electrical resistance. For conduction: R_th = L/(kA) (°C/W). For convection: R_th = 1/(hA) (°C/W). The heat transfer rate is Q = ΔT/R_th, just like Ohm's law (I = V/R). For series thermal resistances (multiple layers), total R = R1 + R2 + R3... For parallel paths, 1/R_total = 1/R1 + 1/R2... The R-value used in building insulation is the thermal resistance per unit area: R-value = L/k (m²·K/W or ft²·°F·h/BTU).
What is the convection heat transfer coefficient?
The convection heat transfer coefficient (h) quantifies how effectively heat transfers between a surface and a fluid. It depends on fluid properties, flow velocity, geometry, and whether convection is natural or forced. Typical values: Natural convection in air = 5-25 W/(m²·K). Forced convection in air = 25-250 W/(m²·K). Natural convection in water = 100-900 W/(m²·K). Forced convection in water = 250-12,000 W/(m²·K). Boiling water = 3,000-100,000 W/(m²·K). The coefficient is often determined empirically using dimensionless correlations involving Nusselt, Reynolds, and Prandtl numbers.
What is the Stefan-Boltzmann Law?
The Stefan-Boltzmann Law governs thermal radiation: Q = εσA(T_hot⁴ - T_cold⁴), where ε is emissivity (0-1, surface property), σ = 5.67×10⁻⁸ W/(m²·K⁴) is the Stefan-Boltzmann constant, A is surface area (m²), and T is absolute temperature in Kelvin. Key points: radiation depends on the fourth power of temperature (doubling temperature increases radiation 16×), all objects above 0K emit radiation, emissivity varies from ~0.02 (polished silver) to ~0.95 (black body paint), and radiation can transfer heat through vacuum (how the Sun heats the Earth).
What is the overall heat transfer coefficient U and how does it combine the three modes?
U bundles every resistance between two fluids into one number, with Q = UAΔT and U = 1/(sum of the area-based resistances). A wall exposed to air on both sides has 1/h_inside + L/k + 1/h_outside. Taking h_inside = 8 W/(m²·K), L/k = 0.15/0.04 = 3.75 m²·K/W and h_outside = 25 W/(m²·K) gives 0.125 + 3.75 + 0.04 = 3.915 m²·K/W, so U = 0.255 W/(m²·K). The largest resistance in the stack dominates the result.
Why do heat exchangers use the log mean temperature difference instead of a simple average?
The temperature gap between the two streams changes along the exchanger, and since the local heat flux is proportional to that gap, an arithmetic average overstates the driving force. The correct value is ΔT_lm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂). With end differences of 50 °C and 20 °C, ΔT_lm = 30/ln(2.5) = 30/0.916 = 32.7 °C rather than 35 °C. Counterflow arrangements keep the gap more uniform and so achieve a higher ΔT_lm than parallel flow for the same terminal temperatures.
When can a heated object be treated as a single uniform temperature?
Check the Biot number, Bi = h·L_c/k, where L_c is volume divided by surface area. When Bi is below about 0.1 the internal conduction resistance is small compared with the surface resistance, so the body stays nearly isothermal and the simple lumped capacitance exponential cooling curve applies. Above that, internal gradients matter and a full conduction solution is needed. How fast those gradients even out is set by the thermal diffusivity, α = k/(ρ·c_p).
At what point does radiation matter as much as natural convection?
Radiation can be written as an equivalent coefficient, h_r = εσ(T_s² + T_∞²)(T_s + T_∞), which follows exactly from factorising the difference of fourth powers. For a surface of emissivity 0.9 at 350 K facing 300 K surroundings, h_r works out at roughly 7 W/(m²·K), squarely inside the 5 to 25 W/(m²·K) band for natural convection in air. So for warm surfaces in still air the two modes are comparable, and radiation pulls ahead as temperature rises.
What is thermal contact resistance and why does it need interface material?
Two solids pressed together touch only at scattered asperities, with trapped air filling the gaps, so an extra resistance appears at the joint that Fourier's law applied to the bulk layers alone misses. Because air has a thermal conductivity around 0.026 W/(m·K), those voids dominate. Thermal paste, pads or solder displace the air with a far more conductive medium, and higher clamping pressure or smoother surfaces increase the real contact area, both cutting the interface resistance substantially.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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