Thermal Resistance Converter
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| Unit | Equivalent Value |
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Absolute Thermal Resistance (symbolized by Rth or θ) measures a component, insulator, or heat sink’s physical resistance to heat flow per unit of power dissipated (Rth = ΔT ÷ P = (T1 - T2) ÷ Q). Governed by the fundamental Thermal Ohm’s Law (ΔT = P · Rth, where temperature difference ΔT acts as voltage V, heat power P in watts acts as electric current I, and Rth acts as electrical resistance R), thermal resistance is critical in semiconductor die junction cooling, computer CPU/GPU heat sink design, power MOSFET packaging, building envelope HVAC insulation, and LED thermal design. Units are categorized across two major engineering standards: SI metric standards (Kelvin per Watt / K/W, Degree Celsius per Watt / °C/W) and Imperial US Customary standards (Degree Fahrenheit Hour per Btu / °F·h/Btu, Degree Fahrenheit Second per Btu / °F·s/Btu).
Our free online Thermal Resistance Converter provides instant, high-precision conversions across all SI metric, Imperial thermal engineering, and semiconductor heat sink units:
- Degree Fahrenheit Hour per Btu (IT) to K/W (°F·h/Btu to K/W): Multiply °F·h/Btu by
1.89563424(1 °F·h/Btu = 1.89563 K/W = 1.89563 °C/W). - Kelvin per Watt to °F·h/Btu (K/W to °F·h/Btu): Multiply K/W by
0.5275279(1 K/W = 1 °C/W = 0.52753 °F·h/Btu = 1,900 °F·s/Btu). - Degree Celsius per Watt to K/W (°C/W to K/W):
1 °C/W = 1.0 K/W(Step-for-step 1:1 identical magnitude). - Degree Fahrenheit Second per Btu (IT) to K/W: Multiply °F·s/Btu by
0.0005265651(1 °F·s/Btu = 0.526565 mK/W = 0.00052657 K/W). - Degree Fahrenheit Hour per Btu (th) [Thermochemical] to K/W: Multiply by
1.89690283(1 °F·h/Btu(th) = 1.89690 K/W).
Master Absolute Thermal Resistance Conversion Table
The table below displays exact mathematical conversion relationships, SI K/W multipliers, and imperial °F·h/Btu equivalents relative to 1 Kelvin per Watt (1 K/W = 1 °C/W):
| Thermal Resistance Unit Name | Symbol | Exact Value in K/W (°C/W) | °F·h/Btu & °F·s/Btu Equivalent | Domain & Technical Application Standard |
|---|---|---|---|---|
| 1 Kelvin per Watt | K/W, K/W |
1.0 K/W (Base SI Unit) |
0.527528 °F·h/Btu (1,899.10 °F·s/Btu / 1.0 °C/W) |
SI Fundamental Electronics Thermal Management Standard |
| 1 Degree Celsius per Watt | °C/W |
1.0 °C/W |
0.527528 °F·h/Btu (1,899.10 °F·s/Btu / 1.0 K/W) |
Global Semiconductor Datasheet Resistance Standard (θJA, θJC) |
| 1 Degree Fahrenheit Hour per Btu (IT) | °F·h/Btu |
1.895634 K/W (1.89563 °C/W) |
1.0 °F·h/Btu (3,600.0 °F·s/Btu) |
US Imperial Building HVAC & Insulation Thermal Barrier |
| 1 Degree Fahrenheit Hour per Btu (th) | °F·h/Btu(th) |
1.896903 K/W (1.89690 °C/W) |
1.00067 °F·h/Btu (3,602.41 °F·s/Btu) | Thermochemical HVAC Heat Barrier Standard |
| 1 Degree Fahrenheit Second per Btu (IT) | °F·s/Btu |
0.00052657 K/W (0.526565 mK/W) |
0.00027778 °F·h/Btu (1.0 °F·s/Btu) | High-Speed Transient Thermal Pulse Analysis |
| 1 Degree Fahrenheit Second per Btu (th) | °F·s/Btu(th) |
0.00052692 K/W (0.526918 mK/W) |
0.00027796 °F·h/Btu (1.00067 °F·s/Btu) | Thermochemical Transient Heat Pulse Analysis |
Step-by-Step Semiconductor Thermal Management Calculation Example
To calculate the die junction temperature (TJ) of a computer CPU dissipating 150 Watts (150 W) in an ambient room temperature of 25°C (TA = 25°C), given:
- Silicon Junction-to-Case Resistance (θJC): 0.10 K/W (0.10 °C/W)
- Thermal Paste Interface Resistance (θCS): 0.05 K/W (0.05 °C/W)
- AIO Liquid Cooler Heat Sink Resistance (θSA): 0.12 K/W (0.12 °C/W)
Step 1 (Total Thermal Resistance Σθ): Σθ = 0.10 + 0.05 + 0.12 = 0.27 K/W (0.27 °C/W)
Step 2 (Imperial °F·h/Btu Conversion): Σθ = 0.27 × 0.5275279 = 0.14243 °F·h/Btu
Step 3 (Junction Temperature Rise ΔT): ΔT = P · Σθ = 150 W × 0.27 °C/W = 40.5 °C
Step 4 (Final Silicon Die Junction Temp TJ): TJ = TA + ΔT = 25°C + 40.5°C = 65.5 °C
Thus, with a total thermal resistance of 0.27 K/W (0.1424 °F·h/Btu), the CPU die runs safely at 65.5°C under a 150W thermal load.
Real-World Electronics Cooling & Heat Sink Resistance Benchmarks
Below is a comparative reference chart showing absolute thermal resistance values (θ) across CPU dies, liquid coolers, desktop air heatsinks, and TO-220 pads:
| Cooling Component / Thermal Barrier | Thermal Resistance in Metric (K/W or °C/W) | Imperial Equivalent (°F·h/Btu) | Electronics Thermal Management Context |
|---|---|---|---|
| High-Performance CPU Silicon Junction-to-Case (θJC) | 0.10 K/W (0.10 °C/W) | 0.05275 °F·h/Btu | Direct heat conduction through soldered copper heat spreader (IHS) |
| 360mm AIO Liquid CPU Cooler Heat Sink (θSA) | 0.10 K/W (0.10 °C/W) | 0.05275 °F·h/Btu | High-airflow liquid cooling radiator heat dissipation to air |
| Dual-Tower Desktop CPU Air Cooler (θSA) | 0.25 K/W (0.25 °C/W) | 0.13188 °F·h/Btu | Heavy aluminum fin tower heat sink with heat pipes |
| TO-220 Power MOSFET Silicone Thermal Pad (θCS) | 1.50 K/W (1.50 °C/W) | 0.79129 °F·h/Btu | Electrical insulating gap pad between transistor flange and heat sink |
| Small Extruded Aluminum PCB Heatsink (Natural Convection) | 12.0 K/W (12.0 °C/W) | 6.33033 °F·h/Btu | Passive IC cooler without forced fan airflow |
History & Physics: 1822 Fourier Law of Conduction vs Semiconductor Thermal Ohm’s Law
1822 Joseph Fourier & Law of Heat Conduction (Rth = L ÷ (k · A))
In 1822, French mathematician Joseph Fourier published Théorie Analytique de la Chaleur, establishing the mathematical equation of heat conduction: Q = -k · A · (dT ÷ dx). Fourier proved that heat flow rate Q is inversely proportional to a material barrier’s thickness and thermal conductivity, defining absolute thermal resistance (Rth = L ÷ (k · A) in K/W), establishing the thermal foundation of thermodynamics.
Semiconductor Junction Temperature & Thermal Ohm’s Law Circuit Modeling
In modern microelectronics, thermal engineers model heat dissipation using Thermal Ohm’s Law circuits. Heat power generated inside a microscopic silicon die (P in Watts) flows through a series network of thermal resistances (θJC junction-to-case + θCS case-to-sink + θSA sink-to-ambient) to ambient air (TA). Computing TJ = TA + P · Σθ prevents thermal runaway and silicon gate destruction.
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Frequently Asked Questions (FAQ)
What is the relationship between K/W and °C/W?
1 Kelvin per Watt (1 K/W) is exactly equal to 1 Degree Celsius per Watt (1 °C/W). Because both units measure temperature difference per watt of power, their numerical values are 1:1 identical.
How do you convert K/W to °F·h/Btu?
To convert Kelvin per Watt (K/W) to Degree Fahrenheit Hour per Btu (°F·h/Btu), multiply K/W by 0.527528. For example, 0.5 K/W × 0.527528 = 0.2638 °F·h/Btu.
How do you convert °F·h/Btu to K/W?
To convert Degree Fahrenheit Hour per Btu (°F·h/Btu) to Kelvin per Watt (K/W), multiply °F·h/Btu by 1.895634. For example, 1.0 °F·h/Btu × 1.895634 = 1.8956 K/W.
Why is a lower K/W rating better for a CPU heat sink?
A lower K/W rating means the heat sink offers less resistance to heat flow. A heat sink rated at 0.10 K/W will experience only a 10°C temperature rise per 100W dissipated, keeping the CPU much cooler than a heat sink rated at 0.50 K/W (which would rise by 50°C per 100W).