Specific Heat Capacity
J/(kg·K)toBTU/(lb·°F)
Convert joules per kilogram-kelvin (J/(kg·K)) to BTUs per pound-Fahrenheit (BTU/(lb·°F)).
Factor1 J/(kg·K) = 0.0002388459 BTU/(lb·°F)
Converter
J/(kg·K)
Accepts numbers or expressions, e.g. 150 + 14.7
Result
BTU/(lb·°F)
Rendered to 6 significant figures.
Formula
Formula
BTU/(lb·°F) = J/(kg·K) × 0.0002388459
Multiply any value in joules per kilogram-kelvin by 0.0002388459 to obtain the value in BTUs per pound-Fahrenheit.
Worked example
Convert 4186.8 J/(kg·K) to BTU/(lb·°F).
- 01Start with 4186.8 J/(kg·K).
- 02Multiply by the conversion factor: 4186.8 × 0.0002388459 = 1 BTU/(lb·°F).
Result4186.8 J/(kg·K) = 1 BTU/(lb·°F)
Conversion table
| J/(kg·K) | BTU/(lb·°F) |
|---|---|
| 1 | 0.00023885 |
| 2 | 0.00047769 |
| 5 | 0.0011942 |
| 10 | 0.0023885 |
| 20 | 0.0047769 |
| 50 | 0.011942 |
| 100 | 0.023885 |
| 200 | 0.047769 |
| 500 | 0.11942 |
| 1000 | 0.23885 |
Reference values rounded to 5 significant figures for display.
FAQ
What is the conversion factor from J/(kg·K) to BTU/(lb·°F)?
1 J/(kg·K) equals 0.0002388459 BTU/(lb·°F). To convert, multiply the value in joules per kilogram-kelvin by 0.0002388459.
How do I convert 1 J/(kg·K) to BTU/(lb·°F)?
1 J/(kg·K) = 0.000238846 BTU/(lb·°F). For any value, multiply by 0.0002388459.
How do I convert BTU/(lb·°F) back to J/(kg·K)?
Divide by the same factor — or equivalently, multiply by 4186.8. So 1 BTU/(lb·°F) = 4186.8 J/(kg·K).
When would I need to convert joule per kilogram-kelvin to BTU per pound-Fahrenheit?
Specific-heat-capacity conversions between J/(kg·K) and BTU/(lb·°F) are routine in thermal engineering, heat-balance work, HVAC design, metallurgy, chemical engineering and materials science. J/(kg·K) and kJ/(kg·K) are the SI standards; cal/(g·°C) is the classic thermochemical convention; BTU/(lb·°F) dominates US process and ASHRAE datasheets. All conversions use fixed multiplicative factors — this category does NOT look up the Cp value of any specific material (water, air, steam, etc.), does NOT perform heat-duty calculations (Q = m·Cp·ΔT), and does NOT convert lumped thermal capacity (J/K) to specific heat capacity (J/(kg·K)) without mass. Temperature units in the denominator represent temperature intervals, not absolute temperatures — a 1 °C interval equals a 1 K interval.
Is the conversion exact?
The factor shown is precise to at least 7 significant figures. For most process-engineering work this is far better than instrument accuracy. For metrology or trade applications, refer to the relevant national standard (NIST, BIPM, ISO 80000).
Related conversions
- BTU/(lb·°F) → J/(kg·K)BTU per pound-Fahrenheit → joule per kilogram-kelvin
- J/(kg·K) → kJ/(kg·K)joule per kilogram-kelvin → kilojoule per kilogram-kelvin
- kJ/(kg·K) → J/(kg·K)kilojoule per kilogram-kelvin → joule per kilogram-kelvin
- J/(kg·K) → J/(g·K)joule per kilogram-kelvin → joule per gram-kelvin
- J/(g·K) → J/(kg·K)joule per gram-kelvin → joule per kilogram-kelvin
- J/(kg·K) → cal/(g·°C)joule per kilogram-kelvin → calorie per gram-Celsius