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Specific Heat Capacity

J/(g·K)toBTU/(lb·°F)

Convert joules per gram-kelvin (J/(g·K)) to BTUs per pound-Fahrenheit (BTU/(lb·°F)).

Factor1 J/(g·K) = 0.2388459 BTU/(lb·°F)

Converter

J/(g·K)

Accepts numbers or expressions, e.g. 150 + 14.7

Result
0.999331BTU/(lb·°F)

Rendered to 6 significant figures.

Formula

Formula
BTU/(lb·°F) = J/(g·K) × 0.2388459

Multiply any value in joules per gram-kelvin by 0.2388459 to obtain the value in BTUs per pound-Fahrenheit.

Worked example

Convert 4.184 J/(g·K) to BTU/(lb·°F).

  1. 01Start with 4.184 J/(g·K).
  2. 02Multiply by the conversion factor: 4.184 × 0.2388459 = 0.999331 BTU/(lb·°F).
Result4.184 J/(g·K) = 0.999331 BTU/(lb·°F)

Conversion table

J/(g·K)BTU/(lb·°F)
10.23885
20.47769
51.1942
102.3885
204.7769
5011.942
10023.885
20047.769
500119.42
1000238.85

Reference values rounded to 5 significant figures for display.

FAQ

What is the conversion factor from J/(g·K) to BTU/(lb·°F)?
1 J/(g·K) equals 0.2388459 BTU/(lb·°F). To convert, multiply the value in joules per gram-kelvin by 0.2388459.
How do I convert 1 J/(g·K) to BTU/(lb·°F)?
1 J/(g·K) = 0.238846 BTU/(lb·°F). For any value, multiply by 0.2388459.
How do I convert BTU/(lb·°F) back to J/(g·K)?
Divide by the same factor — or equivalently, multiply by 4.1868. So 1 BTU/(lb·°F) = 4.1868 J/(g·K).
When would I need to convert joule per gram-kelvin to BTU per pound-Fahrenheit?
Specific-heat-capacity conversions between J/(g·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).

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