Specific Heat Capacity Converter
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| Unit | Equivalent Value |
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Specific Heat Capacity (symbolized by c or cp) measures the thermal energy required to raise the temperature of a unit mass of a substance by one degree (Q = m · c · ΔT ⇒ c = Q ÷ (m · ΔT)). Across solar thermal energy storage, HVAC hydronic heating calculations, engine cooling system design, metallurgical heat treatment, thermal runaway prevention in lithium-ion batteries, and chemical reactor design, specific heat capacity is categorized across three major engineering unit families: International System of Units (SI metric fundamental: Joules per Kilogram Kelvin / J/(kg·K), Joules per Kilogram °C / J/(kg·°C), Kilojoules per Kilogram Kelvin / kJ/(kg·K), Joules per Gram °C / J/(g·°C)), Imperial / US Customary standards (Btu per Pound °F / Btu/(lb·°F), Btu per Pound °R / Btu/(lb·°R), Centigrade Heat Units per Pound °C / CHU/(lb·°C)), and CGS laboratory metrics (Calories per Gram °C / cal/(g·°C), Kilocalories per Kilogram °C / kcal/(kg·°C)).
Our free online Specific Heat Capacity Converter provides instant, high-precision conversions across all SI metric, Imperial HVAC, thermal storage, and laboratory specific heat units:
- Remarkable Imperial-Metric Identity:
1 Btu/(lb·°F) = 1.0 cal/(g·°C) = 1.0 kcal/(kg·°C) = 1.0 CHU/(lb·°C) = 4,186.8 J/(kg·K) = 4.1868 kJ/(kg·K). - Btu (IT) per Pound °F to J/(kg·K): Multiply Btu/(lb·°F) by
4186.8(1 Btu/(lb·°F) = 4,186.8 J/(kg·K) = 4.1868 kJ/(kg·K) = 4.1868 J/(g·°C)). - Joules per Kilogram Kelvin to Btu/(lb·°F): Multiply J/(kg·K) by
0.0002388459(1 J/(kg·K) = 0.00023885 Btu/(lb·°F) = 0.001 kJ/(kg·K)). - Joules per Gram °C [J/(g·°C)] to J/(kg·K): Multiply by
1,000.0(1 J/(g·°C) = 1,000 J/(kg·K) = 1.0 kJ/(kg·K) = 0.23885 Btu/(lb·°F)). - Pound-Force Foot per Pound °R [lbf·ft/(lb·°R)] to J/(kg·K): Multiply by
5.3803205(1 lbf·ft/(lb·°R) = 5.38032 J/(kg·K)). - Kilogram-Force Meter per Kilogram K to J/(kg·K):
1 kgf·m/(kg·K) = 9.80665 J/(kg·K).
Master Specific Heat Capacity Conversion Table
The table below displays exact mathematical conversion relationships, SI J/(kg·K) multipliers, and imperial Btu/(lb·°F) equivalents relative to 1 Joule per Kilogram Kelvin (1 J/(kg·K) = 1 J/(kg·°C)):
| Specific Heat Capacity Unit Name | Symbol | Exact Value in J/(kg·K) | Btu/(lb·°F) & cal/(g·°C) Equivalent | Domain & Technical Application Standard |
|---|---|---|---|---|
| 1 Joule per Kilogram Kelvin | J/(kg·K), J/(kg·°C) |
1.0 J/(kg·K) (Base SI Unit) |
0.000238846 Btu/(lb·°F) (0.000238846 cal/(g·°C) / 0.001 kJ/(kg·K)) |
SI Fundamental Specific Heat Capacity Standard |
| 1 Btu (IT) per Pound °F | Btu/(lb·°F) |
4,186.8 J/(kg·K) (4.1868 kJ/(kg·K)) |
1.0 Btu/(lb·°F) (1.0 cal/(g·°C) / 1.0 kcal/(kg·°C)) |
US Imperial HVAC & Liquid Thermal Storage Standard |
| 1 Calorie (IT) per Gram °C | cal/(g·°C) |
4,186.8 J/(kg·K) (4.1868 kJ/(kg·K)) |
1.0 Btu/(lb·°F) (1.0 cal/(g·°C) / 4.1868 J/(g·°C)) |
Laboratory Calorimetry & Water Reference Benchmark |
| 1 Kilocalorie (IT) per Kilogram °C | kcal/(kg·°C), kcal/(kg·K) |
4,186.8 J/(kg·K) |
1.0 Btu/(lb·°F) (1.0 cal/(g·°C)) | European Thermal Engineering & Food Process Standard |
| 1 Kilojoule per Kilogram Kelvin | kJ/(kg·K), kJ/(kg·°C) |
1,000.0 J/(kg·K) |
0.238846 Btu/(lb·°F) (0.238846 cal/(g·°C) / 1.0 J/(g·°C)) | Industrial Gas Thermodynamics & Energy Engineering |
| 1 Joule per Gram °C | J/(g·°C) |
1,000.0 J/(kg·K) |
0.238846 Btu/(lb·°F) (1.0 kJ/(kg·K)) | Small-Sample Differential Scanning Calorimetry (DSC) |
| 1 Calorie (Thermochemical) per Gram °C | cal(th)/(g·°C) |
4,184.0 J/(kg·K) (4.1840 kJ/(kg·K)) |
0.999333 Btu/(lb·°F) (0.999333 cal/(g·°C)) | Thermochemical Reaction Engineering Standard |
| 1 Centigrade Heat Unit per Pound °C | CHU/(lb·°C) |
4,186.8 J/(kg·K) |
1.0 Btu/(lb·°F) (1.0 cal/(g·°C)) | Historical British Power Plant Engineering Standard |
Step-by-Step Thermal Energy Storage Calculation Example
To calculate the heat energy (Q) required to heat 500 Kilograms (500 kg) of liquid water from 20°C to 80°C (ΔT = 60°C = 60 K), given water’s specific heat capacity of 4,184.0 Joules per Kilogram Kelvin (4.184 kJ/(kg·K) or 1.000 Btu/(lb·°F)):
Step 1 (SI Metric Heat Energy Q): Q = m · c · ΔT = 500 kg × 4,184 J/(kg·K) × 60 K = 125,520,000 Joules (125.52 MJ)
Step 2 (Kilowatt-Hour Electrical Equivalent): Q = 125.52 MJ ÷ 3.6 = 34.867 kWh (34.87 kWh)
Step 3 (Imperial Verification): Q = 1,102.31 lb × 1.0 Btu/(lb·°F) × 108°F = 119,050 Btu (125.5 MJ)
Thus, heating 500kg of water across a 60°C span absorbs 125.52 Megajoules (34.87 kWh) of thermal energy.
Real-World Material Specific Heat Capacity Benchmarks
Below is a comparative reference chart showing specific heat capacities (c) across water, ice, steam, air, aluminum, steel, copper, and lead:
| Substance / Engineering Material | Specific Heat in SI (J/(kg·K) / kJ/(kg·K)) | Imperial & CGS Equivalent (Btu/(lb·°F) / cal/(g·°C)) | Thermal & Physical Engineering Context |
|---|---|---|---|
| Liquid Water (at 20°C, 1 atm) | 4,184.0 J/(kg·K) (4.184 kJ/(kg·K)) | 1.000 Btu/(lb·°F) (1.000 cal/(g·°C)) | High heat capacity stabilizing Earth’s climate & hydronic HVAC |
| Solid Water Ice (at -10°C) | 2,110.0 J/(kg·K) (2.110 kJ/(kg·K)) | 0.504 Btu/(lb·°F) (0.504 cal/(g·°C)) | Solid phase water specific heat (roughly half liquid water) |
| Saturated Water Vapor Steam (at 100°C) | 2,080.0 J/(kg·K) (2.080 kJ/(kg·K)) | 0.497 Btu/(lb·°F) (0.497 cal/(g·°C)) | Gaseous steam constant-pressure specific heat (cp) |
| Dry Air (at 20°C, 1 atm Sea Level) | 1,005.0 J/(kg·K) (1.005 kJ/(kg·K)) | 0.240 Btu/(lb·°F) (0.240 cal/(g·°C)) | Ambient atmospheric air constant-pressure specific heat |
| Pure Aluminum Metal (Element 13) | 900.0 J/(kg·K) (0.900 kJ/(kg·K)) | 0.215 Btu/(lb·°F) (0.215 cal/(g·°C)) | High specific heat lightweight structural metal |
| Carbon Structural Steel / Iron | 470.0 J/(kg·K) (0.470 kJ/(kg·K)) | 0.112 Btu/(lb·°F) (0.112 cal/(g·°C)) | Standard civil engineering structural beam & rebar heat capacity |
| Pure Copper Metal (Element 29) | 385.0 J/(kg·K) (0.385 kJ/(kg·K)) | 0.092 Btu/(lb·°F) (0.092 cal/(g·°C)) | Electrical busbar & CPU heat spreader thermal storage |
| Heavy Lead Metal (Element 82) | 128.0 J/(kg·K) (0.128 kJ/(kg·K)) | 0.0305 Btu/(lb·°F) (0.0305 cal/(g·°C)) | Low mass specific heat due to high atomic mass (Dulong-Petit) |
History & Physics: 1760s Joseph Black Specific Heat Discovery vs 1819 Petit-Dulong Law
1760s Joseph Black & the Discovery of Specific Heat
In the 1760s, Scottish physician and chemist Joseph Black performed landmark calorimetry experiments at the University of Glasgow, proving that equal masses of different materials require unequal quantities of heat to achieve the same temperature rise. Black introduced the concept of Specific Heat Capacity (c) and latent heat, disproving the old caloric theory that heat was a uniform weightless fluid.
1819 Dulong-Petit Law of Molar Heat Capacity (Cm ≈ 3R ≈ 24.9 J/(mol·K))
In 1819, French scientists Pierre Louis Dulong and Alexis Thérèse Petit discovered that the molar specific heat capacity of solid elements is approximately constant: Cm = M · c ≈ 3R ≈ 24.9 J/(mol·K) (where R is the universal gas constant). This law explains why heavy atomic elements like lead (M = 207 g/mol) have a low mass specific heat (128 J/(kg·K)), whereas light atomic elements like aluminum (M = 27 g/mol) have a high mass specific heat (900 J/(kg·K)).
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Frequently Asked Questions (FAQ)
Why does 1 Btu/(lb·°F) equal 1.0 cal/(g·°C)?
Both 1 Btu/(lb·°F) and 1 cal/(g·°C) are defined relative to the specific heat capacity of liquid water. By definition, 1 Btu/(lb·°F) = 1.0 cal/(g·°C) = 1.0 kcal/(kg·°C) = 4,186.8 J/(kg·K).
How do you convert J/(kg·K) to Btu/(lb·°F)?
To convert Specific Heat Capacity from J/(kg·K) to Btu/(lb·°F), multiply by 0.0002388459 (or divide by 4,186.8). For example, aluminum’s 900 J/(kg·K) × 0.0002388459 = 0.215 Btu/(lb·°F).
What substance has the highest specific heat capacity?
Liquid water has an unusually high specific heat capacity of 4,184 J/(kg·K) (1.00 Btu/(lb·°F)) due to extensive intermolecular hydrogen bonding. Pure liquid ammonia is higher at ~4,700 J/(kg·K), and hydrogen gas at high temperature reaches ~14,300 J/(kg·K).
What is the formula to calculate heat energy using specific heat?
The heat energy required is calculated as Q = m · c · ΔT (where m is mass in kg, c is specific heat in J/(kg·K), and ΔT is the temperature difference in K or °C).