Specific Heat Capacity Converter
Convert specific heat between J/(kg·K), cal/(g·°C), BTU/(lb·°F) and more.
Result
4,186.8J/(kg·K)
1 Calorie (IT) per gram °C = 4,186.8 Joule per kilogram kelvin
1 cal/(g·°C) in every unit
| Joule per kilogram kelvin | 4,186.8 J/(kg·K) |
| Kilojoule per kilogram kelvin | 4.1868 kJ/(kg·K) |
| Joule per gram °C | 4.1868 J/(g·°C) |
| Calorie (IT) per gram °C | 1 cal/(g·°C) |
| Calorie (th) per gram °C | 1.000669216 cal_th/(g·°C) |
| Kilocalorie (IT) per kilogram °C | 1 kcal/(kg·°C) |
| BTU (IT) per pound °F | 1 BTU/(lb·°F) |
Show the work
Find the conversion factor from Calorie (IT) per gram °C to Joule per kilogram kelvin
1 cal/(g·°C) = 4,186.8 J/(kg·K)
factor = 4,186.8 ÷ 1 = 4,186.8 J/(kg·K) per cal/(g·°C)
Multiply by the factor: cal/(g·°C) cancels, leaving J/(kg·K)
J/(kg·K) = cal/(g·°C) × (J/(kg·K) per cal/(g·°C))
1 cal/(g·°C) × 4,186.8 J/(kg·K) per cal/(g·°C) = 4,186.8 J/(kg·K)
How it works
1 cal/(g·°C) = 4186.8 J/(kg·K), so J/(kg·K) = cal/(g·°C) × 4186.8
Calorie (IT) per gram °C to Joule per kilogram kelvin table
| Calorie (IT) per gram °C | Joule per kilogram kelvin |
|---|---|
| 1 cal/(g·°C) | 4,186.8 J/(kg·K) |
| 2 cal/(g·°C) | 8,373.6 J/(kg·K) |
| 5 cal/(g·°C) | 20,934 J/(kg·K) |
| 10 cal/(g·°C) | 41,868 J/(kg·K) |
| 20 cal/(g·°C) | 83,736 J/(kg·K) |
| 50 cal/(g·°C) | 209,340 J/(kg·K) |
| 100 cal/(g·°C) | 418,680 J/(kg·K) |
| 1,000 cal/(g·°C) | 4,186,800 J/(kg·K) |
Questions
- How do I convert calorie (IT) per gram °C to joule per kilogram kelvin?
- Multiply by 4186.8. For example, 10 cal/(g·°C) = 41,868 J/(kg·K).
- Which units does the specific heat capacity converter support?
- Joule per kilogram kelvin (J/(kg·K)), Kilojoule per kilogram kelvin (kJ/(kg·K)), Joule per gram °C (J/(g·°C)), Calorie (IT) per gram °C (cal/(g·°C)), Calorie (th) per gram °C (cal_th/(g·°C)), Kilocalorie (IT) per kilogram °C (kcal/(kg·°C)), BTU (IT) per pound °F (BTU/(lb·°F)).
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