Specific Heat Capacity Calculator
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Try an example
Heat needed
334.7 kJ
334.7 kJ · 80 kcal · 317.3 BTU · 0.09298 kWh
Q = m × c × ΔT = 1,000 g × 4.184 J/(g·°C) × 80 °C = 334,700 J
takes at least 3.72 minutes; heat lost to the surroundings makes it longer.
Calculate the heat needed to warm or cool a material with Q = mcΔT, or solve for the mass, the specific heat or the final temperature, in joules, calories, BTU or kWh, with specific heats for water, metals and other common materials and the time a heater of a given power needs.
Formula
Worked Example
1,000 g × 4.184 J/(g·°C) × 80 °C = 334,720 J. A 1,500 W element delivers 1,500 J a second, so it needs 223 seconds even if no heat escapes; boiling the water away would take another 2,257 J per gram.
Frequently Asked Questions
What is specific heat capacity?
The energy needed to raise one gram of a material by one degree Celsius, or one kilogram by one kelvin. Water's is 4.184 J/(g·°C); most metals are far lower, such as aluminum at 0.897 and copper at 0.385, so they heat up and cool down much faster for the same energy.
Is a change of 1 °C the same as 1 K?
Yes. The two scales have the same size of degree and differ only in where zero is, so a temperature change is the same number in both. A change of 1 °C equals 1.8 °F.
Does Q = mcΔT work when ice melts or water boils?
No. During a phase change the temperature stays constant while the material absorbs latent heat: 334 J per gram to melt ice at 0 °C and 2,257 J per gram to boil water at 100 °C. Work out each stage separately, using the specific heat of ice, water or steam for the temperature changes.