Specific heat capacity is how much energy it takes to warm one unit of mass by one degree. It's why a pot of water takes minutes to heat while the air in a room warms quickly, and it's the physics behind every water heater and HVAC calculation.
Q = m × c × ΔT: energy equals mass times specific heat times the temperature change. Water's specific heat is 4.186 J/(g·°C), 1 BTU per pound per °F by the BTU's definition, so heating a 50-gallon tank of water (about 417 lb) by 65°F takes about 27,105 BTU, or 7.94 kWh: about $1.43 at 18 cents per kWh.
Specific Heat Calculator
Q = mcΔT for heating, cooling, and energy storage.
The formula
- Q = m × c × ΔT, in joules with mass in grams and c in J/(g·°C), or in BTU with pounds and BTU/(lb·°F).
- Solve for temperature: ΔT = Q ÷ (m × c).
- Solve for mass: m = Q ÷ (c × ΔT).
One BTU warms one pound of water by 1°F; one kWh is 3,412 BTU.
Specific heats of common materials
The values the calculator uses:
| Material | Specific heat, J/(g·°C) | Density, kg/m³ |
|---|---|---|
| Water | 4.186 | 1,000 |
| Ethanol | 2.44 | 789 |
| Ice | 2.09 | 917 |
| Oak wood | 2.01 | 750 |
| Engine oil | 1.88 | 900 |
| Air (dry) | 1.005 | 1.225 |
| Aluminum | 0.903 | 2,700 |
| Concrete | 0.88 | 2,400 |
| Glass | 0.84 | 2,500 |
| Steel | 0.49 | 7,850 |
| Copper | 0.385 | 8,960 |
Values for wood and concrete vary with species, moisture and mix; treat them as typical.
Why water dominates HVAC
Per unit of volume, water holds about 3,400 times more heat than air, because it's both denser and has a higher specific heat. That's why hydronic systems move heat through small pipes while forced-air systems need large ducts.
Volumetric heat capacity is density times specific heat, so multiplying each row of the table above shows how much heat a cubic meter of each material stores per degree.
| Material | Volumetric heat capacity, kJ/(m³·°C) | Times dry air |
|---|---|---|
| Water | 4,186 | 3,400 |
| Steel | 3,847 | 3,124 |
| Copper | 3,450 | 2,802 |
| Aluminum | 2,438 | 1,980 |
| Concrete | 2,112 | 1,716 |
| Glass | 2,100 | 1,706 |
| Ethanol | 1,925 | 1,564 |
| Ice | 1,917 | 1,557 |
| Engine oil | 1,692 | 1,374 |
| Oak wood | 1,507 | 1,224 |
| Air (dry) | 1.23 | 1 |
Water stores 4,186 kJ per cubic meter per degree, about 3,400 times dry air's 1.23. Dense solids such as steel and concrete also store a lot per volume, which is why slabs and masonry act as thermal mass. Per pound, though, water still leads steel by about 8.5 to 1, since steel's specific heat is only 0.49 against water's 4.186.
Worked examples
- Water heater: 50 gallons is about 417 lb; heating it 65°F takes 417 × 65 = 27,105 BTU, or 7.94 kWh, about $1.43 at 18 cents per kWh with a resistance element.
- Antifreeze: a 50% propylene glycol mix has a specific heat of about 0.81 BTU/(lb·°F) and a density of about 8.80 lb per gallon, so a gallon carries 7.15 BTU per °F against water's 8.34: about 14% less, which a solar or hydronic system makes up with more flow.
- Air: heating air takes far less energy per pound but far more volume; see BTU to CFM for ductwork.
- A room of air: a 12 by 15 ft room with an 8 ft ceiling holds 1,440 cubic feet, about 40.8 cubic meters, which at air's 1.225 kg/m³ is 50 kg or 110 lb. Air's specific heat is 0.24 BTU/(lb·°F), from 1.005 ÷ 4.186, so warming it 20°F takes only 110 × 0.24 × 20 = 529 BTU, about 0.15 kWh.
- Cooling a concrete slab: a 20 by 20 ft slab 4 inches thick is 133 cubic feet, about 3.78 cubic meters, or 9,060 kg (19,980 lb) at concrete's 2,400 kg/m³. Concrete's specific heat is 0.21 BTU/(lb·°F), from 0.88 ÷ 4.186, so dropping it 10°F releases 19,980 × 0.21 × 10 = about 42,000 BTU, or 12.3 kWh. That stored energy is why a slab floor smooths out indoor temperature swings.
- Heating a pool: 15,000 gallons weighs about 125,100 lb at 8.34 lb per gallon, so a 10°F rise needs 125,100 × 1 × 10 = 1,251,000 BTU. That is 367 kWh, roughly $66 of resistance electricity at 18 cents, or 12.5 therms of delivered heat (about $16.89 at an assumed $1.35 per therm) before a gas heater's efficiency losses.
- Reheating after a setback: the air in that 12 by 15 ft room reheats for only about 0.15 kWh, so the real energy goes into rewarming the walls, floor and furnishings and replacing heat the envelope lost while the house sat cool. That envelope loss is what the heat-loss model behind the heating cost calculator estimates. Setting a thermostat back 7–10°F for 8 hours a day still saves up to 10% a year, per DOE, because the building fabric cools slowly and demand drops while you are away.
Converting between joules, BTU, kWh and calories
The same energy can be written four ways, and this calculator moves between them.
- One kilowatt-hour is 3,600,000 joules by definition (1,000 watts for 3,600 seconds), and 3,412 BTU per EIA.
- One BTU is therefore 3,600,000 ÷ 3,412 = about 1,055 joules.
- One calorie is 4.186 joules, the energy that warms one gram of water by 1°C, which is water's specific heat by definition.
- One food Calorie (kcal) is 1,000 of those calories, or 4,186 joules.
Worked conversion: warming 10 kg of water by 20°C takes 10,000 × 4.186 × 20 = 837,200 J. Dividing by 4.186 gives 200,000 cal (200 kcal), by 1,055 gives 794 BTU, and by 3,600,000 gives 0.23 kWh, about 4 cents at 18 cents per kWh.
When Q = mcΔT does not apply
Q = mcΔT only holds while a material stays in one phase. During melting or boiling the temperature holds steady while the material absorbs or releases latent heat, so ΔT is zero and the formula returns zero even though real energy is flowing.
Ice warming from 20°F to 32°F and water cooling toward 212°F follow Q = mcΔT, but melting the ice at 32°F or boiling the water at 212°F does not. A full ice-to-steam problem is the sum of three mcΔT segments plus two latent-heat terms, one for melting and one for boiling.
This calculator reports only the within-phase energy, so it understates any process that crosses a freezing or boiling point.
Frequently asked questions
What is the specific heat of water?
4.186 J/(g·°C), or 1 BTU/(lb·°F).
How do I calculate the energy needed to heat water?
Multiply pounds of water by the temperature rise in °F: the answer is in BTU. Divide by 3,412 for kWh.
Why does water take so long to heat?
Its specific heat is high: about four times that of air per unit mass, and about 3,400 times per unit volume.
What is the formula Q = mcΔT?
Heat energy = mass × specific heat × temperature change.
How much energy does it take to heat a pool?
Multiply the water weight by the temperature rise in °F to get BTU: a 15,000-gallon pool (about 125,100 lb) raised 10°F needs 1,251,000 BTU. That is 367 kWh, roughly $66 of resistance electricity at 18 cents per kWh, before any cover or evaporation losses.
What is the difference between specific heat and heat capacity?
Specific heat is per unit mass, such as water's 4.186 J/(g·°C), and is a fixed property of the material. Heat capacity belongs to a specific object and equals its mass times the specific heat, in J/°C or BTU/°F. A full 50-gallon tank (417 lb) has a heat capacity of 417 BTU/°F, so each degree of rise takes 417 BTU.
Why does Q = mcΔT not work when water boils?
At the boiling point the added energy turns liquid into vapor instead of raising the temperature, so ΔT stays at zero and the formula gives zero. Boiling and melting are latent-heat processes that sit outside Q = mcΔT, which only covers temperature change within a single phase.
Sources & References
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