Heat energy from Q = mcΔT (solve for heat, mass, final temperature or specific heat), latent heat Q = mL for melting or boiling, and Carnot efficiency.
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Adding 334.88 kJ to 1 kg of material with c = 4,186 J/(kg·K) changes its temperature by 80 K, from 20 °C to 100 °C.
A change of 1 °C equals a change of 1 K.
Warming or cooling a material takes heat Q = mcΔT, where m is the mass, c the specific heat capacity and ΔT the temperature change; the calculator solves for any one of heat, mass, final temperature or specific heat. Melting or boiling takes Q = mL at constant temperature, where L is the latent heat. The Carnot mode gives the upper limit on any heat engine's efficiency, 1 − Tc/Th with both temperatures in kelvin.
The default, 1 kg of water heated from 20 °C to 100 °C, needs 334,880 J (0.093 kWh), what a 2 kW kettle delivers in 2 minutes 47 seconds with no losses. Boiling that water away takes a further 2,256 kJ, almost seven times as much.
Specific heats are room-temperature values from OpenStax University Physics (Table 1.3). In reality c varies with temperature, and the calculation assumes no melting or boiling between the two temperatures.
Multiply mass, specific heat capacity and temperature change: Q = m × c × ΔT. Heating 1 kg of water (c = 4,186 J/(kg·K)) from 20 °C to 100 °C takes 1 × 4,186 × 80 = 334,880 J, or 0.093 kWh. The formula holds while the material stays in one phase; melting and boiling need latent heat.
Rearrange for whichever quantity is unknown:
Adding 9 kJ to 500 g of aluminum (c = 900 J/(kg·K)) raises it by 9,000 ÷ (0.5 × 900) = 20 K, so 20 °C becomes 40 °C. If 3,870 J warms 1 kg of an unknown metal by 10 K, its c is 387 J/(kg·K), which matches copper.
Heat removed is negative. Cooling 1 kg of water from 80 °C to 20 °C gives Q = 1 × 4,186 × (20 − 80) = −251,160 J.
Near room temperature, in J/(kg·K), from OpenStax Table 1.3:
Specific heat drifts with temperature, so a single value suits a moderate range. Over a wide range, split the calculation or use a temperature-dependent value from a property table.
A change of 1 °C is a change of 1 K, so ΔT is the same number on both scales. A change of 1 °F is 5/9 K. The aluminum example above rises 20 K, which is 36 °F: from 68 °F to 104 °F.
In US units water needs about 1 BTU per pound per °F. Heating 10 lb of water from 50 °F to 120 °F, a 38.89 K rise, takes 699.9 BTU.
During a phase change the temperature stays put while the heat changes the phase. Latent heats from OpenStax Table 1.4, with melting and boiling points:
Melting 2 kg of ice takes 2 × 334,000 = 668,000 J. Boiling away 0.5 kg of water takes 0.5 × 2,256,000 = 1,128,000 J. Boiling points fall with pressure, so water boils below 100 °C at altitude.
Taking 1 kg of ice at −10 °C to steam at 100 °C needs four separate steps:
The total is 3,029,500 J, about 3.03 MJ, and boiling alone is 74% of it.
No heat engine working between a hot reservoir Th and a cold one Tc can beat η = 1 − Tc/Th, with both temperatures in kelvin.
Between 500 K and 300 K the limit is 40%, so 1,000 J taken from the hot side yields at most 400 J of work. Between boiling and freezing water it is 26.80%.
Run backward, the same pair sets the best possible coefficient of performance: Tc/(Th − Tc) for a refrigerator and Th/(Th − Tc) for a heat pump, 1.5 and 2.5 for 500 K and 300 K.
A COP above 1 counts heat moved per unit of work, not an efficiency over 100% (OpenStax on the Carnot cycle).
The rows are the 1 kg water example (Default), 9 kJ into 500 g of aluminum from 20 °C (Solve final), 2 kg of ice melted at 0 °C (Melt water) and the 500 K to 300 K engine (Carnot).
| Scenario | Selected result |
|---|---|
| Default | 334,880 J |
| Solve final | 40 °C |
| Melt water | 668,000 J |
| Carnot | 40% |
For how a gas's pressure and volume respond to temperature, use the ideal gas law calculator. To convert readings between scales, use the temperature converter.
Checked against: 1 × 4186 × 80 = 334880 J (OpenStax c_water = 4186)
Checked against: 1 × 4186 × (20 − 80)
Checked against: ΔT = 9000/(0.5 × 900) = 20 K
Checked against: 3870/(1 × 10) = 387 J/(kg·K), copper in OpenStax Table 1.3
4,186 J per kilogram per degree Celsius, water's specific heat capacity. Heating 1 litre (1 kg) from 20 °C to 100 °C takes 1 × 4,186 × 80 = 334,880 J, or 0.093 kWh. In US units that is about 1 BTU per pound per °F, which is how the BTU was originally defined.
The heat needed to raise 1 kg of a substance by 1 K (the same as 1 °C), in J/(kg·K). Water's is 4,186, among the highest of common substances, while copper's is 387 and lead's 128. The same 10 kJ warms 1 kg of water by 2.4 °C but 1 kg of copper by 25.8 °C, which is why water is used for cooling and heat storage.
The energy absorbed or released during a phase change at constant temperature, Q = mL. For water the latent heat of fusion is 334 kJ/kg at 0 °C and of vaporisation 2,256 kJ/kg at 100 °C (OpenStax Table 1.4). Melting 2 kg of ice takes 668 kJ, enough to heat the same 2 kg of water by about 80 °C.
η = 1 − Tc/Th, the largest fraction of heat that any engine can turn into work between a hot reservoir at Th and a cold one at Tc, both in kelvin. Between 500 K and 300 K it is 40%; between boiling and freezing water, 26.8%. Real engines fall short of it: coal-fired power stations typically convert about 37% of their fuel's heat into electricity.
COP = Th/(Th − Tc) with temperatures in kelvin, the Carnot limit on heat delivered per unit of work. Pumping heat from 0 °C outdoors into a 35 °C heating loop allows at most 308.15/35 ≈ 8.8. At −10 °C outside the limit drops to 6.8, and real machines stay well below it because of compressor and heat-exchanger losses.
This calculator includes 10 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.