How much energy does it take to boil a kettle? Enter the mass, specific heat capacity and the starting and final temperatures to get the heat energy in joules, kilojoules and kWh.
Enter the mass and specific heat capacity.
Enter the starting and final temperatures.
Read the heat energy.
Heat: Q = m × c × ΔT
When you heat water for tea or cool a hot iron rod, energy flows in or out. This calculator finds how much, using Q = mcΔT. You enter the mass in kg, the specific heat capacity c in J/kg·°C, and the starting and final temperatures. It returns the heat in kilojoules, joules and kilowatt-hours, and labels it as absorbed when the temperature rises or released when it falls.
The kWh figure is handy in India, because electricity bills are charged per unit, and one unit is one kWh. So you can relate a physics answer to what heating a geyser full of water might cost. Class 11 thermal physics, Class 10 and 11 chemistry calorimetry, and engineering heat balance all rest on this same equation.
1. Write the mass of the substance in kilograms.
2. Look up its specific heat capacity c in J/kg·°C. Water is about 4186, aluminium about 900, iron about 450 and copper about 385.
3. Find the temperature change ΔT = final − initial. A kelvin step and a Celsius step are the same size, so either scale works for ΔT.
4. Multiply: Q = m × c × ΔT, in joules.
5. Divide by 1000 for kilojoules, or by 3,600,000 for kilowatt-hours.
6. If Q is negative, the substance released that much heat to its surroundings.
Specific heat capacity is the energy needed to raise 1 kg of a material by 1 °C. It depends on how many ways the particles can store energy. Water's value of about 4186 J/kg·°C is very high because of hydrogen bonding between its molecules, so it warms and cools slowly. Metals have low values, which is why a steel spoon in hot tea heats up quickly. Heat needed is proportional to mass, to c and to the temperature change, giving Q = mcΔT.
Q = mcΔT covers only sensible heat, the heat that changes temperature. At a melting or boiling point, energy goes into breaking bonds instead, and the temperature stays constant until the change of state is complete. This latent heat is Q = mL. Boiling away 1 kg of water at 100 °C takes about 2.26 MJ, far more than the roughly 0.31 MJ needed to heat it from 25 °C to 100 °C. The calculator does not include latent heat, so stop the range at the boiling or melting point.
When a hot object is dropped into cooler water in an insulated container, the heat lost by the hot object equals the heat gained by the water and the container, if no heat escapes. Writing m₁c₁(T₁ − T) = m₂c₂(T − T₂) lets you find a final temperature T or an unknown specific heat. Real setups lose some heat to the air, so measured values usually differ a little from textbook figures, and good experiments keep the time short and the container covered.
Sunita's 15-litre electric geyser holds 15 kg of water that she wants to heat from a winter-morning 18 °C to 60 °C for her bath in Shimla.
Temperature change: ΔT = 60 − 18 = 42 °C
Q = m c ΔT: = 15 × 4186 × 42 = 26,37,180 J
In other units: = 2,637.18 kJ = 0.73255 kWh
Answer: Heat absorbed 2,637.18 kJ; In joules 26,37,180 J; In kWh 0.73255 kWh
Using mass in grams with a specific heat quoted per kilogram.
Running the formula across a boiling or melting point and ignoring latent heat.
Converting both temperatures to kelvin and then adding 273 again to the difference.
Using the specific heat of water for milk, oil or other liquids with different values.
Expecting the electricity used to equal Q, when real heaters also lose heat to the surroundings.
Estimating electricity units needed to heat water in a geyser or kettle.
Class 11 physics and chemistry calorimetry experiments.
Sizing heaters, chillers and heat exchangers in engineering.
Understanding cooking times and why heavy metal pans heat evenly.
Explaining why coastal cities like Chennai have milder temperature swings than inland ones.
Does this include boiling the water away?
No. Changing state needs extra latent heat, 2.26 MJ/kg for boiling water.
Why is my kettle slower than predicted?
Real kettles lose heat to the surroundings, so they need more energy than Q.