Sensible heat Q = mcΔT: convert kJ to kWh and estimate ideal heating time

Bridge thermodynamics (kJ) to electrical engineering (kWh & kW) via 3,600!

Heating mass m (kg) with specific heat capacity Cp (water = 4.184 kJ/(kg*°C)) from T_in to T_out requires Sensible Heat Q_kJ = m * Cp * (T_out - T_in). Because 1 Watt = 1 Joule/sec, 1 Kilowatt-Hour equals 1 kJ/s * 3,600 s = 3,600 kJ!

Heating a 150 kg (150 L) tank of cold 15°C water up to 55°C (dT = 40°C) absorbs 150 * 4.184 * 40 = 25,104 kJ = 25,104 / 3,600 = 6.9733 kWh—taking 6.9733 / 4.5 kW = 1.55 hours with a 4.5 kW heating element!

The example assumes constant specific heat, no phase change, no heat loss and a uniformly heated mass. Treat 150 kg as approximately 150 litres of water; density and heat capacity depend on temperature.

The kWh result is heat absorbed by the water, not automatically metered electrical or fuel consumption. Divide by an appropriate efficiency when input power is specified, or use delivered thermal power directly. Real tank heating time also depends on losses and control behavior.

Thermal energy (kWh) and heater recovery-time formulas

Divide thermal kJ by 3,600 to express the same energy in kWh.

=ROUND((A2 * B2 * (D2 - C2)) / 3600, 4)

Heater recovery hours required at Power Rating E2 (kW)

E2 is constant delivered thermal power in kW. It must be positive; the heating-time example requires target temperature >= initial temperature.

=IF(OR(E2<=0,D2<C2),"Check power and temperatures",ROUND(((A2 * B2 * (D2 - C2)) / 3600) / E2, 4))

Further reading

• Microsoft Support — ROUND function