Heat Transfer Calculator
Calculate the energy, power and flow needed to change an object's temperature.
Formula and sources updated · Data sources · 1View algorithm, constants and default inputs (JSON)
About this calculator
Use mass in kg, specific heat in J/(kg·K), time in seconds, and both temperatures in the same scale (the page inputs are °C). Temperature differences in °C and K have the same numeric value. Energy is m·c·ΔT; power is energy divided by time, and input power is divided by efficiency.
Worked example
calculated with the default values below
Mass 2 kg · Specific heat capacity 4,186 J/(kg·K) · Initial temperature 20 °C · Final temperature 80 °C · Time allowed 300 s · Heating efficiency 100 % Energy required 502,320 J · Power needed 1,674.4 W · Temperature change 60.00 K · Energy required 502.32 kJ · Energy required 120.06 kcal · Power drawn at this efficiency 1,674.4 W · Heat flow Energy must be added
Method and assumptions
Formula basis and scope
Q = mcΔT in joules; average power = Q/time; input power divides by the efficiency fraction.
Constant entered specific heat, no phase transition. Negative energy denotes cooling; heat loss is represented only by the entered efficiency.
Data sources
Disclaimer
These calculators give general mathematical results. They suit study, planning and estimation — verify anything that feeds a graded submission, a published figure or a financial commitment.
Implementation source
This record describes the implementation used by this page. It is not independent validation. External references, when available, support only the stated formula or definition; estimates and custom rules retain their own assumptions.
Formula and variables
Q=m·c·(Tf−Ti); P=Q/t; Pin=P/(efficiency/100)
- m
- Mass, kg
- c
- Specific heat capacity, J/(kg·K)
- T_i
- Initial temperature
- T_f
- Final temperature
- t
- Time, s
- Q
- Heat transferred, J
Q=m·c·ΔT; P=Q/t; input power=P/(efficiency/100). Use m in kg, c in J/(kg·K), t in seconds, and consistent temperature units; Δ°C and ΔK have equal numerical size.
Frequently asked questions
What does the default water example calculate?
With 2 kg of water, specific heat 4,186 J/(kg·K), a 60°C rise, and 300 seconds, the model gives 502,320 J (502.32 kJ) and 1,674.4 W at 100% efficiency. Lower efficiency requires more input power. Use SI units and one temperature scale for both temperature fields.