Editorially reviewedReviewed by Agarapu Ramesh, science educator (chemistry). LinkedIn
Last reviewed: May 2026|Standard temperature conversion formulas
What this calculator does
This mixing temperature calculator estimates the final equilibrium temperature of combined materials using a weighted thermal energy balance. It is especially useful for liquid blends and simple classroom thermodynamics where mixing losses are small.
When two or more bodies at different temperatures are mixed, heat flows from the warmer material to the cooler material until equilibrium is reached. If you know the mass, specific heat, and starting temperature of each stream, you can estimate the final mixture temperature without solving a full transient model.
Inputs explained
Mass of each stream: Enter the amount of each component that is being mixed.
Specific heat of each stream: Enter the heat capacity for each component so the energy weighting is correct.
Initial temperature of each stream: Enter the starting temperature for each component before mixing.
How it works / method
The page sums the thermal energy capacity of each input stream and computes a weighted average temperature using m x c x T terms. The result is the equilibrium temperature predicted by a simple conservation-of-energy model. It is a clean first estimate when heat losses are small and no phase change occurs.
Formula used
Tfinal = sum(m_i c_i T_i) / sum(m_i c_i)
This formula is valid when all streams remain in the same phase and the system can be treated as adiabatic over the mixing period. If all components have the same specific heat, the relationship simplifies to a mass-weighted average.
Practical note: Real mixing problems can lose heat to the vessel and surroundings, and some materials change specific heat with temperature. The calculated result is therefore an estimate based on an idealized energy balance.
Mass (kg/g)Spec. Heat (cp)Temp
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Final Equilibrium Temperature
Step-by-step example
Suppose you mix 2 kg of water at 60 C with 3 kg of water at 20 C. Because the specific heats are the same, the final result becomes a mass-weighted average.
Enter the first stream mass, specific heat, and temperature.
Enter the second stream values and any additional streams if needed.
The page multiplies each stream's mass, specific heat, and temperature.
It divides the total energy-weighted temperature term by the total heat-capacity term.
The final temperature lands between the hottest and coldest starting streams if no phase change occurs.
Use cases
Estimating blended water temperature in tanks, tubs, or process lines.
Teaching energy conservation with multiple thermal masses.
Checking whether a planned temperature blend is realistic before a trial run.
Comparing how different specific heats shift the final equilibrium temperature.
Assumptions and limitations
The page assumes no heat loss to the surroundings and no latent heat effects.
It does not account for chemical reaction, incomplete mixing, or thermal stratification.
Specific heat is treated as constant for each stream over the relevant range.
If a stream boils, freezes, or changes phase, this simple balance is incomplete.
Use the result as a quick equilibrium estimate. Add vessel losses, phase change, or time-dependent transport if the real system is more complex.
Frequently Asked Questions
Conservation of energy: heat lost by the hotter liquid equals heat gained by the colder one. So m1c1(T1 − Tf) = m2c2(Tf − T2). Solve for Tf: Tf = (m1c1T1 + m2c2T2)/(m1c1 + m2c2). For two different liquids you keep the c terms; for the same liquid they cancel out. Mix 1 kg of water at 80°C with 2 kg at 20°C and you get Tf = (80 + 40)/3 = 40°C. Assumes no heat lost to the container or air.
Same substance simplifies to a mass-weighted average: Tf = (m1T1 + m2T2)/(m1 + m2). Specific heats cancel because both samples are water. Mix 3 litres at 70°C with 5 litres at 15°C and Tf = (3×70 + 5×15)/8 = 35.6°C. Treat litres as kilograms for water — the density is close enough at room temperature. The formula assumes a perfectly insulated mixing vessel, which is fine for thermos work and a rough estimate elsewhere.
Rearrange the mixing equation. Set Tf as your target, T1 as the hot, T2 as the cold, and solve for m1/m2 = (Tf − T2)/(T1 − Tf). Want 40°C from 80°C and 20°C? Ratio = (40−20)/(80−40) = 20/40 = 0.5, so half as much hot as cold — one part hot to two parts cold. This is the maths behind every shower mixing valve and bathtub fill. Memorise the lever rule and bath drawing becomes intuitive.
For two water samples of mass m1 at T1 and m2 at T2, Tf = (m1T1 + m2T2)/(m1 + m2). With equal masses it is just the average: 1 kg at 90°C plus 1 kg at 30°C gives Tf = 60°C. With unequal masses it leans toward whichever is heavier. The same formula works for any same-substance mix. The implicit assumption is no heat exchange with the surroundings — perfect insulation. In practice you lose a few degrees to the container.
Equilibrium temperature is the steady value the whole system reaches once heat exchange has finished. Use Tf = (Σ mi ci Ti) / (Σ mi ci), summing over every substance involved. Each term is mass times specific heat times initial temperature; the denominator is the total heat capacity. Add the calorimeter itself as a "mass equivalent of water" if you want lab accuracy. At equilibrium, every part of the system is at the same temperature and net heat flow is zero.
Use the lever rule. Given m1 kg of hot water at T1 and you want a final temperature Tf using cold water at T2, the cold mass needed is m2 = m1 × (T1 − Tf)/(Tf − T2). To bring 2 kg of 90°C water down to 40°C using 20°C tap water: m2 = 2 × (90−40)/(40−20) = 2 × 50/20 = 5 kg. So you add five litres of cold to two of hot. Always check the target is between the two source temperatures.
With equal masses of the same liquid, the final temperature is exactly the arithmetic mean — no weighting needed. So 1 kg at 80°C plus 1 kg at 20°C gives 50°C. The total heat is conserved; the hot half loses 30°C of "thermal value" and the cold half gains 30°C. This only works for the same substance with the same specific heat. Mix 1 kg of water with 1 kg of oil and you must keep the c terms in the equation, because oil has roughly half the specific heat of water.
Apply heat lost = heat gained. For a hot mass m1c1 cooling from T1 to Tf, the heat released is Q = m1c1(T1 − Tf). The cold side absorbs the same Q, raising its temperature: Q = m2c2(Tf − T2). Set them equal and solve for Tf or for whichever quantity is unknown. If a phase change is involved — ice melting, water boiling — add the latent heat term mL on the appropriate side. Otherwise the basic balance is enough.
