Contents
How the water-mixing calculator works
Choose common volume and temperature units, then enter the amount and temperature of each portion. The calculator returns the ideal final temperature, total volume, and the share supplied by each portion.
A volume-weighted temperature
For two portions of liquid water with no heat loss, thermal balance reduces to a weighted average:
The larger portion has more influence. Two liters at 80 °C mixed with one liter at 20 °C produce 60 °C, not 50 °C, because the warm portion supplies two thirds of the total water.
Three checks for the result
- 2 L at 80 °C plus 1 L at 20 °C: 3 L at 60 °C. The shares are 66.7% and 33.3%.
- 1 L at 100 °C plus 1 L at 0 °C: the ideal midpoint is 50 °C because the volumes are equal.
- 8 L at 70 °C plus 12 L at 15 °C: 20 L at 37 °C. The larger cold portion pulls the answer closer to 15 °C.
Metric and US customary input
Liters, milliliters, US gallons, and US fluid ounces are available. Both portions always use the selected unit, and changing it converts the entered amounts. Celsius and Fahrenheit work the same way; 80 °C and 20 °C become 176 °F and 68 °F without changing the physical mixture.
The real container also absorbs heat
The model assumes liquid water, complete mixing, additive volumes, and no heat exchanged with the vessel or air. Ice, steam, dissolved substances, evaporation, and a cold metal container require more than a weighted average.
Use a thermometer whenever temperature affects a person, food, or equipment. The calculated number is a planning estimate, not a safety measurement.
Questions about mixing water
Why can I not simply average the two temperatures?
A simple average works only when the two volumes are equal. Otherwise the larger portion carries more thermal capacity and receives more weight.
Can the final temperature lie outside the two inputs?
Not in this ideal two-water model. A result outside the interval indicates invalid data, measurement error, or another heat source or phase change.
May one portion have zero volume?
Yes. Then it contributes nothing and the result equals the temperature of the nonzero portion. Both volumes cannot be zero.
Does this work with ice?
No. Melting ice requires latent heat, so the ordinary weighted-temperature formula is incomplete.
How do I mix three portions?
Add all volume-times-temperature products and divide by the sum of all volumes, provided every portion is liquid water and the same assumptions hold.
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