Boiling Water Evaporation Rate Calculator Formula

This calculator estimates the average mass of water evaporated per hour while a pot boils, using the water’s starting and ending mass and elapsed time. Enter the measured masses in kilograms and the boiling duration in hours; the result is an average evaporation rate in kilograms per hour.

Formula

The average boiling-water evaporation rate is the lost water mass divided by the elapsed boiling time:

r = (m₁ − m₂) / t

• r is average evaporation rate, in kilograms per hour (kg/h).
• m₁ is the starting water mass, in kilograms (kg).
• m₂ is the ending water mass, in kilograms (kg).
• t is the elapsed time, in hours (h).

The calculator also reports grams per minute (g/min): multiply kg/h by 1,000 grams per kilogram and divide by 60 minutes per hour. The zero-times-square-root terms in the calculator expression enforce the physically sensible condition that ending mass cannot exceed starting mass and elapsed time must be greater than zero; otherwise it cannot produce a valid rate.

Worked Example

Using the calculator’s default values, the starting water mass is 1 kg, the ending mass is 0.9 kg, and the boiling time is 0.25 h.

1. Find the mass lost: 1 − 0.9 = 0.1 kg.
2. Divide by elapsed time: 0.1 kg ÷ 0.25 h = 0.4 kg/h.
3. Convert to grams per minute: 0.4 × 1,000 ÷ 60 = 6.666… g/min, or 6.667 g/min rounded to three decimal places.

The headline result is therefore 0.400 kg/h. This is an average over the measured interval, not a claim that evaporation proceeded at precisely the same rate at every moment.

What the Rate Means

Boiling water evaporates as liquid molecules escape and become water vapor. The rate describes how much liquid mass disappears over a period; it does not directly describe the visible size of the steam plume. Much of the vapor is invisible water gas, while the white cloud commonly called steam consists largely of tiny liquid droplets formed as vapor cools and condenses.

Mass is usually the most practical basis for a home measurement. A kitchen scale can compare the pot and its contents before and after heating, provided the same pot is weighed both times and the scale can accommodate it. If you measure volume instead, remember that the formula requires mass: convert volume to mass using water density appropriate to its temperature. Near ordinary boiling conditions, a liter of water has a mass close to, but not exactly, one kilogram. Weighing avoids that conversion.

This calculation assumes the observed mass loss is water that evaporated. Spills, splashing, water carried away on a utensil, or a change in what is included on the scale can all look like evaporation in the result.

How to Measure It

1. Choose a stable, heat-resistant setup and record the mass of the pot plus water just as the measurement period begins. If the pot is included, keep it in both weighings.
2. Bring the water to a steady boil, then start the timer for the interval you want to measure. Do not count heating from cold as boiling time unless that is intentionally part of the experiment.
3. After the interval, stop heating and weigh the same pot and remaining water using the same arrangement. Follow the scale’s instructions and avoid placing a hot pot where it could damage the scale.
4. Subtract the ending mass from the starting mass. Divide the loss by time expressed in hours for kg/h, or in minutes for kg/min.

For a more repeatable comparison, use the same pot, water quantity, burner setting, lid position, and measurement duration. Take care that neither weighing includes a wet surface or a different utensil. Use a reasonably long interval: a very short test can be dominated by scale resolution, timing error, or small splashes.

What Changes Evaporation

The formula converts measured loss into a rate; it does not predict a universal rate for a particular pot or stove. The actual result depends on the rate at which heat reaches the water and on how readily vapor escapes.

• Heat input: More energy delivered to the boiling water can support more vapor production. Burner settings alone do not establish how much energy reaches the pot.
• Exposed surface area: A wider uncovered surface generally offers more area for vapor to leave than a narrow opening, although the full result also depends on heat transfer and airflow.
• Lid and airflow: A lid can reduce net water loss by limiting vapor’s escape. Air movement can carry vapor away from the surface.
• Altitude and pressure: Water boils at a lower temperature when atmospheric pressure is lower. This changes boiling conditions, but it does not by itself specify an evaporation rate.
• Measurement interval: The rate may change as heat input, boiling intensity, or remaining water changes. A longer average can conceal those variations.

Common Errors and Limits

A frequent mistake is mixing time units. If a mass loss of 0.1 kg occurs in 15 minutes, the rate is 0.1 ÷ 0.25 = 0.4 kg/h; dividing by 15 while labeling the result kg/h would be wrong. Convert minutes to hours by dividing by 60, or report the rate in kg/min consistently. Likewise, 1 kg is 1,000 g, and 1 US fluid quart is a volume—not a mass. US customary volume measures need a density conversion before they can be used as mass.

Another mistake is treating a negative mass loss as a negative evaporation rate. If ending mass is higher, check for measurement differences, condensation added to the pot, or an incorrect entry; this calculator flags that input as invalid rather than interpreting it as evaporation. A zero time interval also cannot be used because division by zero is undefined.

The formula gives a measured average, not a complete physical model. It does not estimate a rate from pot diameter, burner power, or water temperature alone. A theoretical energy estimate would need the heat reaching the water and the energy needed to vaporize it, among other factors; appliance input ratings do not equal heat absorbed by the water. The simple mass-loss method is generally more direct when the goal is to know what happened in a particular setup.

Frequently Asked Questions

What is the formula for boiling water evaporation rate?
Subtract ending water mass from starting water mass, then divide by elapsed boiling time: rate = (starting mass − ending mass) ÷ time. Keep mass and time units consistent.

How do I calculate evaporation in grams per minute?
Measure mass loss in grams and divide by elapsed minutes. If your result is in kg/h, multiply by 1,000 and divide by 60 to convert to g/min.

Can I calculate the rate from pot size or burner power?
Not reliably from those values alone. Pot geometry and burner rating do not reveal how much heat reaches the water or how much vapor escapes. Measure the mass change for a direct average.

Does a lid change the calculated rate?
A lid can reduce net water loss from the pot by restricting vapor escape, so the measured rate may be lower. Condensation that drips back into the water is not water lost from the pot.

Why is my measured rate negative?
A negative result means the ending mass was greater than the starting mass. Check the weighing setup and entries; added condensation or inconsistent measurements can produce that pattern.

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in m1 | Starting water mass | kg | 1 | 0 | 1000
in m2 | Ending water mass | kg | 0.9 | 0 | 1000
in t | Boiling time | h | 0.25 | 0 | 1000
out r | Average evaporation rate | kg/h | (m1 - m2) / t + 0 * sqrt(m1 - m2) + 0 * sqrt(t) | 3
out g | Average evaporation rate | g/min | r * 1000 / 60 | 3