Altitude Boiling Point Calculator

This online calculator calculates the boiling temperature of water based on the atmospheric pressure in millimeters of mercury or the altitude above the sea level.

This page exists due to the efforts of the following people:

Timur

Timur

Maxim Tolstov

Created: 2015-07-30 09:13:10, Last updated: 2023-03-07 16:32:20
Creative Commons Attribution/Share-Alike License 3.0 (Unported)

This content is licensed under Creative Commons Attribution/Share-Alike License 3.0 (Unported). That means you may freely redistribute or modify this content under the same license conditions and must attribute the original author by placing a hyperlink from your site to this work https://planetcalc.com/275/. Also, please do not modify any references to the original work (if any) contained in this content.

After creating a pressure calculator (Pressure units converter) and an atmospheric pressure calculator using the barometric formula (Altitude difference by barometric formula), I wanted to determine how to calculate the boiling point according to altitude. It is commonly known that water boils at a lower temperature at higher altitudes, but what is that temperature? Find below the calculators to answer this question, along with the relevant formulas and theory.

Calculator to find boiling point temperature from altitude:

PLANETCALC, Boiling point dependence on the altitude above sea level

Boiling point dependence on the altitude above sea level

Digits after the decimal point: 1
Boiling point
 

Calculator to find boiling point temperature from pressure:

PLANETCALC, Boiling point dependence on the atmospheric pressure

Boiling point dependence on the atmospheric pressure

Digits after the decimal point: 1
Boiling temperature
 

Altitude Boiling Point

Determination of the boiling point involves two stages: establishing the dependence of atmospheric pressure on altitude and the dependence of boiling point on pressure.

Boiling is a phase transition of the first order, in which water changes its physical state from liquid to gas. This phase transition is described by the Clapeyron equation:
\frac{dP}{dT}=\frac{q_{12}}{T(v_2-v_1)},
where
q_{12} - the specific heat of the phase transition, which is numerically equal to the amount of heat received by a unit of mass for the phase transition.
T - phase transition temperature
v_2 - v_1 - change of the specific volume in the transition

Clausius simplified the Clapeyron equation for the case of evaporation and sublimation by assuming that the vapor follows the ideal gas law and that the specific volume of the fluid is much smaller than that of the steam. As a result, the state of the vapor can be described using the Mendeleev-Clapeyron equation.
PV=\frac{M}{\mu} RT,
and the specific volume of fluid v_1can be neglected.
Thus, Clapeyron equation takes the following form
\frac{dP}{dT}=\frac{q_{12}}{Tv},
where the specific volume can be expressed through
v=\frac{V}{M}=\frac{RT}{P\mu},
and finally
\frac{dP}{dT}=\frac{q_{12}\mu P}{RT^2}
separating the variables, we obtain
\frac{1}{P}dP=\frac{q_{12}\mu }{RT^2}dT

By integrating the left part P_1 to P_2 and the right part from T_1 to T_2 i.e. from one point (P_1,T_1) to another (P_2,T_2), lying on the line liquid-vapor equilibrium, we obtain the following equation
lnP_2-lnP_1=\frac{q_{12}\mu }{R}(\frac{1}{T_1}-\frac{1}{T_2})
called the Clausius-Clapeyron equation.

Actually, this is the desired dependence of the boiling temperature of the pressure.

Here are some more transformations
ln\frac{P_2}{P_1}=\frac{q_{12}\mu }{R}(\frac{1}{T_1}-\frac{1}{T_2})
\frac{P_2}{P_1}=e^{\frac{q_{12}\mu }{R}(\frac{1}{T_1}-\frac{1}{T_2})},
where
\mu - molar mass of the water, 18 gram/mol
R -universal gas constant 8.31 J/(mol K)
q_{12} - specific heat of water vaporisation 2.3
10^6 J / kg

Now we have to to establish the dependence of the altitude to the atmospheric pressure. Here we will use the barometric formula (we don't have any other anyway):
P=P_0e^{\frac{-\mu gh}{RT}}
or
\frac{P}{P_0}=e^{\frac{-\mu gh}{RT}},
where
\mu - molar mass of the air , 29 gram/mol
R - universal gas constant, 8.31 J/(mol K)
g - acceleration of gravity, 9.81 m/(s
s)
T - air temperature

We will mark the value relating to the air with index v and relating to the water with index h.
By equating and getting rid of the exponent, we will get
-\frac{\mu_v gh}{RT_v}=\frac{q_{12}\mu_h }{R}(\frac{1}{T_0}-\frac{1}{T_h})

And the final formula is
T_h=\frac{T_0T_vq_{12}\mu_h}{q_{12}\mu_hT_v+\mu_vghT_0}

Of course, the actual air pressure doesn't follow the barometric formula as that with high altitude difference; air temperature can not be considered permanent. Moreover, the gravitational acceleration depends on the geographical latitude, atmospheric pressure, and water vapor concentration. So we can have only the estimated results with this formula. Due to this, I also included another calculator which finds boiling point temperature given the atmospheric pressure, according to ln\frac{P_2}{P_1}=\frac{q_{12}\mu }{R}(\frac{1}{T_1}-\frac{1}{T_2}) formula.

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PLANETCALC, Altitude Boiling Point Calculator

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