Does anyone know the boiling temperature of water?carefull its not as simple as it sounds?

Answer 1

100 degrees celsius at sea level.

Answer 2

The Celsius (sometimes called Centigrade, though use of that term is no longer considered correct) temperature scale was originally defined so that the freezing point and boiling point of pure water, both at one atmosphere pressure, were 0 and 100 degrees, respectively. This definition ceased to be valid with the adoption of a new International Temperature Scale in 1990.

The thermodynamic definition of temperature is based solely on the behavior of an ideal gas; also one fixed point is needed to set the size of the degree. The fixed point used is the "triple point" of water, which is the pressure/temperature condition where solid, liquid, and vapor all coexist. This is used because the triple point is a unique condition that can be precisely reproduced; water's triple point is specifically chosen because it is relatively convenient to realize in the laboratory. The temperature of the triple point of water is defined to be exactly 273.16 kelvins (where 0 K is the absolute zero of temperature). While this completely determines the thermodynamic temperature scale, temperature measurements require approximating the thermodynamic temperature by a "practical" scale that contains other fixed points at which instruments can be calibrated. Temperatures are assigned to these points based on the best scientific estimate of their true thermodynamic temperatures, and procedures are specified for interpolating between the fixed points. While previous temperature scales used the atmospheric boiling point of water as a fixed point (assigning it 373.15 K, which is 100 degrees Celsius), the reproducibility of that point is not as good as other choices. The new International Temperature Scale adopted in 1990 (known as ITS-90) covers this region with the solid/liquid equilibrium (melting/freezing) points of gallium (302.9146 K) and indium (429.7485 K). On ITS-90, the atmospheric boiling temperature of water turns out to be approximately 373.124 K (99.974 degrees Celsius).

It turns out that the true temperature of water's boiling point is not quite what people thought it was when the Celsius scale was first defined long ago.

Answer 3

Here is a pretty accurate formula:

D = log(P / 6.1078)

T = (237.3 D) / (7.5 - D) + .0002352(P - 6.0178)

for which:

log is base 10 logarithm
P is atmospheric pressure in millibars (mb)
T is the boiling point in °C

Example 1: P = 1013.25 mb - standard atmospheric pressure. Then:

D = log(1013.25 / 6.1078) = 2.21983

T = (237.3*2.21983) / (7.5 - 2.21983) + .0002352(1013.25 - 6.1078) = 100.00

As expected, the boiling point at standard atmospheric pressure is 100 °C which is 212 °F.

°F = 9 / 5 (°C) + 32

Example 2: P = 800 mb (on a mountain about 2 km up). Then:

D = log(800 / 6.1078) = 2.11721

T = (237.3*2.11721) / (7.5 - 2.11721) + .0002352(800 - 6.1078) = 93.52 °C = 200.3 °F

So in this case the boiling point is about 12 °F lower than its average at sea level - have to go pretty high to make much of a difference.

I used the formula for the dew point at a given saturation vapor pressure in the link below:

http://www.faqs.org/faqs/meteorology/temp-dewpoint/

The boiling point is the temperature at which the saturation vapor pressure equals atmospheric pressure (all water vaporizes). At the triple point of water, P = 6.107 mb and T = .01 °C - which is close to what the formula gives. I added the extra term .0002352*(P - 6.1078) at the end because the original formula becomes worse at higher temperatures. Because I want the formula to be best at the boiling point, that term makes it exactly correct there. The boiling point is 50 °C at an atmopsheric pressure of 123.39 mb - use the formula and you'll see it is still very close there (gives about 50.1 °C).

Most of the time you don't know the station pressure (i.e., the actual atmospheric pressure where you are) but the pressure corrected to sea level. In that case, you have to uncorrect it - an approximate formula for which is:

P = P0 exp(-z / 8.314)

for which:

exp is the exponential function - e.g., exp(X) = 2.71828...^X
P is (actual) station pressure
P0 = sea level pressure
z is altitude in km

Example 3: Suppose you are at an altitude of 2 km and the sea level pressure is 1005 mb (a little lower than standard sea level pressure). Then:

P = 1005 exp(-2 / 8.314) = 790.12 mb

Using this in the first formula yields T = 93.19 °C = 199.7 °F.

Pretty close to example 2 above with slightly different conditions.

The number 8.314 used is the scale height in km for the standard atmosphere - it varies with temperature and humidity, but is a good average value to use to make the formula easier.

Answer 4

at standard atmospheric pressure water boils at 212F or 100C. As you go up in elevation the temperature that water boils at will decrease.

Answer 5

212 degrees Fahrenheit

Answer 6

100 Celsius at sea level in 100% purity

Answer 7

Depends on the atmospheric pressure where you are at, but it is approximately 373 Kelvin

Answer 8

212 degrees Farenheit, at sea level

Answer 9

212 degrees

Answer 10

100 C at 1.00 atm