Methods Of Ascertaining The Exact Boiling Temperature


The normal boiling temperature of water all nations have tacitly agreed to fix under a normal barometric pressure of 29·922 inches of mercury, having the temperature of melting ice, in the latitude of 45°, and at the sea-level. If the atmospheric pressure at the time or place of graduating a thermometer does not equal this, the boiling temperature will be higher or lower according as the pressure is greater or less. Hence a reading must be taken from a reliable barometer, which must
also be corrected for errors and temperature, and reduced for latitude, in order to compare the actual atmospheric pressure at the time with the assumed normal pressure. Tables of vapour tension, as they are termed, have been computed from accurate experimental investigations and theory,—giving the temperatures of the vapour of water for all probable pressures; Regnault’s, the most recent, is considered the most accurate; and his investigations are based upon the standard pressure given above, and are for the same latitude. His Table, therefore, will give the temperature on the thermometric scale corresponding to the pressure.



The Commissioners appointed by the British Government to construct standard weights and measures, decided that the normal boiling-point, 212°, on the thermometer should represent the temperature of steam generated under an atmospheric pressure equal in inches of mercury, at the temperature of freezing water, to 29·922 + (cos. 2 latitude × ·0766) + (·00000179 × height in feet above the sea-level). Hence, at London, lat. 51°30´ N., we deduce 29·905 as the barometric pressure representing the normal boiling point of water,—the trifling correction due to height being neglected. If then, in the latitude of London, the barometric pressure, at the time of fixing the boiling point, be not 29·905 inches, that point will be higher or lower, according to the difference of the pressure from the normal. Near the sea-level about 0·59 inch of such difference is equivalent to 1° Fahrenheit in the boiling point.



Suppose, then, the atmospheric pressure at London to be 30·785 inches, the following calculation gives the corresponding boiling temperature for Fahrenheit’s scale:—











































Observed pressure   30·785
Normal "   29·905
Difference   ·880


As 0·59 is to 0·88, so is 1° to 1°·5.



That is, the water boils at 1°·5 above its normal temperature; so that, in this case, the normal temperature to be placed on the scale, viz. 212°, must be 1°·5 lower than the mark made on the tube at the height at which the mercury stood under the influence of the boiling water.



The temperature of the vapour of boiling water may be found, at any time and place, as follows:—Multiply the atmospheric pressure by the factor due to the latitude, given in the annexed Table V., and with the result seek the temperature in Table VI.

















































































































































































































































































































































Table V. Table VI.
Latitude. Factor. Temperature


of Vapour.
Tension. Temperature


of Vapour.
Tension.
Degrees.   Degrees. Inches. Degrees. Inches.
0 0·99735 179 14·934 197 22·036
5 0·99739 180 15·271 198 22·501
10 0·99751 181 15·614 199 22·974
15 0·99770 182 15·963 200 23·456
20 0·99797 183 16·318 201 23·946
25 0·99830 184 16·680 202 24·445
30 0·99868 185 17·049 203 24·952
35 0·99910 186 17·425 204 25·468
40 0·99954 187 17·808 205 25·993
45 1·00000 188 18·197 206 26·527
50 1·00046 189 18·594 207 27·070
55 1·00090 190 18·998 208 27·623
60 1·00132 191 19·409 209 28·185
65 1·00170 192 19·828 210 28·756
70 1·00203 193 20·254 211 29·335
75 1·00230 194 20·688 212 29·922
80 1·00249 195 21·129 213 30·515
    196 21·578 214 31·115


How to use the Tables.—When the temperature is known to decimals of a degree, take out the tension for the degree, and multiply the difference between it and the next tension by the decimals of the temperature, and add the product to the tension, for the degree.



Required the tension corresponding to 197°·84.











































































°
197   = 22·036   ·465 × ·84 = ·391
198   = 22·501   197° = 22·036
  Difference   ·465   197·84 = 22·427


When the tension is given, take the difference between it and the next less tension in the Table, and divide this difference by the difference between the next less and next greater tensions. The quotient will be the decimals to add to the degree opposite the next less tension.



Thus, for 23·214 inches, required the temperature.















































































































Given 23·214       Next greater   23·456
  22·974       Next less   22·974
  ·240   Difference   ·482
  And   ·240   = ·5
·482
Temperature opposite next less   199·0
Temperature required   199·5


A similar method of interpolation in taking out numerical quantities is applicable to almost all tables; and should be practised with all those given in this work.



Example.—Thus, in Liverpool, lat. 53° 30´ N., the barometer reading 29·876 inches, its attached thermometer 55°, and the correction of the instrument being + ·015 (including index error, capillarity and capacity), what temperature should be assigned for the boiling point marked on the thermometer?













































































































Observed barometer   29·876
Correction   + ·015
    29·891
Correction for temperature   - ·074
Reduced reading   29·817
Factor from Table V.   1·00077
    208719
    208719
    29817
Equivalent for lat. 45°   29·83995909


In Table VI., 29·84 gives temperature 211°·86.








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