More Thermometerscale

THE THERMOMETERSCALE OF OLE RØMERErling Poulsen  

In his notebook he describes one of his thermometers: the internal diameter of the bore in the tube of the thermometer is 0,43 lin. (measured with a drop of mercury), the length of the tube is 18″, the diameter of the spherical bulb is 6,30 lin., and an increase in 10°Rø in temperature will rise the liquid, “Spiritus vini”, 12,17 lin. (1″ = 12 lin. = 2,62 cm). The fixpoints are the boiling point and the freezing point of water. Then he divides the volume between the fixpoints in seven equal parts. He places one of the parts under the freezing mark and here he has 0°Rø, 7½°Rø is at the freezing point and 60°Rø is at the boiling point. This gives an average of 1,9°Rø/°C. From this you can find the expansion-coefficient of the liquid and from that you find the strength of app. 39%(vol). He finds that the human bodytemperature is 22½°Rø, which he uses as a fixpoint in some shorter thermometers (it was one of these which Fahrenheit learned to make). The purpose of the very low point zero was to make all readings positive.

The figures which has been analysed come from EEC’s tables of ethanol from July 27, 1976, published in 1978, ISBN 92-825-0147-7 and from E. W. Washburn et al., International Critical Tables of Numerical Data, Physics, Cemestry and Technology (McGraw Hill, 1933), the expansion of the thermometerglass has been taken into account (vol. expansioncoefficient for glass = 3*10-5 C -1).

MATHEMATICA 2.2 has been used in the analysis.

The values in the table below shows the volume of 1000 l. 39%(vol) at the temperature when heated to 20°C.

°C tabel 1 ml 20°C 1 ml 0°C glass expansion taken into account.
-20 1023 0,9775171 0,9892473 0,9898
-19 1022 0,9784735 0,9902152 0,9908
-18 1022 0,9784735 0,9902152 0,9908
-17 1021 0,9794319 0,9911851 0,9917
etc. etc. etc. etc. etc.
0 1012 0,9881423 1 1,0000
etc. etc. etc. etc. etc.
20 1000 1 1,012 1,0114
etc. etc. etc. etc. etc.

engtem~0

Analysis:

Fit[{{-20,0.9898}, {-19,0.9908}, {-18,0.9908}, {-17,0.9917}, {-16,0.9917}, {-15,0.9926}, {-14,0.9926}, {-13,0.9935}, {-12,0.9935}, {-11,0.9944}, {-10,0.9944}, {-9,0.9954}, {-8,0.9953}, {-7,0.9963}, {-6,0.9962}, {-5,0.9972}, {-4,0.9981}, {-3,0.9981}, {-2,0.9991}, {-1,0.9990}, {0,1.0000}, {1,1.0000}, {2,1.0009}, {3,1.0019}, {4,1.0019}, {5,1.0028}, {6,1.0028}, {7,1.0038}, {8,1.0037}, {9,1.0047}, {10,1.0057}, {11,1.0056}, {12,1.0066}, {13,1.0076}, {14,1.0075}, {15,1.0085}, {16,1.0085}, {17,1.0095}, {18,1.0104}, {19,1.0104}, {20,1.0114}, {21,1.0124}, {22,1.0124}, {23,1.0133}, (24,1.0143}, {25,1.0143}, {26,1.0153}, {27,1.0163}, {28,1.0162}, {29,1.0172}, {30,1.0182}, {31,1.0182}, {32,1.0192}, {33,1.0202}, {34,1.0212}, {35,1.0212}, {36,1.0222}, {37,1.0232}, {38,1.0232}, {39,1.0242}, {40,1.0252}, {50,1.0322}, {60,1.0404}}, {1,C,C^2}, C]

This gives the parabola:

Volt°C = 0.999789 + 0.000533672*t + 2.38614*10-6*t2

Vol100°C = 1.07702; expansion is 0,07702

Vol0 °Rø = 1 – 0.07702/7 = 0.988997; Vol22,5 °Rø = 1 + 2*0.07702/7 = 1.02201

Solution for 0°Rø: Solve [0.999789 + .000533672*C + 2.38614*10^-6*C^2 == .988997,C]

{{C -> -201.173}, {C -> -22.4821}}. 0°Rø = -22,5°C

Solution for 22,5°Rø: Solve [0.999789 + .000533672*C + 2.38614*10^-6*C^2 == 1.02201,C]

{{C -> -259.536}, {C -> 35.8814}}. 22,5°Rø = 35,9°C

If you approximates the relationship between the two scales with a parabola and use the fixpoints (Rø, C): (60100), (22 ½, 36) and (7 ½, 0) it gives the following result (which does not fit completely with the above based on table values for thermometer liquid):
 rcskala