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THE THERMOMETERSCALE OF OLE RØMER
Erling Poulsen printerfriendly
version
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. |
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
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