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RØMERS TERMOMETRE Tallene der er analyserede stammer fra EF's brugstavler over ethanol fra 27. juli, 1976, udgivet 1978, ISBN 92-825-0147-7
samt E. W. Washburn et al., International Critical Tables of Numerical Data, Physics, Cemestry and Technology (McGraw Hill, 1933),
der er dog blevet taget højde for termometerbeholderens egenudviddelse (vol. udviddelse for glas = 3*10-5 C-1).
Der er blevet analyseret ved hjælp af programmet MATHEMATICA 2.2
Behandlingen af tabelværdierne, som angiver hvor meget 1000 l ved den pågældende temperatur bliver til ved 20°C:
| ° C | tabel | 1 ml v. 20°C | 1 ml v. 0°C | mht.beholder udvid. |
|---|
| -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 |
| osv. | osv. | osv. | osv. | osv. |
| 0 | 1012 | 0,9881423 | 1 | 1,0000 |
| osv. | osv. | osv. | osv. | osv. |
| 20 | 1000 | 1 | 1,012 | 1,0114 |
| osv. | osv. | osv. | osv. | osv. |
Analyse:
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]
Nærmeste parabel:
Vol°C = 0.999789 + 0.000533672*C + 2.38614*10-6*C2
Vol100 C = 0.999789 + .000533672*100 + 2.38614*10^-6*1002 = 1.07702
Tilvækst i volumen fra 0°C til 100°C er 0,07702
Vol0 °Rø = 1 - .07702/7 = 0.988997
Vol22,5 °Rø = 1 + 2*.07702/7 = 1.02201
Løsning af ligning der giver Celsius temp. For 0 °Rø:
Solve [0.999789 + .000533672*C + 2.38614*10^-6*C^2 == .988997,C]
{{C = -201.173}, {C = -22.4821}}.
Løsning : -22,5°C
Løsning af ligning der giver Celsius temp. For 22,5 °Rø:
Solve [0.999789 + .000533672*C + 2.38614*10^-6*C^2 == 1.02201,C]
{{C = -259.536}, {C = 35.8814}}.
Løsning 35,9°C
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