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:
° Ctabel1 ml v. 20°C1 ml v. 0°Cmht.beholder udvid.
-201023 0,97751710,98924730,9898
-191022 0,97847350,99021520,9908
-181022 0,9784735 0,99021520,9908
-171021 0,9794319 0,99118510,9917
osv. osv. osv. osv. osv.
0 1012 0,9881423 11,0000
osv. osv. osv. osv. osv.
20 1000 1 1,0121,0114
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