Question

1. The thermoelectric potential of a thermocouple is given: p = arº + be + c. where is the temperature, o is the potential, and a, b, and c are empirical constants. Properly calibrated, a ferromagnetic material may be used as a thermometer. Susan, a mathematics and physics major, calibrates her ferromagnetic thermometer at three temperatures: Standard Material Liquid Nitrogen Dry Ice Triple Point of H20 Temperature (K) 77.36 194.65 273.16 Potential (mV) 0.0 0.06287 0.1126 Ligudna) Calculate empirical constant a. 0=2171.36"+b(71.36)+C. © = 5984.541 +99.366+2 +691 Dry Ice 0.06284. = 2(194.65)+61194.65)+e 0.06284 = 37888-628+194.65b+e es Triple point Ho 0.1126 = 2(293-6 + b(248.16+e 0.1126= 44616.39 2 +978-16 640- 93 UVODNE 393 b) Calculate empirical constant b.

Answer 1

Answer The theme electric potentiad a thermocouple $ = att bttc liquid Nitrogen Temperatun Yu) = 77:36 = a T + 7 +C . osa (77.36) + 77:36btc 35984.57 477.36b+C =0 0 Dry Ice potentical (mv) 3 0.06287 192.659 +194.65b7c @ 3785662a + 194.65btc : Triple Pouit Ho = (K) = 0.1126 a/ 27.3.163 + 27 3.16 btc - 74616 ha. La + 213.166 @ Empirical constant a - 2.974

(②) The Empirial Constant b b =0.0004 @ The empirical costante C -0.034 Solening 0 6 and a=0.0049715 er en b-0.00040071046 C=0.033 975 dj ¢ -0.03731 Toriple posit of liquid Trepo? = 0.03731 = 0:6000004 731 060 0000697615 T 10.000600710467 - 0.03397598534 cm EquationsI T2 129.976 k . I.-955.505k

Negatin absoluke temperation does not suit T = 149.976 k.