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Question 17.4: A Platinum Resistance Thermometer GOAL Apply the temperature...

A Platinum Resistance Thermometer

GOAL Apply the temperature dependence of resistance.

PROBLEM A resistance thermometer, which measures temperature by measuring the change in resistance of a conductor, is made of platinum and has a resistance of 50.0 Ω at 20.0°C. (a) When the device is immersed in a vessel containing melting indium, its resistance increases to 76.8 Ω. From this information, find the melting point of indium. (b) The indium is heated further until it reaches a temperature of 235°C. What is the ratio of the new current in the platinum to the current I_{\mathrm{mp}}  at the melting point?

STRATEGY For part (a), solve Equation 17.7 for T-\,T_{\mathrm{0}} and get a for platinum from Table 17.1, substituting known quantities. For part (b), use Ohm’s law in Equation 17.7.

R=R_{\mathrm{0}}[1+\alpha(T-T_{\mathrm{0}})]                              [17.7]

Table 17.1 Resistivities and Temperature Coefficients of Resistivity for Various Materials (at 20°C)
 

Material

 

Resistivity (Ω · m)

Temperature Coefficient of Resistivity [(°C)^{-1}]
Silver 1.59\times10^{-8} 3.8\times10^{-3}
Copper 1.7\times10^{-8} 3.9\times10^{-3}
Gold 2.44\times10^{-8} 3.4\times10^{-3}
Aluminum 2.82\times10^{-8} 3.9\times10^{-3}
Tungsten 5.6\times10^{-8} 4.5\times10^{-3}
Iron 10.0\times10^{-8} 5.0\times10^{-3}
Platinum 11\times10^{-8} 3.92\times10^{-3}
Lead 22\times10^{-8} 3.9\times10^{-3}
Nichrome^{a} 150\times10^{-8} 0.4\times10^{-3}
Carbon 3.5\times10^{-5} -0.5\times10^{-3}
Germanium 0.46 -48\times10^{-3}
Silicon 640 -75\times10^{-3}
Glass 10^{10}{-}10^{14}
Hard rubber \approx10^{13}
Sulfur 10^{15}
Quartz (fused) 75\times10^{16}
\mathbf{}^{a}A nickel-chromium alloy commonly used in heating elements.
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