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. | ||
(a) Find the melting point of indium.
Solve Equation 17.7 for T-\ T_{0} :
T-\ T_{0}={\frac{R~-~R_{0}}{\alpha R_{0}}}={\frac{76.8\ \Omega~-~50.0\ \Omega}{[3.92~\times~10^{-3}\ {{}\ {(}°C)^{-1}][50.0\ \Omega]}}}
= 137° C
Substitute {\mathbf{}}T_{0} = 20.0°C and obtain the melting point of indium:
T = 157°C
(b) Find the ratio of the new current to the old when the temperature rises from 157°C to 235°C.
Write Equation 17.7, with R_{\mathrm{0}} and T_{\mathrm{0}} replaced by R_{\mathrm{mp}} and T_{\mathrm{mp}}, the resistance and temperature at the melting point.
R=R_{\mathrm{mp}}[1+\alpha(T-T_{\mathrm{mp}})]
According to Ohm’s law, R = ΔV/I and R_{\mathrm{mp}} = ΔV/I_{\mathrm{mp}}.
Substitute these expressions into Equation 17.7:
{\frac{\Delta V}{I}}={\frac{\Delta V}{I_{\mathrm{mp}}}}\,[\,1\,+\,\alpha(\,T-\,T_{\mathrm{mp}})\,]
Cancel the voltage differences, invert the two expressions, and then divide both sides by I_{\mathrm{mp}}:
{\frac{I}{I_{\mathrm{mp}}}}={\frac{1}{1~+~\alpha(T~-~\,T_{\mathrm{mp}})}}
Substitute T = 235°C, T_{\mathrm{mp}} = 157°C, and the value for α, obtaining the desired ratio:
{\frac{I}{I_{\mathrm{mp}}}}=~{{0.766}}
REMARKS As the temperature rises, both the rms speed of the electrons in the metal and the resistance increase.