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_{ mp } at the melting point?
Strategy In part (a), solve Equation 17.7 for T-T_{0} and get α for platinum from Table 17.1, substituting known quantities. For part (b), use Ohm’s law in Equation 17.7.
R=R_{0}\left[1+\alpha\left(T-T_{0}\right)\right]
TABLE 17.1
| Resistivities and Temperature Coefficients of Resistivity for Various Materials (at 20°C) |
| \begin{array}{ccc} & &\begin{array}{c}\text { Temperature Coefficient } \\\text { of Resistivity }\end{array} \\\text { Material } & \text { Resistivity } & {\left[(\Omega \cdot C )^{-1}\right]} \\\end{array} |
| \begin{array}{lcc} \text { Silver } & 1.59 \times 10^{-8} & 3.8 \times 10^{-3} \\\text { Copper } & 1.7 \times 10^{-8} & 3.9 \times 10^{-3} \\\text { Gold } & 2.44 \times 10^{-8} & 3.4 \times 10^{-3} \\\text { Aluminum } & 2.82 \times 10^{-8} & 3.9 \times 10^{-3} \\\text { Tungsten } & 5.6 \times 10^{-8} & 4.5 \times 10^{-3} \\\text { Iron } & 10.0 \times 10^{-8} & 5.0 \times 10^{-3} \\\text { Platinum } & 11 \times 10^{-8} & 3.92 \times 10^{-3} \\\text { Lead } & 22 \times 10^{-8} & 3.9 \times 10^{-3} \\\text { Nichrome }^{a} & 150 \times 10^{-8} & 0.4 \times 10^{-3} \\\text { Carbon } & 3.5 \times 10^{5} & -0.5 \times 10^{-3} \\\text { Germanium } & 0.46 & -48 \times 10^{-3} \\\text { Silicon } & 640 & -75 \times 10^{-3} \\\text { Glass } & 10^{10}-10^{14} & \\\text { Hard rubber } & \approx 10^{13} & \\\text { Sulfur } & 10^{15} & \\\text { Quartz (fused) } & 75 \times 10^{16} & \\\end{array} |
| { }^{a} \text { A nickel-chromium alloy commonly used in heating elements. } |
(a) Find the melting point of indium.
Solve Equation 17.7 for T-T_{0} :
\begin{aligned}T-T_{0} &=\frac{R-R_{0}}{\alpha R_{0}}=\frac{76.8 \Omega-50.0 \Omega}{\left[3.92 \times 10^{-3}\left({ }^{\circ} C \right)^{-1}\right][50.0 \Omega]} \\&=137^{\circ} C\end{aligned}
Substitute T_{0}=20.0^{\circ} C and obtain the melting point of indium:
T_{ mp }=157^{\circ} 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_{0} \text { and } T_{0} \text { replaced by } R_{ mp } and T_{ mp } , the resistance and temperature at the melting point.
R=R_{ mp }\left[1+\alpha\left(T-T_{ mp }\right)\right]
According to Ohm’s law, R=\Delta V / I \text { and } R_{ mp }=\Delta V / I_{ mp } . Substitute these expressions into Equation 17.7:
\frac{\Delta V}{I}=\frac{\Delta V}{I_{ mp }}\left[1+\alpha\left(T-T_{ mp }\right)\right]
Cancel the voltage differences, invert the two expressions, and then divide both sides by I_{ mp } :
\frac{I}{I_{ mp }}=\frac{1}{1+\alpha\left(T-T_{ mp }\right)}
Substitute T=235^{\circ} C , T_{ mp }=157^{\circ} C , and the value for α, obtaining the desired ratio:
\frac{I}{I_{ mp }}=0.766