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 \Omega at 20.0^{\circ} \mathrm{C}. (a) When the device is immersed in a vessel containing indium at its melting point, its resistance increases to 76.8 \Omega. From this information, find the melting point of indium. (b) The indium is heated further until it reaches a temperature of 235^{\circ} \mathrm{C}. Estimate the ratio of the new current in the platinum to the current I_{\mathrm{mp}} at the melting point, assuming the coefficient of resistivity for platinum doesn’t change significantly with temperature.
STRATEGY For part (a), solve Equation 17.7
R=R_{0}[1+\alpha (T-T_{0})] [17.7]
for T-T_{0} and get \alpha for platinum from Table 17.1,
Table 17.1 resistivities and temperature coefficients of resistivity for various Materials (at 20°C)
\begin{matrix} \hline && \textbf{Temperature Coefficient} \\ &\textbf{Resistivity}& \textbf{of Resistivity} \\ \textbf{Material }& \mathbf{(\Omega \cdot m)} & \mathbf{[(^{\circ}C)^{-1}]} \\ \hline \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}^{\mathbf{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{matrix}
^{\mathrm{a}}A nickel-chromium alloy commonly used in heating elements.
substituting known quantities. For part (b), use Ohm’s law in Equation 17.7.
(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}{[3.92 \times 10^{-3} \left({ }^{\circ} \mathrm{C}\right)^{-1}][50.0 \Omega]} \\&=137^{\circ} \mathrm{C}\end{aligned}
Substitute T_{0}=20.0^{\circ} \mathrm{C} and obtain the melting point of indium:
T=157^{\circ} \mathrm{C}
(b) Estimate the ratio of the new current to the old when the temperature rises from 157^{\circ} \mathrm{C} to 235^{\circ} \mathrm{C}.
Write Equation 17.7, with R_{0} and T_{0} replaced by R_{\text {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=\Delta V / I and R_{\mathrm{mp}}=\Delta V / I_{\mathrm{mp}}. Substitute these expressions into Equation 17.7:
\frac{\Delta V}{I}=\frac{\Delta V}{I_{\mathrm{mp}}}[\begin{array}{ll}\end{array}1 +\alpha(\begin{array}{ll} T-T_{\mathrm{mp}})] \end{array}
Cancel the voltage differences, invert the two expressions, and then divide both sides by I_{\text {mp }} :
\frac{I}{I_{\mathrm{mp}}}=\frac{1}{1+\alpha\left(T-T_{\mathrm{mp}}\right)}
Substitute T=235^{\circ} \mathrm{C}, T_{\text {mp }}=157^{\circ} \mathrm{C}, and the value for \alpha, obtaining the desired ratio:
\frac{I}{I_{\mathrm{mp}}}=0.766
REMARKS The answer to part (b) is only an estimate because the temperature coefficient is, in fact, temperature dependent. As the temperature rises, both the rms speed of the electrons in the metal and the resistance increase.