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Chapter 21: Problem 7

A resistance thermometer is a thermometer in which the electrical resistance changes with temperature. We are free to define temperatures measured by such a thermometer in kelvins (K) to be directly proportional to the resistance \(R\), measured in ohms \((\Omega) .\) A certain resistance thermometer is found to have a resistance \(R\) of \(90.35 \Omega\) when its bulb is placed in water at the triple-point temperature \((273.16 \mathrm{~K}) .\) What temperature is indicated by the thermometer if the bulb is placed in an environment such that its resistance is \(96.28 \Omega ?\)

Short Answer

Expert verified
The temperature indicated by the thermometer when its resistance is \(96.28 \Omega\) is approximately \(290.82 K\).

Step by step solution

01

Understand the Direct Proportionality

First, we need to establish the relationship between temperature and resistance. Given that temperature in Kelvin is directly proportional to the resistance in ohms, we can write an equation like this: \(T = k \cdot R\) where \(T\) is the temperature in Kelvin, \(R\) is the resistance in ohms, and \(k\) is the proportionality constant we need to find.
02

Calculate the Proportionality Constant

To find \(k\), we employ the given values of \(T\) and \(R\) at one known state. The problem provides the values at the triple-point temperature of water which is \(273.16K\) and \(90.35\Omega\). Insert these values into our equation to calculate \(k = T / R = 273.16K / 90.35\Omega = 3.022 K/\Omega\) (approximately).
03

Calculate the Unknown Temperature

Using the value of \(k\) we just calculated, we can now determine the unknown temperature corresponding to a resistance of \(96.28\Omega\). Substitute \(k\) and \(R\) into the proportion equation to find the temperature \(T = k \cdot R = 3.022 K/\Omega \cdot 96.28\Omega = 290.82K\) (approximately)

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Most popular questions from this chapter

Density is mass divided by volume. If the volume \(V\) is temperature dependent, so is the density \(\rho\). Show that the change in density \(\Delta \rho\) with change in temperature \(\Delta T\) is given by $$\Delta \rho=-\beta \rho \Delta T$$ where \(\beta\) is the coefficient of volume expansion. Explain the minus sign.

(a) Using the ideal gas law and the definition of the coefficient of volume expansion (Eq. \(21-12\) ), show that \(\beta=1 / T\) for an ideal gas at constant pressure. \((b)\) In what units must \(T\) be expressed? If \(T\) is expressed in those units, can you express \(\beta\) in units of \(\left(\mathrm{C}^{\circ}\right)^{-1} ?(c)\) Estimate the value of \(\beta\) for an ideal gas at room temperature.

Oxygen gas having a volume of \(1130 \mathrm{~cm}^{3}\) at \(42.0^{\circ} \mathrm{C}\) and a pressure of \(101 \mathrm{kPa}\) expands until its volume is \(1530 \mathrm{~cm}^{3}\) and its pressure is \(106 \mathrm{kPa}\). Find \((a)\) the number of moles of oxygen in the system and ( \(b\) ) its final temperature.

A \(1.28-\mathrm{m}\) -long vertical glass tube is half-filled with a liquid at \(20.0^{\circ} \mathrm{C}\). How much will the height of the liquid column change when the tube is heated to \(33.0^{\circ} \mathrm{C}\) ? Assume that \(\alpha_{\text {glass }}=1.1 \times 10^{-5} / \mathrm{C}^{\circ}\) and \(\beta_{\text {liquid }}=4.2 \times 10^{-5} / \mathrm{C}^{\circ}\)

A quantity of ideal gas at \(12.0^{\circ} \mathrm{C}\) and a pressure of \(108 \mathrm{kPa}\) occupies a volume of \(2.47 \mathrm{~m}^{3}\). (a) How many moles of the gas are present? \((b)\) If the pressure is now raised to \(316 \mathrm{kPa}\) and the temperature is raised to \(31.0^{\circ} \mathrm{C}\), how much volume will the gas now occupy? Assume there are no leaks.
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