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A copper-constantan thermocouple generates a voltage of \(4.75 \times\) \(10^{-3}\) volts when the temperature of the hot junction is \(110.0^{\circ} \mathrm{C}\) and the reference junction is kept at a temperature of \(0.0^{\circ} \mathrm{C} .\) If the voltage is proportional to the difference in temperature between the junctions, what is the temperature of the hot junction when the voltage is \(1.90 \times 10^{-3}\) volts?

Short Answer

Expert verified
The temperature of the hot junction is approximately \( 44.0^{\circ}C \).

Step by step solution

01

Understand the Proportional Relationship

The voltage generated by a thermocouple is proportional to the temperature difference between the hot and reference junctions. This can be expressed as: \( V = k \Delta T \), where \( V \) is the voltage, \( k \) is the proportionality constant, and \( \Delta T \) is the temperature difference.
02

Determine the Proportionality Constant

Using the given values, we find the proportionality constant \( k \). When the voltage is \( 4.75 \times 10^{-3} \) volts, the temperature difference \( \Delta T \) is \( 110^{\circ}C - 0^{\circ}C = 110^{\circ}C \). Thus, \( k = \frac{4.75 \times 10^{-3}}{110} \approx 4.3182 \times 10^{-5} \) volts per degree Celsius.
03

Calculate the Temperature Difference for New Voltage

With the constant \( k \) known, we calculate the temperature difference for the new voltage \( 1.90 \times 10^{-3} \) volts: \( \Delta T = \frac{1.90 \times 10^{-3}}{4.3182 \times 10^{-5}} \approx 43.997 \) degrees Celsius.
04

Find the Temperature of the Hot Junction

Since \( \Delta T = T_{\text{hot}} - 0^{\circ}C \), the temperature of the hot junction \( T_{\text{hot}} \) is approximately \( 44.0^{\circ}C \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proportionality Constant
In thermocouples, a **proportionality constant** links the voltage to the temperature difference sensed by the device. Think of it as a conversion factor. It tells us how much voltage will be generated for each degree of temperature difference. In the given problem, when the voltage generated is known, and the temperature difference is provided, the proportionality constant allows us to relate these two quantities. For example, with a voltage of \(4.75 \times 10^{-3}\) volts generated by a temperature difference of \(110^{\circ}C\), we calculate the constant \(k\) as follows:\[ k = \frac{4.75 \times 10^{-3}}{110} \approx 4.3182 \times 10^{-5} \text{ volts per degree Celsius} \]This means every degree change produces roughly \(4.3182 \times 10^{-5}\) volts. This proportionality helps simplify finding unknown temperatures if the voltage is altered and still fits this linear pattern.
Temperature Difference
The **temperature difference** between the two junctions in a thermocouple is crucial for voltage determination. In simpler terms, it's the gap in temperature that creates an electric potential or voltage. By measuring this temperature gap, or \(\Delta T\), we can determine the generated voltage.For instance, starting with a known voltage of \(4.75 \times 10^{-3}\) volts due to a \(110^{\circ}C\) temperature difference, the task becomes finding the new temperature difference when the voltage is reduced to \(1.90 \times 10^{-3}\) volts. Using the previously found proportionality constant \(k\), we calculate:\[ \Delta T = \frac{1.90 \times 10^{-3}}{4.3182 \times 10^{-5}} \approx 43.997 \text{ degrees Celsius} \]Thus, when the voltage decreases, the temperature difference narrows. Understanding this relationship is vital for accurate temperature sensing.
Voltage Generation
**Voltage generation** in a thermocouple comes from the unique properties of materials like copper and constantan working together. When two different metals are joined and exposed to differing temperatures at their junctions, a voltage arises. This voltage, called a thermoelectric voltage, is a direct result of the temperature difference between the junctions. In our case, when adjusting conditions led to a voltage of \(1.90 \times 10^{-3}\) volts, we calculated the impact on temperature measurements:1. Used the known proportionality constant \(k\) from the original setup.2. Directly linked the drop in voltage to a lower temperature difference.Such thermocouple voltage outputs are integral for practical applications, especially in environments where knowing temperature changes is crucial for operations.

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Most popular questions from this chapter

A constant-volume gas thermometer (see Figures 12.3 and 12.4 ) has a pressure of \(5.00 \times 10^{3}\) Pa when the gas temperature is \(0.00^{\circ} \mathrm{C} .\) What is the temperature (in \({ }^{\circ} \mathrm{C}\) ) when the pressure is \(2.00 \times 10^{3} \mathrm{Pa} ?\)

You and your team are testing a device that is to be submerged in the cold waters of Antarctica. It is designed to rotate a small wheel at a very precise rate. One component of the device consists of a steel wheel (diameter \(D_{\text {sieel }}=3.0000 \mathrm{cm}\) at \(\left.T=78.400^{\circ} \mathrm{F}\right)\) and a large aluminum wheel that drives it (diameter \(D_{\mathrm{A}}=150.000 \mathrm{cm}\) at \(T=78.400^{\circ} \mathrm{F}\) ). (a) If the aluminum wheel rotates at 35.0000 RPM, at what rate does the smaller, steel wheel turn while both wheels are at \(T=78.400^{\circ} \mathrm{F} ?\) (b) You submerge the device in a large vat of water held at \(32.800^{\circ} \mathrm{C},\) simulating its working environment in the Antarctic seas. Assuming the larger, aluminum driving wheel still rotates at \(35.0000 \mathrm{RPM},\) at what rate does the smaller, steel wheel rotate when the device is submerged in the cold water? (c) How should you adjust the rate of the aluminum driving wheel so that the steel wheel rotates at the same rate as it had at \(T=78.400^{\circ} \mathrm{F} ?\) Note: \(\alpha_{\mathrm{Al}}=23 \times 10^{-6}\) and \(\alpha_{\text {steel }}=12 \times 10^{-6}\)

A 0.35-kg coffee mug is made from a material that has a specific heat capacity of \(920 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) and contains \(0.25 \mathrm{kg}\) of water. The cup and water are at \(15^{\circ} \mathrm{C}\). To make a cup of coffee, a small electric heater is immersed in the water and brings it to a boil in three minutes. Assume that the cup and water always have the same temperature and determine the minimum power rating of this heater.

A steel bicycle wheel (without the rubber tire) is rotating freely with an angular speed of \(18.00 \mathrm{rad} / \mathrm{s}\). The temperature of the wheel changes from \(-100.0 \mathrm{to}+300.0^{\circ} \mathrm{C} .\) No net external torque acts on the wheel, and the mass of the spokes is negligible. (a) Does the angular speed increase or decrease as the wheel heats up? Why? (b) What is the angular speed at the higher temperature?

(a) Objects A and B have the same mass of \(3.0 \mathrm{kg}\). They melt when \(3.0 \times 10^{4} \mathrm{J}\) of heat is added to object \(\mathrm{A}\) and when \(9.0 \times 10^{4} \mathrm{J}\) is added to object B. Determine the latent heat of fusion for the substance from which each object is made. (b) Find the heat required to melt object A when its mass is \(6.0 \mathrm{kg}\).

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