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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 \(44.01^{\circ} \mathrm{C}\).

Step by step solution

01

Understand the relationship

According to the problem, the voltage generated by the thermocouple is proportional to the temperature difference (\(\Delta T\)) between the hot junction and the reference junction. We can express this relationship as \( V = k \times \Delta T \), where \( V \) is the voltage and \( k \) is the proportionality constant.
02

Find the proportionality constant

We are given that a voltage of \(4.75 \times 10^{-3}\) volts is generated when \(\Delta T = 110^{\circ} \mathrm{C}\). Using \( V = k \times \Delta T \), we can find \( k \) as follows:\[ k = \frac{V}{\Delta T} = \frac{4.75 \times 10^{-3}}{110} \approx 4.318 \times 10^{-5} \text{ volts per degree Celsius} \].
03

Calculate the new temperature difference

Now, we use the given new voltage of \(1.90 \times 10^{-3}\) volts to find the new temperature difference \(\Delta T\):\[ \Delta T = \frac{V}{k} = \frac{1.90 \times 10^{-3}}{4.318 \times 10^{-5}} \approx 44.01^{\circ} \mathrm{C} \].
04

Determine the hot junction temperature

Since the reference junction is at \(0^{\circ} \mathrm{C}\), the temperature of the hot junction is the same as \(\Delta T\). Therefore, the temperature of the hot junction is approximately \(44.01^{\circ} \mathrm{C}\).

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Key Concepts

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

Proportionality Constant
When working with thermocouples, one important aspect is the concept of the proportionality constant. The proportionality constant, denoted as 'k', is a factor that relates the voltage generated by a thermocouple to the temperature difference between its hot and reference junctions.
This relationship is linear, which means that the voltage is directly proportional to the temperature difference. In simpler terms, if you know the temperature difference, you can determine the voltage, and vice versa, using this constant.

The equation used is:
  • \( V = k \times \Delta T \)
where:
  • \( V \) is the voltage
  • \( \Delta T \) is the temperature difference
  • \( k \) is the proportionality constant
For example, in the exercise provided, a copper-constantan thermocouple has a known proportionality constant, \( k \), of approximately \( 4.318 \times 10^{-5} \) volts per degree Celsius. Knowing \( k \) helps us understand how much voltage a specific temperature difference will create.
Temperature Difference
The temperature difference, \( \Delta T \), is a central component of thermocouple functionality. When we talk about temperature difference, we're looking at the variation between the hot junction and the reference junction.
In our thermocouple example, the exercise provides a reference junction at \( 0^{\circ} C \). This makes it easier to consider the temperature at the hot junction directly as the temperature difference.

Understanding \( \Delta T \) is crucial because it is the driving force that prompts the generation of voltage in a thermocouple. Using the formula:
  • \( \Delta T = \frac{V}{k} \)
allows us to calculate the new temperature difference based on given voltage measurements and our previously calculated \( k \).For instance, with a new voltage of \( 1.90 \times 10^{-3} \) volts, we find that \( \Delta T \) equals approximately \( 44.01^{\circ} C \). This tells us the difference in temperature driving the voltage production for that particular measurement.
Voltage Generation
Voltage generation in thermocouples occurs due to the Seebeck effect, where two dissimilar metals, like copper and constantan, produce a voltage when exposed to a temperature gradient.
The fundamental concept here is that a larger temperature difference between the two junctions results in a higher voltage.

In the context of the given exercise, the voltage generated is directly proportional to the temperature difference at the junctions. Using our calculated proportionality constant, \( k \), we can compute the voltage for any given temperature difference by employing:
  • \( V = k \times \Delta T \)
For example, if the temperature at the hot junction changes, resulting in a measured voltage of \( 1.90 \times 10^{-3} \) volts, it means the change in temperature difference, \( \Delta T \), was significant enough to cause this new output. With an understanding of this, it becomes clear how essential such calculations are in applications relying on precise thermal measurements.

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Most popular questions from this chapter

Suppose you are hiking down the Grand Canyon. At the top, the temperature early in the morning is a cool \(3{ }^{\circ} \mathrm{C} .\) By late afternoon, the temperature at the bottom of the canyon has warmed to a sweltering \(34^{\circ} \mathrm{C} .\) What is the difference between the higher and lower temperatures in (a) Fahrenheit degrees and (b) kelvins?

A piece of glass has a temperature of \(83.0^{\circ} \mathrm{C} .\) Liquid that has a temperature of \(43.0^{\circ} \mathrm{C}\) is poured over the glass, completely covering it, and the temperature at equilibrium is \(53.0^{\circ} \mathrm{C} .\) The mass of the glass and the liquid is the same. Ignoring the container that holds the glass and liquid and assuming that the heat lost to or gained from the surroundings is negligible, determine the specific heat capacity of the liquid.

Multiple-Concept Example 4 reviews the concepts that are in volved in this problem. A ruler is accurate when the temperature is \(25^{\circ} \mathrm{C}\). When the temperature drops to \(-14^{\circ} \mathrm{C},\) the ruler shrinks and no longer measures distances accurately. However, the ruler can be made to read correctly if a force of magnitude \(1.2 \times 10^{3} \mathrm{N}\) is applied to each end so as to stretch it back to its original length. The ruler has a cross-sectional area of \(1.6 \times 10^{-5} \mathrm{m}^{2},\) and it is made from a material whose coefficient of linear expansion is \(2.5 \times 10^{-5}\left(\mathrm{C}^{\circ}\right)^{-1} .\) What is Young's modulus for the material from which the ruler is made?

During an all-night cram session, a student heats up a one-half liter \(\left(0.50 \times 10^{-3} \mathrm{m}^{3}\right)\) glass (Pyrex) beaker of cold coffee. Initially, the temperature is \(18^{\circ} \mathrm{C},\) and the beaker is filled to the brim. A short time later when the student returns, the temperature has risen to \(92^{\circ} \mathrm{C}\). The coefficient of volume expansion of coffee is the same as that of water. How much coffee (in cubic meters) has spilled out of the beaker?

A \(0.200-\mathrm{kg}\) piece of aluminum that has a temperature of \(-155^{\circ} \mathrm{C}\) is added to \(1.5 \mathrm{kg}\) of water that has a temperature of \(3.0^{\circ} \mathrm{C}\). At equilibrium the temperature is \(0.0^{\circ} \mathrm{C}\). Ignoring the container and assuming that the heat exchanged with the surroundings is negligible, determine the mass of water that has been frozen into ice.

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