Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A thermocouple is formed from two different metals, joined at two points in such a way that a small voltage is produced when the two junctions are at different temperatures. In a particular iron-constantan thermocouple, with one junction held at \(0^{\circ} \mathrm{C}\), the output voltage varies linearly from 0 to \(28.0 \mathrm{mV}\) as the temperature of the other junction is raised from 0 to \(510^{\circ} \mathrm{C}\). Find the temperature of the variable junction when the thermocouple output is \(10.2 \mathrm{mV}\).

Short Answer

Expert verified
The temperature of the variable junction when the thermocouple output is \(10.2 \, \mathrm{mV}\) is approximately \(186^{\circ} \mathrm{C}\)

Step by step solution

01

Identify the Variation

Understand that the relationship between temperature and voltage is linear, which means it can be written in the form y = mx + c. Here, 'y' represents the voltage, 'x' is the temperature, 'm' is the slope, and 'c' is the y-intercept.
02

Calculate the Slope

Calculate the slope of the linear equation by using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\), which results in \(m = \frac{28 - 0}{510 - 0} = 0.0549 \, \mathrm{mV/°C}\). Record this value as it will be used to calculate the temperature.
03

Apply the Thermocouple Output

When the thermocouple output is \(10.2 \, \mathrm{mV}\), substitute 'y' in the equation with '10.2' and the value of 'm' as '0.0549' to solve for 'x' (the temperature in °C). This results in \(10.2 = 0.0549x + c\), since the output voltage is '0' at '0°C', it means that 'c' is '0'.
04

Solve the Equation

Solve for 'x' (the temperature of the variable junction) using the rearranged equation from step 3, \(x = \frac{10.2}{0.0549}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Understanding Linear Relationship in Thermocouples
Thermocouples are fascinating devices that exploit the linear relationship between temperature and voltage. They consist of two different metals which produce a voltage when exposed to a temperature differential at their junctions.
This relationship is essential because it allows us to measure unknown temperatures using known voltages. The term "linear relationship" indicates that the changes in temperature are directly proportional to changes in voltage. This forms a straight line on a graph plotted with voltage on the y-axis and temperature on the x-axis.
In mathematical terms, we represent this relationship with the linear equation: \[ y = mx + c \]where:
  • \( y \) is the voltage output,
  • \( x \) is the temperature of the unknown junction,
  • \( m \) is the slope or the rate of change in voltage with respect to temperature,
  • \( c \) is the y-intercept, which is the voltage when the temperature is zero (in our case, \( c \) is 0 because the voltage is 0 at \( 0^{\circ} \mathrm{C}\)).
Understanding this relationship is crucial because it forms the basis of interpreting thermocouple data. If the relationship weren’t linear, calculations would become much more complex. Linear relationships simplify our tasks significantly.
Voltage Measurement with Thermocouples
Voltage measurement in thermocouples is a key method used to determine temperature differences. These small voltage signals are generated due to the Seebeck effect, which involves the differing thermoelectric properties of the metals used at the junctions. The measurable output voltage from a thermocouple is typically in the millivolt range.
It’s important to have precise instruments to measure these voltages accurately, as the values are often very small (in our example, up to 28.0 mV).
Measuring voltage allows for a clear and direct interpretation of the corresponding temperature, thanks to the linear characteristics mentioned earlier. The quality and accuracy of the voltage measurement process directly influence the reliability of any temperature readings deduced from these measurements.
Modern digital voltmeters are common tools used for such measurements, providing high resolution and accuracy necessary to read the tiny voltages produced by thermocouples, thus ensuring that we can accurately translate them to meaningful temperature readings.
Converting Voltage to Temperature
Converting voltage output from a thermocouple into temperature involves applying the linear equation derived from the experimental data. The conversion process hinges on understanding that each specific thermocouple's metals and configuration results in a unique slope value (\( m \)).
In the given exercise, with the slope calculated as 0.0549 mV/°C, it becomes straightforward to find out the temperature from any measured voltage. For instance, for a voltage of 10.2 mV, one recalculates using the formula:\[ x = \frac{y}{m} \]Substituting the given values:\[ x = \frac{10.2}{0.0549} \]This computation yields the temperature of the variable junction as approximately 185.6°C.
This conversion is critical, as it allows engineers, scientists, and technicians to make temperature measurements in various industrial and laboratory settings efficiently. Through this method, precise temperature readings are achieved for otherwise challenging environments. The straightforwardness of the process lies in the consistency of the linear relationship once you have the initial slope value and intercept, making thermocouples an invaluable tool for temperature monitoring.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(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.

When the temperature of a copper penny (which is not pure copper) is raised by \(100 \mathrm{C}^{\circ}\), its diameter increases by \(0.18 \%\). Find the percent increase in \((a)\) the area of a face, \((b)\) the thickness, ( \(c\) ) the volume, and \((d)\) the mass of the penny. \((e)\) Calculate its coefficient of linear expansion.

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 \(\beta_{\text {liquid }}=4.2 \times 10^{-5} / \mathrm{C}^{\circ}\)

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 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 \(96.28 \Omega ?\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free