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The temperature of the cold junction of a thermocouple is \(0{ }^{\circ} \mathrm{C}\) and the temperature of hot iunction is \(T^{\circ} \mathrm{C}\). The emf is \(E=16 T-0.04 T^{2} \mu \mathrm{V}\). The inversion temperature \(T_{i}\) is (a) \(200^{\circ} \mathrm{C}\) (b) \(400^{\circ} \mathrm{C}\) (c) \(100^{\circ} \mathrm{C}\) (d) \(300^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The inversion temperature is \( 400^{°C} \). Option (b) is correct.

Step by step solution

01

Understand the Problem

We need to find the inversion temperature \( T_i \) for the given emf equation \( E = 16T - 0.04T^2 \). Inversion temperature is the temperature \( T_i \) at which the emf \( E \) changes sign.
02

Set the EMF to Zero

At the inversion temperature, the emf \( E \) is zero. Thus, we need to solve the equation \( 16T - 0.04T^2 = 0 \) for \( T \).
03

Solve the Quadratic Equation

Factor out \( T \) from the equation to get: \( T(16 - 0.04T) = 0 \). This gives us two potential solutions: \( T = 0 \) or \( 16 - 0.04T = 0 \).
04

Solve for Non-zero Temperature

We ignore \( T = 0 \) since it is not the inversion temperature (this corresponds to no temperature difference). Solve \( 16 - 0.04T = 0 \) by rearranging to find \( T \).
05

Calculate Inversion Temperature

Rearrange \( 16 = 0.04T \) to get \( T = \frac{16}{0.04} \). Calculate this to find the inversion temperature.
06

Interpret the Calculation

Perform the division: \( T = \frac{16}{0.04} = 400 \). Thus the inversion temperature \( T_i \) is \( 400^{ ext{°C}} \).
07

Verify with Answer Options

Check the options: (a) \( 200^{\circ}C \), (b) \( 400^{\circ}C \), (c) \( 100^{\circ}C \), (d) \( 300^{\circ}C \). The correct answer is \( 400^{\circ}C \).

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Key Concepts

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

Thermoelectric Effect
The thermoelectric effect is a fascinating phenomenon where a voltage (emf) is generated across a circuit formed by two different conductors or semiconductors when there is a temperature difference between them. This effect is at the core of thermocouple operations.
Understanding the thermoelectric effect involves exploring three main components:
  • Seebeck Effect: When different parts of a thermocouple are subjected to different temperatures, an emf is generated, which is known as the Seebeck voltage. This is why the emf in the given problem depends on the temperature.
  • Peltier Effect: The reverse of the Seebeck effect, whereby an emf applied across a circuit causes a temperature difference. Though not directly used in this problem, it helps in understanding the broader concept.
  • Thomson Effect: It involves heating or cooling of a current-carrying conductor in a temperature gradient, adding more complexity to practical applications.
In our exercise, the emf is influenced by temperature differences, demonstrating the direct relationship delineated by the Seebeck effect. By understanding these effects, one can see how temperature variations translate to electrical voltage.
EMF Calculation
Electromotive force, or emf, is an important tool in electromagnetism and thermodynamics. It represents the work done by a generator or a cell in driving a unit charge through the circuit. In terms of a thermocouple, the emf calculation is crucial for determining the temperature difference.
In the given problem, the emf is defined as a function of temperature: \[ E = 16T - 0.04T^2 \] where \( E \) is given in microvolts (\( ext{μV} \)) and \( T \) in degrees Celsius.
To find the inversion temperature, where the emf is zero, we set \( E = 0 \) and solve the quadratic equation. This approach helps us identify the point where the voltage changes direction, a critical factor in thermocouple design. Understanding how to calculate the emf and its significance assists in predicting temperature variations accurately along with their effects in thermal systems.
Quadratic Equations in Physics
Quadratic equations frequently appear in physics, especially when dealing with forces, energy, and motion. They are incredibly useful for continuously changing systems like thermocouples.
In the context of our exercise, we use the quadratic equation to determine the inversion temperature by setting the emf to zero:\[ 16T - 0.04T^2 = 0 \]We factor this equation, acknowledging the potential solutions. The physics concept here is that this mathematical tool helps to describe a changing system where the variables (like temperature and emf) interact in non-linear ways.

This leads us to the interpretation of roots in the polynomial equation:
  • A solution of \( T = 0 \) might imply no change, but is irrelevant in the exercise context as it denotes no temperature difference.
  • The non-zero solution, often achieved through rearranging and solving, gives the pivotal inversion temperature.

Quadratic equations allow us to model these physical scenarios effectively, turning complex relationships into manageable calculations. Knowing how to solve these equations provides insight into predicting and understanding systems within physics.

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Most popular questions from this chapter

The current flowing through a wire depends on time as \(I=3 t^{2}+2 t+5\). The charge flowing the cross section of the wire in time from \(t=0\) to \(t=2 \mathrm{~s}\) in (a) \(21 \mathrm{C}\) (b) \(10 \mathrm{C}\) (c) \(22 \mathrm{C}\) (d) \(1 \mathrm{C}\)

Two wires of same metal have the same length but their cross-sections are in the ratio \(3: 1\). They are joined in series. The resistance of the thicker wire is \(10 \Omega\). The total resistance of the combination is (a) \(5 / 2 \Omega\) (b) \(40 / 3 \Omega\) (c) \(40 \Omega\) (d) \(100 \Omega\)

The thermo emf of a thermo-couple is found to depend on temperature \(T\) (in degree celsius) as \(E=4 T-\frac{T^{2}}{200}\), where \(T\) is the temperature of the hot junction. The neutral and inversion temperatures of the thermocouple are (in degree celsius) (a) 100,200 (b) 200,400 (c) 300,600 (d) 400,800

A dry cell of emf \(1.5 \mathrm{~V}\) and internal resistance \(0.10 \Omega\) is connected across a resistor in series with a very low resistance ammeter. When the circuit is switched on, the ammeter reading settles to a steady rate of \(2 \mathrm{~A}\). Find (i) chemical energy consumption of the cell (ii) energy dissipation inside the cell (iii) energy dissipation inside the resistor (iv) power output of source is (a) (i) 3 W (ii) 0.4 W (iii) 2.6 W (iv) 2.6 W (b) (i) 0.4 W (ii) 3 W (iii) 2.6 W (iv) 2.6 W (c) (i) 2.6 W (ii) 0.4 W (iii) 9 W (iv) 1 W (d) None of the above

A heating element using nichrome connected to a \(230 \mathrm{~V}\) supply draws an initial current of \(3.2 \mathrm{~A}\) which settles after a few seconds to a steady value of \(2.8 \mathrm{~A}\). What is the steady temperature of the heating element, if the room temperature is \(27.0^{\circ} \mathrm{C}\) ? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is \(1.70 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1} .\) (a) \(967^{\circ} \mathrm{C}\) (b) \(867^{\circ} \mathrm{C}\) (c) \(853^{\circ} \mathrm{C}\) (d) \(937^{\circ} \mathrm{C}\)

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