
In a thermocouple, the thermo emf produced is $3\mu V$ per one degree difference of temperature between two junctions. When the cold junction of such a thermocouple is at ${20^ \circ }C,$ the thermo emf is $0.3mV$. Then the temperature of the hot junction is:
A) ${80^ \circ }C$
B) ${100^ \circ }C$
C) ${120^ \circ }C$
D) $100K$
Answer
224.1k+ views
Hint: A thermocouple has two junctions, a hot junction, and a cold junction. The temperature difference will give rise to an EMF. The EMF produced per one-degree difference in temperature is $3\mu V$. This gives the relation between emf and temperature in this particular thermocouple. Use this relation to find the temperature at the hot end.
Complete step by step solution:
From the question, we know that an emf $3\mu V$ is produced per one degree difference of temperature between the hot junction and the cold junction.
Let the temperature difference between the two junctions be ${t^ \circ }C$
The emf per unit change in temperature can be written as,
$e = at$ Where \[a = 3 \times {10^{ - 6}}\dfrac{V}{{^ \circ C}}\] ($1\mu V=$${10^{ - 6}}$$V$, To convert $\mu V$ into $V$ multiply with ${10^{-6}}$)
$a$ is the EMF produced per one-degree difference of temperature.
When the cold junction is at ${20^ \circ }C,$the emf is given as $0.3mV$ that is,
$e = 0.3 \times {10^{ - 3}}V$
$e = 3 \times {10^{ - 4}}V$
$\therefore t = \dfrac{{3 \times {{10}^{ - 4}}}}{{3 \times {{10}^{ - 6}}}} = {100^ \circ }C$
The temperature difference between the hot junction and the cold junction is found to be ${100^ \circ }C$
$\therefore $The temperature at the cold junction is $20 + t$$ = 20 + 100 = {120^ \circ }C$
The correct answer is option (C), ${120^ \circ }C.$
Note: Thermocouples are used to measure temperature. It will have two legs made of metals. At one end, there will be a junction made by welding the two legs together. The temperature is measured at this junction. There will be an emf whenever the temperature in this junction changes. Thermocouples have unique characteristics. There are various types of thermocouples available. While solving this problem one thing to be taken care of is while converting the power of emf. To convert $\mu V$ into Volt multiply with ${10^{ - 6}}$. To convert $mV$into Volt, multiply with${10^{ - 3}}$.
Complete step by step solution:
From the question, we know that an emf $3\mu V$ is produced per one degree difference of temperature between the hot junction and the cold junction.
Let the temperature difference between the two junctions be ${t^ \circ }C$
The emf per unit change in temperature can be written as,
$e = at$ Where \[a = 3 \times {10^{ - 6}}\dfrac{V}{{^ \circ C}}\] ($1\mu V=$${10^{ - 6}}$$V$, To convert $\mu V$ into $V$ multiply with ${10^{-6}}$)
$a$ is the EMF produced per one-degree difference of temperature.
When the cold junction is at ${20^ \circ }C,$the emf is given as $0.3mV$ that is,
$e = 0.3 \times {10^{ - 3}}V$
$e = 3 \times {10^{ - 4}}V$
$\therefore t = \dfrac{{3 \times {{10}^{ - 4}}}}{{3 \times {{10}^{ - 6}}}} = {100^ \circ }C$
The temperature difference between the hot junction and the cold junction is found to be ${100^ \circ }C$
$\therefore $The temperature at the cold junction is $20 + t$$ = 20 + 100 = {120^ \circ }C$
The correct answer is option (C), ${120^ \circ }C.$
Note: Thermocouples are used to measure temperature. It will have two legs made of metals. At one end, there will be a junction made by welding the two legs together. The temperature is measured at this junction. There will be an emf whenever the temperature in this junction changes. Thermocouples have unique characteristics. There are various types of thermocouples available. While solving this problem one thing to be taken care of is while converting the power of emf. To convert $\mu V$ into Volt multiply with ${10^{ - 6}}$. To convert $mV$into Volt, multiply with${10^{ - 3}}$.
Recently Updated Pages
JEE Main 2026 Session 1 Correction Window Started: Check Dates, Edit Link & Fees

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Trending doubts
JEE Main 2026: City Intimation Slip Releasing Today, Application Form Closed, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Hybridisation in Chemistry – Concept, Types & Applications

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

