
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
244.5k+ 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 Admit Card OUT LIVE Soon| Session 2 Direct Download Link

JEE Main 2026 Session 2 City Intimation Slip Expected Soon: Check How to Download

JEE Main 2026 Session 2 Application Form: Reopened Registration, Dates & Fees

JEE Main 2026 Session 2 Registration (Reopened): Last Date, Fees, Link & Process

WBJEE 2026 Registration Started: Important Dates Eligibility Syllabus Exam Pattern

Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Understanding the Angle of Deviation in a Prism

Understanding Differential Equations: A Complete Guide

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

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

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

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

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

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

CBSE Notes Class 11 Physics Chapter 14 - Waves - 2025-26

