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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) $100K$

Last updated date: 05th Mar 2024
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IVSAT 2024
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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${10^{ - 3}}$.