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Question: The temperature of the hot junction of a thermocouple changes from \( {80^ \circ }C \) to \( {100^ \...

The temperature of the hot junction of a thermocouple changes from 80C{80^ \circ }C to 100C{100^ \circ }C .The percentage change in thermoelectric power is:
(A) 8%8\%
(B) 10%10\%
(C) 20%20\%
(D) 25%25\%

Explanation

Solution

Use the relation between thermoelectric power and thermoelectric emf to solve the sum. The formulae used to solve this problem are listed as:
E=kθ2E = k{\theta ^2} , where EE represents thermoelectric emf, kk represents a constant and θ\theta represents temperature.
P=dEdθP = \dfrac{{dE}}{{d\theta }} , where PP represents thermoelectric power, dEdE represents change in thermoelectric emf and dθd\theta represents change in temperature.

Complete step by step answer
In the question it is given that the temperature of the hot junction of the thermocouple changes from 80C{80^ \circ }C to 100C{100^ \circ }C . So, we need to calculate the percentage by which the thermoelectric power changes during this change of temperature.
So, firstly we will find out the thermoelectric electro-motive force(emf). For a thermocouple, the thermoelectric emf can be written as :
E=kθ2E = k{\theta ^2} (eqn.1)(eqn.1)
where all the terms have the same physical meaning as stated above.
Now, we will calculate the thermoelectric power. We know that the thermoelectric power is defined as the rate of change of thermoelectric emf with temperature. So, mathematically we can represent it as:
P=dEdθP = \dfrac{{dE}}{{d\theta }} ; where all the terms have the same physical meaning as stated above.
Substituting the value of EE from (eqn.1)(eqn.1) in the previous equation we get:
P=dEdθ=d(kθ2)dθ=2kθ\Rightarrow P = \dfrac{{dE}}{{d\theta }} = \dfrac{{d(k{\theta ^2})}}{{d\theta }} = 2k\theta
As 2k2k is a constant we can write,
P=Kθ\Rightarrow P = K\theta ; where K=2kK = 2k
Or, PθP \propto \theta
Now, from the previous equation we can conclude that thermoelectric power is directly proportional to the temperature and hence the change in power we can calculate as:
ΔPP×100%=P100P80P100×100%\dfrac{{\Delta P}}{P} \times 100\% = \dfrac{{{P_{100}} - {P_{80}}}}{{{P_{100}}}} \times 100\%
ΔPP×100%=1008080×100%\Rightarrow \dfrac{{\Delta P}}{P} \times 100\% = \dfrac{{100 - 80}}{{80}} \times 100\%
ΔPP×100%=2080×100%=14×100%=25%\Rightarrow \dfrac{{\Delta P}}{P} \times 100\% = \dfrac{{20}}{{80}} \times 100\% = \dfrac{1}{4} \times 100\% = 25\%
Therefore, when there is a temperature change of hot junction in a thermocouple from 80C{80^ \circ }C to 100C{100^ \circ }C the thermoelectric power changes by 25%25\% .
So, option (D) is a correct answer to this question.

Note
In these types of problems, the information about change in temperature may be given in two ways:
-One may be given in the same way as given in this question i.e., the two temperatures will be given. So, we have to calculate the change by taking the difference between the temperatures.
-Another way of giving the information will be like “the temperature increased/decreased by “ a certain amount i.e., here the difference/change itself is given. So, we just need to substitute the value while solving the sum.