How Do You Calculate Temperature Difference Using a Thermocouple?

[Puny]

Homework Statement

The one end of the thermocouple is placed in 45C water, the other end in oil of unknown temperature. Of the voltmeter connected to it's end I got 1.35mV. If the sensitivity of the thermocouple is 0.05mV/C, what will the difference of temperature of the liquids be?

Homework Equations

This is really what I'm looking for. Not sure at all what equation I would use.

The Attempt at a Solution

Would doing something like:
1.35mV / 0.05mV/C = 27C
Would 27C be the difference

Thanks for any input

 

[Shooting Star]

I don't see what else could it be. What does your textbook say?

 

[ogb p]

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I would first identify the type of thermocouple used in this experiment and its corresponding temperature range. This information is important in determining the accuracy and reliability of the results.

Next, I would use the thermocouple sensitivity equation, which is V = αΔT, where V is the voltage, α is the sensitivity, and ΔT is the temperature difference. Plugging in the given values, we get:

1.35mV = (0.05mV/C)ΔT

Solving for ΔT, we get a temperature difference of 27°C.

However, it is important to note that this result is only an estimate and may not be completely accurate due to factors such as temperature gradients, thermal conductivity, and calibration errors. To improve the accuracy of the results, multiple readings should be taken and averaged. Additionally, the experiment should be repeated with different temperatures to confirm the sensitivity of the thermocouple.