Question

A thermocouple gives a reading of 4.2 millivolts with one junction in steam and the other in melting ice. When one junction is placed in a hot liquid, the other remaining in the melting ice, the thermocouple reads 2.8 millivolts. Determine the temperature of the hot liquid, assuming that the thermoelectric e.m.f is proportional to the temperature difference between the two junctions.

Answer 1

Since the thermoelectric e.m.f is proportional to the temperature difference between the two junctions, we can write:

$$\dfrac{V_1}{\Delta T_1} = \dfrac{V_2}{\Delta T_2} = const$$

where $V_1 = 4.2 mV$ is the first reading, $V_2 = 2.8 mV$ is the second reading, $\Delta T_1 = 100\degree C$ is the  temperature difference between steam and ice and $\Delta T_2$ is the temperature difference between the hot liquid and ice. Expressing $\Delta T_2$, obtain:

$$\Delta T_2 = \dfrac{V_2\Delta T_1}{V_1} = \dfrac{2.8\cdot100}{4.2} = 66.67\degree C$$

Since the temperature of the ice is $0\degree C$, the temperature of the hot liquid is:

$$T_2 = \Delta T_2 = 66.67\degree C$$

Answer. 66.67 degree C.