Wires of two dissimilar metals, A and B, develop an open-circuit voltage of 45 $$muV$$ per °C of temperature difference above a reference temperature. A thermopile consists of 50 thermocouples connected in series as shown in the figure. The reference junctions are maintained at 60°C, and $$V_(ab)$$, is measured to be 60 mV. The temperature (°C) of the bot junctions is most nearly: 

A.  27
B.  87
C.  1,330
D.  1,390

Since the open circuit voltage across 50 thermocouples, $V_{ab}$, is known to be 60mV, we can divide this voltage by 50 in order to get the open circuit voltage across each individual thermocouple:

$\frac{V_{ab}}{50}=\frac{60mV}{50}=1.2mV$

Now that we have this value, we can simply plug it in to the following equation in order to find the temperature:

$$\begin{gathered} \alpha (T-T_0)=V \\ \alpha =open circuit voltage per degree Celsius difference above a reference temp. \\ T = hot junction temp. \\ {T}_0=cold junction temp. \\ \left(45\mu V\right)(T-60)=1.2mV \\ T = 86.67 \end{gathered}$$

Therefore, the temperature of the hot junction is most nearly 87 degrees Celsius, choice B.