(c) Given, Thomson’s coefficient,$\ \ \sigma = 3\mu$V/$℃$
$\sigma = \frac{dV}{dT}$
$or\ d\ V = \sigma dT$
so $\ V = \sigma(T_{2} - T_{1})$
or $3 \times 10^{- 4} = 3 \times 10^{- 6}(T_{2} - 20)$
or $100 + 20 = T_{2}$
or ${\ T}_{2} = 120{^\circ}\ $C