The temperature gradient across a copper rod of thickness 0.02m, maintained at two temperature junctions of 20º and 80ºC respectively is?
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Correct Answer: Option B
Explanation:
\(\Delta\) = \(\frac{\bigtriangleup\theta}{\delta} = \frac{80-20}{0.02}\)
= 3.0 x 10\(^{3}\)km\(^{-1}\)
\(\Delta\) = \(\frac{\bigtriangleup\theta}{\delta} = \frac{80-20}{0.02}\)
= 3.0 x 10\(^{3}\)km\(^{-1}\)