The temperature gradient across a copper rod of thickness 0.02m, maintained at two temperature junctions of 20°C and 80°C respectively is
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Correct Answer: Option B
Explanation:
Temperature Gradient = \(\frac{\Delta {\text{temp.}}}{\text{thickness}}\)
= \(\frac{80 - 20}{0.02}\)
= \(\frac{60}{0.02}\)
= 3000Km-1
= 3.0 x 103Km-1
Temperature Gradient = \(\frac{\Delta {\text{temp.}}}{\text{thickness}}\)
= \(\frac{80 - 20}{0.02}\)
= \(\frac{60}{0.02}\)
= 3000Km-1
= 3.0 x 103Km-1