The temperature gradient across a copper rod of thickness 0.02m, maintained at two temperature junctions of 20°C and 80°C respectively is
- 3.0 x 102Km-1
-
3.0 x 103Km-1
- 5.0 x 103Km-1
- 3.0 x 104Km-1
The correct answer is: 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