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