Question on: JAMB Physics - 1999
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
Ask EduPadi AI for a detailed answer
Correct Option: B
Temperature Gradient = \(\frac{\Delta {\text{temp.}}}{\text{thickness}}\)
= \(\frac{80 - 20}{0.02}\)
= \(\frac{60}{0.02}\)
= 3000Km-1
= 3.0 x 103Km-1
= \(\frac{80 - 20}{0.02}\)
= \(\frac{60}{0.02}\)
= 3000Km-1
= 3.0 x 103Km-1
Add your answer
Please share this, thanks!
No responses