The temperature gradient across a copper rod of thickness 0.02m, maintained at two temperature junctions of 20º and 80ºC respectively is?
\(\Delta\) = \(\frac{\bigtriangleup\theta}{\delta} = \frac{80-20}{0.02}\)
= 3.0 x 10\(^{3}\)km\(^{-1}\)
There is an explanation video available below.
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}