288KJ is conducted across two opposite faces of a 3m cube of temperature gradient 90ºCm\(^{-1}\) in 7200s. Calculate the thermal conductivity.
k = \(\frac{Q.d}{A. ΔT .t}\)
d = thickness = 3m
A = 9m\(^2\)
Q = 288kJ = 288000J, t = 7200s, ΔT = 90ºC/m → 90K/m x 3 = 270K
k = \(\frac{288000 \times 3}{9 \times 270 \times 7200}\)
k = 4.93 x 10\(^{-2}\)Wm\(^{-1}\)k\(^{-1}\)
There is an explanation video available below.
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}