Thermal Conduction Quantified- Physics?
Here is y problem:
"A temperature difference of 31 K is maintained between the sides of a 1.3-cm thick slab of plywood. What is the rate of heat transfer through a 58-cm2 area of the slab?"
Here is what the answers are not: 1.52e2, 1.19e3, -1.19e3, 1.5e3, 1.5e2
I believe the answer needs to be in Watts. All I am certain of is it is required to be in W.
- billrussell42Lv 710 months agoFavorite Answer
you need the Thermal Conductivity coefficient for the plywood. There are many kinds of plywood.
I found this, 0.13 W/m•K but that may not match your number.
58-cm2 ?? guessing that thus is 58 cm² (the dash confuses things)
58 cm² x (1m/100cm)² = 58e-4 m²
P = 0.13 W/m•K x 31K x 58e-4 m² / 0.013 m = 1.8 watts
P = K x ∆T x A / d
K is thermal conductivity in W/m•K
P is power in watts transferred through the material
K is thermal conductivity of the material in W/m•K
∆T is change in temperature across the material
A is cross-sectional area in m²
d is thickness in m