here,
for the total surface area
A = 6 * 0.397^2
A = 0.946 m^2
let the thermal conductivity of the material is a
for the heat flow rate
Q/time = a * Area * difference in temperature/thickness
3.46 *10^6/(24 * 3600) = a * 0.946 * (16.3 - (-81.3))/(4.36 *10^-2)
solving for a
a = 0.0189 W/(m.K)
the thermal conductivity of the material is 0.0189 W/(m.K)
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