k*6L^2*(99.5 C)/d = Q = 3*10^6 J/86400 s
Solve for k. You have to divide Q by the number of seconds in a day, to get k in Watts/meter degree C
I used L = 0.35 m for the side length, and L^2 for the side area. There are six sides, hence the 6
d = the plate thickness (0.0300 m)
A closed box is filled with dry ice at a temperature of -78.5 degree C, while the outside temperature is 21.0 degree C. The box is cubical, measuring 0.350m on a side, and the thickness of the walls is 3.00*10^-2 m. In one day, 3.10*10^6 J of heat is conducted through the six walls. Find the thermal conductivity of the material from which the box is made.
Here is how I do it:
I use the equation: k=(HL)/A*deltaT
H=3.10*10^6J
A=(0.350m)^2
delta T=99.5 degree C
L=3.00*10^-2m
However, I could not get the right answer. Please correct me. Thanks a lot!
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