Question

A closed box is filled with dry ice at a temperature of -78.5 degrees C, while the outside temperature is 21.0 degrees C. The box is cubical, measuring 0.350 m on a side, and the thickness of the walls is 3..00 x 10^-2 m. In one day, 3.10 x10^6 J of heat is conducted through the six walls. Find the thermal conductivity of the material from which the box is made.

Answer 1

let thermal conductivity of the material is k.

power of heat conduction=k*A*dT/dx

where A=cross sectional area

dT=temperature difference

dx=thickness

thickness on each side=0.03 m

area=side^2=0.35*0.35=0.1225 m^2

so total power of heat conduction=6*k*A*tempeature difference/thickness

=6*k*0.1225*(21-(-78.5))/0.03

=2437.8*k W

heat conducted per day=power*time

=2437.8*k*24*3600 joules

as total heat conducted is 3.1*10^6 J

3.1*10^6=2437.8*k*24*3600

==>k=3.1*10^6/(2437.8*24*3600)=0.014718 J/(m.degree celcius)

so thermal conductivity of the material is 0.014718 J/(m.degree celcius).