Question

A closed box is filled with dry ice at a temperature of -94.7°C, while the outside temperature is 26.8°C. The box is cubical, measuring 0.285 m on a side, and the thickness of the walls is 3.75 x 10^-2 m. In one day, 3.64 x 10^6 J of heat is conducted through the six walls. Find the thermal conductivity of the material from which the box is made.

Answer 1

Answer:

Explanation:

3.64 x 10⁶ J passes through 6 walls

heat energy passing through 1 wall = 0.606 x 10⁶ J

Surface Area of 1 wall A = .285² = 0.081225 m²

Temperature Difference = T₁ - T₂ = 26.8 + 94.7 = 121.5

Thickness of wall d = 3.75 x 10⁻² m

Rate of heat flow per second R = $\frac{0.606 \times10^6}{24\times60\times60}$

=7.01 J per s.

Formula for rate of heat flow

R = $\frac{KA(T_1-T_2)}{d}$

Where K is thermal conductivity.

7.01 = $\frac{K\times121.5\times.081225}{3.75\times10^{-2}}$

K = 2.66 X 10⁻² W m⁻¹s⁻¹