The drawing showing the block is missing, so i have attached it.
Answer:
Q_p = 19575 J
Q_o = 8700 J
Q_g = 78300 J
Explanation:
We are given L0 = 0.5 m
From the image, we can see that;
Length; L = 3L0 = 3 × 0.5 = 1.5 m
Width; W = L0 = 0.5 m
Height; H = 2L0 = 2 × 0.5 = 1 m
Temperature of warmer surface; T_h = 37 ˚C
Temperature of cooler surface; T_l = 8 ˚C
thermal conductivity; k = 300 J/(s·m·C˚)
Time for which the heat flows; t = 3 s
Now, from Fourier's law of conduction, the rate of heat flow is given by;
Q/t = kA(dT/dx)
Where;
A is area
dt is change in temperature = 37 - 8 = 29°C
dx is perpendicular distance
Now, for the amount heat flow perpendicular to the pink surface, from the image dx will be the height 1 m.
Thus;
Q_p/3 = 300 × (1.5 × 0.5) × (29/1)
Q_p/3 = 6525
Q_p = 3 × 6525
Q_p = 19575 J
for the amount heat flow perpendicular to the orange surface, from the image dx will be the length 1.5 m.
Thus;
Q_o/3 = 300 × (1 × 0.5) × (29/1.5)
Q_o/3 = 2900
Q_o = 2900 × 3
Q_o = 8700 J
for the amount heat flow perpendicular to the green surface, from the image dx will be the width 0.5 m.
Thus;
Q_g/3 = 300 × (1 × 1.5) × (29/0.5)
Q_g/3 = 26100
Q_g = 26100 × 3
Q_g = 78300 J
