The block in the drawing has dimensions L0×2L0×3L0,where L0 =0.5 m. The block has a thermal conductivity of 300 J/(s·m·C˚). In drawings A, B, and C, heat is conducted through the block in three different directions; in each case the temperature of the warmer surface is 37 ˚C and that of the cooler surface is 8 ˚C Determine the heat that flows in 3 s for each case.

The drawing showing the block is missing, so i have attached it.

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

lol

explanation: