Fundamentals of Macroscopic and Microscopic Thermodynamics

week 1 Assignment :

PROMPT

Consider two equal 1000-cm$^3$ cubes of copper. Initially separated, one has a temperature of 20$^\circ$C and the other is at 100$^\circ$C. They are then brought into contact along one wall, but otherwise isolated from their surroundings. Estimate how long it will take for the two cubes to come into equilibrium.

Given,
volume of copper cubes = 1000 cm^3 . That mean each
side of cubes are 10 cm and the cross-sectional area of
heat flow is 100cm^2 or 0.01 m^2 .
From Fourier's law of conduction,
Q/t = KA (T2 -T1) / d
where,
Q = heat transfer
t = time taken for heat transfer
K = thermal conductivity of Copper cubes = 390 (approx)
A = cross-sectional area of heat flow
T1, T2 = temperatures of two cubes
d = thickness of two cubes = 10 + 10 = 20 cm 0.2 m
Substituting the values we get,
Q/ t = 390 * 0.01 * 80 / 0.2 = 1560 Joules/sec ---
equation 1
But, we also have
Q = T1 - T2 / R1 + R2
where,
R1 = R2 = resistance = thickness/K = 0.1 / 390
Q = 80 / (2* 0.1/390) --- equation 2
Equating equations 1 and 2 we get,
t = 100 sec