**Fundamentals of Macroscopic and Microscopic Thermodynamics**

week 1 Assignment :

PROMPT

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

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