Consider the state that obtains when the temperature T of the joint is constant. Let CA and CB be the thermal capacities of the cylinders, let kA and kB be their coefficients of thermal conductivity, and TA and TB be the temperatures of their ends. Since the cylinders are homogeneous and heat is not given up through their sides, the temperature in the state that is set up changes along each of the cylinders according to a linear equation and the temperature T of the joint can be found from the equality of the flow of heat through any two cross-sections of the two cylinders
from which
HEAT AND CAPILLARY PHENOMENA 203
T = kATA + kBTB.
kA + kB (1)
The amount of heat which flows through each cylinder can be found by replacing the variable temperature along the whole cylinder by the average temperature (since the distribution of temperature along the cylinder changes according to a linear equation). The total amount of heat which flows through the cylinders
or
Qt = cATA/2 + cBTB/2 + (CA + CB)T/2. (2)
Substituting the value for T from expression (1), we shall get:
Qt = cATA/2 + cBTB/2 + (CA + CB)/2 · (kATA + kBTB)/(kA + kB).
Bringing the right-hand side over a common denominator and collecting terms containing TA and TB, we finally find that:
Qt = (2kAcA + kAcB + kBcA)/(2(kA + kB)) TA + (2kBcB + kAcB + kBcA)/(2(kA + kB)) TB.
And this expression gives us the answer to the problem posed. Since the coefficients are, in general, different at temperatures TA and TB, Q will have a different value if we replace TA by TB, i.e, the sum total of heat flowing through the cylinders will in general depend on which of the ends is heated and which is cooled. And since the coefficients at temperatures TA and TB differ only in the values of the terms kAcA and kBcB, the amount of heat will be greater if we heat the end of that cylinder for which the product of thermal capacity and thermal conductivity is greater. This is quite clear physically: the role of thermal capacity is obvious; and the greater the thermal conductivity, the less will
204 SOLUTIONS
be the fall of temperature along the cylinders and the higher will be the temperature of all points.
Only in the special case when kAcA = kBcB, as is laid down by our problem, will the amount of heat flowing through the cylinders be independent of which end is heated and which is cooled.