Ratio of Thermal Conductivities in Series Cylinders

To find the ratio of the thermal conductivities $K_1 / K_2$, we consider the steady-state heat conduction through the two cylinders connected in series.

Let:

• $L$ be the length of each cylinder.
• $r_1$ be the radius of the smaller cylinder, so its cross-sectional area is $A_1 = \pi r_1^2$.
• $r_2 = 2r_1$ be the radius of the larger cylinder, so its cross-sectional area is: $A_2 = \pi r_2^2 = \pi (2r_1)^2 = 4\pi r_1^2 = 4A_1$
• $T_1 = 300\text{ K}$ be the temperature of the heat bath connected to the smaller cylinder.
• $T_2 = 100\text{ K}$ be the temperature of the heat bath connected to the larger cylinder.
• $T_j = 200\text{ K}$ be the steady-state temperature at the junction.

Since the sides are thermally insulated, there is no lateral heat loss. In the steady state, the rate of heat flow $\left(\frac{dQ}{dt}\right)$ through the smaller cylinder must equal the rate of heat flow through the larger cylinder: $\left(\frac{dQ}{dt}\right)_1 = \left(\frac{dQ}{dt}\right)_2$

Using Fourier's law of thermal conduction: $\frac{K_1 A_1 (T_1 - T_j)}{L} = \frac{K_2 A_2 (T_j - T_2)}{L}$

Substitute the given temperatures: $T_1 - T_j = 300\text{ K} - 200\text{ K} = 100\text{ K}$ $T_j - T_2 = 200\text{ K} - 100\text{ K} = 100\text{ K}$

Thus, the equation simplifies to: $K_1 A_1 (100) = K_2 A_2 (100)$ $K_1 A_1 = K_2 A_2$

Now, substituting $A_2 = 4A_1$: $K_1 A_1 = K_2 (4A_1)$

Dividing both sides by $K_2 A_1$: $\frac{K_1}{K_2} = 4$