Question 963: Thermodynamics & KTG

Question

Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity $2 K$ while that in the middle has thermal conductivity $K$. The left end of the combination is maintained at temperature $3 T$ and the right end at $T$. The rods are thermally insulated from outside. In steady state, temperature at the left junction is $T_1$ and that at the right junction is $T_2$. The ratio $T_1 / T_2$ is

Accepted Answer

First, consider the equivalent thermal resistance $R_{	ext{eq}}$ of the series arrangement:
$ R_{	ext{eq}} = R_1 + R_2 + R_3 $
Calculating each resistance:
$ R_1 = \frac{1}{2KA}, \quad R_2 = \frac{1}{KA}, \quad R_3 = \frac{1}{2KA} $
Summing these gives:
$ R_{	ext{eq}} = \frac{1}{2KA} + \frac{1}{KA} + \frac{1}{2KA} = \frac{2}{KA} $
In a series arrangement, the rate of heat flow is constant. Thus:
$ \frac{3T - T_1}{R_1} = \frac{3T - T}{R_{	ext{eq}}} $
Substituting the resistances:
$ \frac{(3T - T_1) \cdot 2KA}{I} = \frac{2T \cdot KA}{2I} $
Solving the equation yields:
$ 6T - 2T_1 = T \implies 2T_1 = 5T \implies T_1 = \frac{5T}{2} \quad \ldots (1) $
Next, examine the heat flow rate in the third section:
$ \frac{T_2 - T}{R_3} = \frac{3T - T}{R_{	ext{eq}}} $
Substitute:
$ \frac{(T_2 - T) \cdot 2KA}{I} = \frac{2T \cdot KA}{2I} $
Solving gives:
$ 2T_2 - 2T = T \implies 2T_2 = 2T + T \implies T_2 = \frac{3T}{2} \quad \ldots (2) $
By substituting equations (1) and (2) into the ratio:
$ \frac{T_1}{T_2} = \frac{5T/2}{3T/2} = \frac{5}{3} $
Thus, the ratio $\frac{T_1}{T_2}$ is $\frac{5}{3}$.

Answer

$\frac{5}{3}$

Answer

$\frac{5}{4}$

Answer

$\frac{3}{2}$

Answer

$\frac{4}{3}$

Thermodynamics & KTG

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