Temperature at midway between 2 rods

Classical Mechanics Level 4

Two long, thin, solid cylinders are identical in size, but they are made of different substances with two different thermal conductivities. The two cylinders are connected in series between a reservoir at temperature T h o t T_{hot} and a reservoir at temperature T c o l d T_{cold} . The temperature at the boundary between the two cylinders is T b T_b . One can conclude that :-

(A) T b T_b is closer to T h o t T_{hot} than it is to T c o l d T_{cold}

(B) T b T_b is closer to T c o l d T_{cold} than it is to T h o t T_{hot}

(C) T b T_b is closer to the temperature of the reservoir that is connected through the cylinder of lower thermal conductivity

(D) T b T_b is closer to the temperature of the reservoir that is connected through the cylinder of higher thermal conductivity

D D C C B B None of These A A

Nishant Rai by Nishant Rai , 25, India

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2 solutions

The Heat Transfer Rate for both cylinder would be same, as they are connected in series . Assume,

Thermal Conductivity of more conductive rod = k h k_h

Thermal Conductivity of less conductive rod = k l k_l

Temp. Difference for more conductive rod = T H o t − T b T_{Hot}-T_b

Temp. Difference for less Conductive rod = T b − T c o l d T_b - T_{cold}

Length of the rods = l l

Area of Cross Section= A

From the equations of Thermal Conduction, k h A ( T h o t − T b ) l = k l A ( T b − T c o l d ) l \frac{k_h\text{A}(T_{hot}-T_b)}{l}=\frac{k_l\text{A}(T_b-T_{cold})}{l} → T b − T c o l d T h o t − T b = k h k l \rightarrow \frac{T_b-T_{cold}}{T_{hot}-T_b}=\frac{k_h}{k_l}

As you mentioned +declared already that, k h > k l k_h>k_l

Then, T b − T c o l d > T h o t − T b T_b-T_{cold}>T_{hot}-T_b So, T b T_b is closer to T h o t T_{hot} . And T h o t − T b T_{hot}-T_b is the Temp. Difference between the ends of the rod of higher thermal conductivity .

Which means, T b T_b is closer to the temperature of the reservoir that is connected through the cylinder of higher thermal conductivity.

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Shuvam Keshari
Sep 27, 2015

convert the entire situation to the situation of current electricity!!

temperature of reservoirs are equivalent to voltages. and heat flow is analogous to current.

and we have 2 different resistances. with the help of 2 circuit diagrams we arrive at the answer immediately!!

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