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One end of thermally insulated rod is ke...

One end of thermally insulated rod is kept at a temperature `T_(1)` and the other at `T_(2)`. The rod is composed of two section of length `l_(1)` and `l_(2)` thermal conductivities `k_(1)` and `k_(2)` respectively. The temerature at the interface of two section is

A

`((k_(1)l_(1)T_(1)+K_(2)l_(2)T_(2)))/((K_(1)l_(1)+K_(2)l_(2)))`

B

`((K_(2)l_(2)T_(1)+K_(1)l_(1)T_(2)))/((K_(1)l_(1)+K_(2)l_(2)))`

C

`((K_(2)l_(1)T_(1)+K_(1)l_(2)T_(2)))/((K_(2)l_(2)+K_(1)l_(2)))`

D

`((K_(1)l_(2)T_(1)+K_(2)l_(2)T_(2)))/((K_(1)l_(2)+K_(2)l_(1)))`

Text Solution

AI Generated Solution

The correct Answer is:
To find the temperature at the interface of the two sections of the thermally insulated rod, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Setup**: - We have a rod divided into two sections. - The left section has length \( l_1 \) and thermal conductivity \( k_1 \). - The right section has length \( l_2 \) and thermal conductivity \( k_2 \). - The temperatures at the ends of the rod are \( T_1 \) (at the left end) and \( T_2 \) (at the right end). 2. **Understand Heat Flow**: - Heat flows from the hotter end \( T_1 \) to the cooler end \( T_2 \). - Let \( T_0 \) be the temperature at the interface between the two sections. 3. **Apply Fourier's Law of Heat Conduction**: - The rate of heat flow (heat current) through each section must be equal at steady state. - For section 1 (length \( l_1 \)): \[ Q_1 = \frac{k_1 A (T_1 - T_0)}{l_1} \] - For section 2 (length \( l_2 \)): \[ Q_2 = \frac{k_2 A (T_0 - T_2)}{l_2} \] 4. **Set the Heat Currents Equal**: - Since \( Q_1 = Q_2 \): \[ \frac{k_1 A (T_1 - T_0)}{l_1} = \frac{k_2 A (T_0 - T_2)}{l_2} \] - The area \( A \) cancels out from both sides. 5. **Rearrange the Equation**: - Cross-multiply to eliminate the fractions: \[ k_1 (T_1 - T_0) l_2 = k_2 (T_0 - T_2) l_1 \] - Expand both sides: \[ k_1 T_1 l_2 - k_1 T_0 l_2 = k_2 T_0 l_1 - k_2 T_2 l_1 \] 6. **Collect Terms Involving \( T_0 \)**: - Rearranging gives: \[ k_1 T_1 l_2 + k_2 T_2 l_1 = T_0 (k_1 l_2 + k_2 l_1) \] 7. **Solve for \( T_0 \)**: - Isolate \( T_0 \): \[ T_0 = \frac{k_1 T_1 l_2 + k_2 T_2 l_1}{k_1 l_2 + k_2 l_1} \] ### Final Result: The temperature at the interface \( T_0 \) is given by: \[ T_0 = \frac{k_1 T_1 l_2 + k_2 T_2 l_1}{k_1 l_2 + k_2 l_1} \]
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