/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 One end of a thermally insulated... [FREE SOLUTION] | 91Ó°ÊÓ

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One end of a thermally insulated rod is kept at a temperature \(T_{1}\) and the other end at \(T_{2}\). The rod is composed of two sections of length \(l_{1}\) and \(l_{2}\) and thermal conductivities \(K_{1}\) and \(K_{2}\), respectively. The temperature at the interface of the two sections is (A) \(\left(K_{2} l_{2} T_{1}+K_{1} l_{1} T_{2}\right) /\left(K_{1} l_{1}+K_{2} l_{2}\right)\) (B) \(\left(K_{2} l_{1} T_{1}+K_{1} l_{2} T_{2}\right) /\left(K_{2} l_{1}+K_{1} l_{2}\right)\) (C) \(\left(K_{1} l_{2} T_{1}+K_{2} l_{1} T_{2}\right) /\left(K_{1} l_{2}+K_{2} l_{1}\right)\) (D) \(\left(K_{1} l_{1} T_{1}+K_{2} l_{2} T_{2}\right) /\left(K_{1} l_{1}+K_{2} l_{2}\right)\)

Short Answer

Expert verified
The short answer is: \(T_i = \frac{(K_1 l_2 T_1 + K_2 l_1 T_2)}{(K_1 l_2 + K_2 l_1)}\).

Step by step solution

01

Understand the problem

We have a thermally insulated rod composed of two sections with different thermal conductivities and lengths, and we need to find the temperature at the interface of the two sections.
02

Use the concept of thermal resistance

Since heat flows through a medium with thermal conductivity K and length l, it experiences a thermal resistance R given by \[R = \frac{l}{K}\] In this case, we have two different thermal resistances for the two sections: \[R_1 = \frac{l_1}{K_1}\] \[R_2 = \frac{l_2}{K_2}\]
03

Apply Ohm's law for heat flow

Ohm's law for heat flow can be used to relate the heat flow (Q) through the two sections to the temperature difference across the sections: \[Q = \frac{ΔT}{R}\] Since the rod is insulated, we can assume that the heat flow through both sections is equal: \[Q_1 = Q_2\]
04

Setup the heat flow equations

Using Ohm's law for heat flow through both sections: \[\frac{T_1 - T_i}{R_1} = \frac{T_i - T_2}{R_2}\] Where \(T_i\) is the temperature at the interface of the two sections.
05

Solve for the interface temperature

Rearrange the equation from step 4 to solve for \(T_i\): \[T_i (R_1 + R_2) = R_1 T_2 + R_2 T_1\] Substitute the values for \(R_1\) and \(R_2\) from step 2: \[T_i\left(\frac{l_1}{K_1} + \frac{l_2}{K_2}\right) = \frac{l_1}{K_1} T_2 + \frac{l_2}{K_2} T_1\] Now, we can solve for \(T_i\): \[T_i = \frac{\frac{l_1}{K_1} T_2 + \frac{l_2}{K_2} T_1}{\frac{l_1}{K_1} + \frac{l_2}{K_2}}\] Comparing the obtained expression for \(T_i\) with the given options, we can see that the answer is option (C): \[T_i = \frac{(K_1 l_2 T_1 + K_2 l_1 T_2)}{(K_1 l_2 + K_2 l_1)}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermal Conductivity
Thermal conductivity is a fundamental material property denoting a material's ability to conduct heat. It's represented by the symbol \( K \) and measures how easily heat passes through a material. When you think about heat transfer, imagine it like water flowing through a pipe: thermal conductivity indicates how "wide" that pipe is, or how straightforward it is for heat to flow.
Imagine two materials, one with high thermal conductivity and one with low. The high thermal conductivity material allows heat to move through it more quickly, akin to how water flows rapidly through a wide, smooth pipe. On the other hand, a material with low thermal conductivity hinders the flow of heat, like water struggling through a narrow, rough pipe.
  • A high \( K \) value means good heat conduction.
  • A low \( K \) value means poor heat conduction.
  • In this exercise, each section of the rod has its own thermal conductivity, affecting the heat transfer rate through each section.
Ohm's Law for Heat Flow
Ohm's law is commonly associated with electrical circuits, but heat flow can also be described using a similar principle. Ohm’s law for heat flow states that the rate of heat transfer (\( Q \)) through a material is proportional to the temperature difference across it and inversely proportional to the thermal resistance (\( R \)).
This relationship can be mathematically represented as:
\[ Q = \frac{\Delta T}{R} \] where \( \Delta T \) is the temperature difference and \( R \) is the thermal resistance, equivalent to the obstacle heat encounters as it flows through a material. This concept helps us understand how processes in the rod's sections relate to each other.
  • In both sections of the rod, heat flow is maintained equal due to being thermally insulated.
  • Thermal resistance forms naturally when heat flows through a medium as defined by the length \( l \) and thermal conductivity \( K \) of the section.
Using Ohm's law for both sections in this exercise, we can derive an equation that allows for solving the interface temperature. This approach ensures the temperature at the interface achieves equilibrium in heat flow across both sections.
Interface Temperature
The interface temperature \( T_i \) in this context is the temperature at the junction where two segments of a rod meet. Each segment differs in length and thermal conductivity, affecting how they conduct heat. Determining \( T_i \) is crucial as it represents the balancing point of heat transfer between these two distinct parts.
To find this temperature, set up an equation derived from equalizing the heat flow through both sections, thanks to the application of Ohm's law for heat flow. Solving this will give us the temperature where the heat loss from one section equals the heat gain in the other.
  • Knowing \( T_i \) helps in understanding how different materials regulate the temperature in thermal systems.
  • It's found using the formula:
    \[ T_i = \frac{(K_1 l_2 T_1 + K_2 l_1 T_2)}{(K_1 l_2 + K_2 l_1)} \]
  • This formula considers both the thermal conductivity and length of each section, ensuring accurate representation of thermal behavior.
Understanding interface temperature is essential for designing systems where junctions between two materials occur, as it affects overall system efficiency.

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