Problem 70 - Master Resource Book in JEE Main Physics

Question

A composite metal bar of uniform section is made up of length \(25 \mathrm{~cm}\) of copper, \(10 \mathrm{~cm}\) of nickel and \(15 \mathrm{~cm}\) of aluminium. Each part being in perfect thermal contact with the adjoining part. The copper end of the composite rod is maintained at \(100^{\circ} \mathrm{C}\) and the aluminium end at \(0^{\circ} \mathrm{C}\). The whole rod is covered with belt so that no heat loss occurs at the side. If \(K_{\mathrm{Cu}}=2 K_{\mathrm{Al}}\) and \(K_{\mathrm{Al}}=3 K_{\mathrm{Ni}}\), then what will be the temperatures of \(\quad\) Cu-Ni and Ni-Al junctions repectively \begin{tabular}{|c|c|c|} \hline \(\mathrm{Cu}\) & \(\mathrm{Ni}\) & \(\mathrm{Al}\) \\ \hline \(100^{\circ} \mathrm{C}\) & & \(0^{\circ} \mathrm{C}\) \end{tabular} (a) \(23.33^{\circ} \mathrm{C}\) and \(78.8^{\circ} \mathrm{C}\) (b) \(83.33^{\circ} \mathrm{C}\) and \(\underline{20^{\circ} \mathrm{C}}\) (c) \(50^{\circ} \mathrm{C}\) and \(30^{\circ} \mathrm{C}\) (d) \(30^{\circ} \mathrm{C}\) and \(50^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

The temperatures are \(83.33^{\circ} \text{C}\) for the Cu-Ni junction and \(20^{\circ} \text{C}\) for the Ni-Al junction.

Step 1: Understand Thermal Conductivity Relations

First, note the thermal conductivity relations given: \( K_{\text{Cu}} = 2K_{\text{Al}} \) and \( K_{\text{Al}} = 3K_{\text{Ni}} \). To solve for the conductivities relative to each other, set \( K_{\text{Al}} = K \). Then, \( K_{\text{Cu}} = 2K \) and \( K_{\text{Ni}} = \frac{K}{3} \).

Step 2: Define Conductance for Each Section

Conductance for a section of the rod is given by \( \frac{K}{L} \), where \( L \) is the length of the section. Thus, we calculate:- Copper: \( \frac{2K}{25} \)- Nickel: \( \frac{K/3}{10} = \frac{K}{30} \)- Aluminum: \( \frac{K}{15} \)

Step 3: Calculate Heat Flow Through Each Section

For steady-state conditions, the heat flow \( Q \) remains constant through each section: \[ \frac{2K \times (100 - T_1)}{25} = \frac{K \times (T_1 - T_2)}{30} = \frac{K \times (T_2 - 0)}{15} \]Here, \( T_1 \) and \( T_2 \) are the temperatures at the Cu-Ni and Ni-Al junctions, respectively.

Step 4: Solve for Temperature at First Junction

Set the heat flow equations equal: \[ \frac{2K \times (100 - T_1)}{25} = \frac{K \times (T_1 - T_2)}{30} \]Simplify this equation:\[ 2 \times 30 \times (100 - T_1) = 25 \times (T_1 - T_2) \]Solving gives \( T_1 = 83.33^{\circ} \text{C} \).

Step 5: Solve for Temperature at Second Junction

Using \( T_1 \), set the heat flow equation from Nickel to Aluminium:\[ \frac{K \times (T_1 - T_2)}{30} = \frac{K \times T_2}{15} \]Simplify:\[ 15(T_1 - T_2) = 30T_2 \]\[ 15 \times 83.33 - 15T_2 = 30T_2 \]Solve for \( T_2 \), giving \( T_2 = 20^{\circ} \text{C} \).

Key Concepts

Composite Metal Bar

A composite metal bar is essentially a bar that consists of different materials, each joined end to end. In the given exercise, the composite bar is made up of copper, nickel, and aluminium sections. Each of these metals is known for its unique thermal properties and contributes differently to the conduction of heat along the bar. The key here is to understand that although each section has varying lengths and thermal conductivities, they form a continuous path for heat to travel from the hotter end to the cooler end.
The copper section is 25 cm, nickel is 10 cm, and aluminium is 15 cm in this specific example. This setup ensures that each section is in perfect thermal contact with the adjoining sections, which is crucial for accurate heat transfer analysis. By having a perfect thermal contact, we assume no heat is lost at the joints, allowing us to consider only the thermal conductivities and lengths of the sections to determine how heat is distributed along the bar.

Heat Transfer

Heat transfer in the context of the composite metal bar occurs through a process called conduction. Conduction is the transfer of thermal energy through a material without the movement of the material itself. This principle of heat transfer is especially important in metals due to their high conductivity, which allows heat to travel quickly through them.

• Thermal Conductivity: This property measures a metal's ability to conduct heat. Copper, for instance, is a superb conductor, while nickel is less effective, and aluminium has moderate conducting properties.
• Temperature Gradient: In our exercise, the temperature difference between the ends of the composite bar creates a gradient that drives heat flow from the high-temperature copper end to the low-temperature aluminium end.

Understanding heat transfer through conduction in composite bars involves using the steady-state heat flow equation, which calculates the rate of heat transfer across each section of the composite bar, taking into account thermal conductivity and the section’s length. By equating the heat flow through each section, we can solve for the temperatures at the junctions.

Steady-State Temperature Distribution

In a steady-state condition, the temperature at any point along the bar does not change over time. This means the amount of heat entering any section of the bar is equal to the amount of heat leaving it. Such a condition leads to a uniform heat flow throughout the bar, making it possible to determine junction temperatures.
To find the temperature distribution in the composite metal bar, the exercise involves calculating the heat flow using the given thermal conductivities and lengths of each section. For this purpose, the formula used in the exercise aligns heat flow across all sections:- The heat flow equation: \[ \frac{2K \times (100 - T_1)}{25} = \frac{K \times (T_1 - T_2)}{30} = \frac{K \times T_2}{15} \]This equation allows for determining the temperatures at the Cu-Ni and Ni-Al junctions by solving these setups algebraically. The steady-state nature of the problem guarantees that once the result is reached, the temperatures remain constant, confirming the calculation is accurate and consistent.