Problem 102 - University Physics with Modern Physics

Question

Rods of copper, brass, and steel-each with crosssectional area of 2.00 cm\(^2\)-are welded together to form a Y-shaped figure. The free end of the copper rod is maintained at 100.0\(^\circ\)C, and the free ends of the brass and steel rods at 0.0\(^\circ\)C. Assume that there is no heat loss from the surfaces of the rods. The lengths of the rods are: copper, 13.0 cm; brass, 18.0 cm; steel, 24.0 cm. What is (a) the temperature of the junction point; (b) the heat current in each of the three rods?

Step-by-Step Solution

Answer

The junction temperature is approximately 36.1°C, with heat currents through copper, brass, and steel calculated individually.

Step 1: Understand the Thermal Conductivity Formula

The heat current, or thermal current, through a rod is given by the equation \( Q = \frac{kA(T_H - T_C)}{L} \), where \( k \) is the thermal conductivity of the material, \( A \) is the cross-sectional area, \( T_H \) and \( T_C \) are the temperatures of the hot and cold ends respectively, and \( L \) is the length of the rod.

Step 2: Identify the Given Values

For each material, we are given the following:- Cross-sectional area \( A = 2.00 \) cm\(^2\)- Temperatures of copper \( T_H = 100.0 \)°C and brass and steel \( T_C = 0.0 \)°C- Lengths: copper \( L = 13.0 \) cm, brass \( L = 18.0 \) cm, steel \( L = 24.0 \) cm- We need the thermal conductivities: copper \( k_{Cu} = 401 \) W/m·K, brass \( k_{Br} = 109 \) W/m·K, and steel \( k_{St} = 50 \) W/m·K.

Step 3: Calculate Heat Current through Copper Rod

Using the formula \( Q = \frac{kA(T_H - T_j)}{L} \) for copper:\[ Q_{Cu} = \frac{401 \times (2 \times 10^{-4}) \times (100 - T_j)}{0.13} \]Keep in mind to convert areas and lengths to meters squared and meters.

Step 4: Calculate Heat Current through Brass and Steel Rods

For the brass rod, use:\[ Q_{Br} = \frac{109 \times (2 \times 10^{-4}) \times (T_j - 0)}{0.18} \]For the steel rod, use:\[ Q_{St} = \frac{50 \times (2 \times 10^{-4}) \times (T_j - 0)}{0.24} \]

Step 5: Establish Equilibrium at Junction Point

Since the junction point is at thermal equilibrium, the total heat entering and exiting should be zero:\[ Q_{Cu} = Q_{Br} + Q_{St} \]Substitute the equations from Steps 3 and 4 to solve for the junction temperature \( T_j \).

Step 6: Solve for Junction Temperature

Equate:\[ \frac{401 \times 2 \times 10^{-4} \times (100 - T_j)}{0.13} = \frac{109 \times 2 \times 10^{-4} \times T_j}{0.18} + \frac{50 \times 2 \times 10^{-4} \times T_j}{0.24} \]Solve the equation to find \( T_j \).

Step 7: Calculate Heat Currents in Individual Rods

Insert the solved \( T_j \) back into individual heat current formulas for copper, brass, and steel rods to find the respective \( Q_{Cu}, Q_{Br}, \) and \( Q_{St} \).

Key Concepts

Heat Transfer

Heat transfer is the process by which thermal energy flows from a hot object to a cold one. This can happen through conduction, convection, or radiation. In heat conduction, like in this exercise, energy passes through a solid material. Here, the copper, brass, and steel rods transfer heat between their ends due to the temperature difference.
For conduction, the heat current is calculated with the formula:

• \( Q = \frac{kA(T_H - T_C)}{L} \)

Where:

• \( k \) is the material's thermal conductivity
• \( A \) is the cross-sectional area
• \( T_H \) and \( T_C \) are the temperatures at the hot and cold ends, respectively
• \( L \) is the length of the rod

By understanding these parameters, we're able to solve for the rate of heat transfer through different materials, just like solving for the heat current in the copper, brass, and steel rods in the given Y-shaped thermal system.

Y-shaped thermal system

In a Y-shaped thermal system, rods are connected at a junction forming a 'Y' configuration, allowing for a mix of materials to be analyzed for heat flow and temperature effects. The copper, brass, and steel rods are joined together, creating a composite system for examining heat transfer efficiency and temperature changes.
Such combinations are common in engineering to optimize thermal properties. Each rod in the Y-shape has its own specific heat conductivity, and their different lengths affect how heat flows through the system. By exploring how each material contributes to the overall heat balance, students can gain insights into design choices in thermal management. The challenge is to find the junction's temperature and rods' heat currents, setting a solid understanding of thermal dynamics in mixed metal systems.

Thermal Equilibrium

Thermal equilibrium occurs when a system reaches a state where there is no net flow of thermal energy between its components. In other words, all parts of the system are at the same temperature. For the Y-shaped system, this means the heat current flowing into the junction equals the heat current flowing out.
The balance equation can be written as:

• \[ Q_{Cu} = Q_{Br} + Q_{St} \]

Where \( Q_{Cu}, Q_{Br}, \) and \( Q_{St} \) are the heat currents through the copper, brass, and steel rods, respectively. Solving this equation determines the equilibrium temperature at the junction, revealing how efficiently heat is distributed across different materials. Understanding thermal equilibrium helps in applications where maintaining uniform temperature distribution is crucial, such as in electronics cooling or building design.

Thermal Conductivities of Metals

The thermal conductivity of a material describes its ability to conduct heat. It's an intrinsic property that varies widely between different metals. For instance, copper has a high thermal conductivity \( k_{Cu} = 401 \) W/m·K, indicating it can quickly transmit heat. Brass (\( k_{Br} = 109 \) W/m·K) and steel (\( k_{St} = 50 \) W/m·K) are less conductive, causing slower heat flow through these materials.
Understanding these properties is essential in selecting materials for heat management applications. Materials with higher thermal conductivity are preferred in applications that require rapid heat dissipation, like heat sinks. Exploring these differences in this exercise, students can grasp why one metal might be favored over another based on their thermal properties and how these choices impact systems' efficiency and effectiveness.