Chapter 17: Problem 102
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?
Short Answer
Step by step solution
Understand the Thermal Conductivity Formula
Identify the Given Values
Calculate Heat Current through Copper Rod
Calculate Heat Current through Brass and Steel Rods
Establish Equilibrium at Junction Point
Solve for Junction Temperature
Calculate Heat Currents in Individual Rods
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Heat Transfer
For conduction, the heat current is calculated with the formula:
- \( Q = \frac{kA(T_H - T_C)}{L} \)
- \( k \) 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
Y-shaped thermal system
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
The balance equation can be written as:
- \[ Q_{Cu} = Q_{Br} + Q_{St} \]
Thermal Conductivities of Metals
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.