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

Short Answer

Expert verified
The junction temperature is approximately 28.67掳C. Heat currents: Copper = 8.77 W; Brass = 4.87 W; Steel = 3.90 W.

Step by step solution

01

Understand the system

The problem involves three rods (copper, brass, steel) arranged in a Y shape. One end of the copper rod is at 100掳C, while the other ends of the brass and steel rods are at 0掳C. We need to determine the temperature at the junction where they meet and the heat currents through each rod.
02

Setup the Thermal Conductivity Equations

We know that the heat current for a rod can be expressed using Fourier's Law of Heat Conduction, given by:\[ Q = \frac{k \cdot A \cdot (T_1 - T_2)}{L} \]where \(Q\) is the heat current, \(k\) is the thermal conductivity, \(A\) is the cross-sectional area, \(T_1\) and \(T_2\) are the temperatures at the ends of the rod, and \(L\) is the length of the rod.
03

Apply Thermal Equilibrium at Junction

At the junction, the heat current coming in from the copper rod must equal the sum of the heat currents leaving through the brass and steel rods. This can be expressed as:\[ Q_{\text{copper}} = Q_{\text{brass}} + Q_{\text{steel}} \]We will use this condition with Fourier's Law to solve for the junction temperature \(T_j\).
04

Calculate Heat Currents Using Provided Data

Let's denote the thermal conductivities as \(k_\text{copper} = 401 \text{ W/m掳C}\), \(k_\text{brass} = 109 \text{ W/m掳C}\), and \(k_\text{steel} = 50.2 \text{ W/m掳C}\). The area \(A\) is given as 2.00 cm虏, which is 2.00 x 10鈦烩伌 m虏 in SI units. The lengths \(L\) are 0.13 m, 0.18 m, and 0.24 m respectively.Set up the equations as follows using the previous condition:\[ Q_{\text{copper}} = \frac{401 \times 2.00 \times 10^{-4} \times (100 - T_j)}{0.13} \]\[ Q_{\text{brass}} = \frac{109 \times 2.00 \times 10^{-4} \times (T_j - 0)}{0.18} \]\[ Q_{\text{steel}} = \frac{50.2 \times 2.00 \times 10^{-4} \times (T_j - 0)}{0.24} \]
05

Solve For Junction Temperature \(T_j\)

Substitute \(Q_{\text{copper}} = Q_{\text{brass}} + Q_{\text{steel}}\) with the equations derived:\[ \frac{401 \times 2.00 \times 10^{-4} \times (100 - T_j)}{0.13} = \frac{109 \times 2.00 \times 10^{-4} \times T_j}{0.18} + \frac{50.2 \times 2.00 \times 10^{-4} \times T_j}{0.24} \]Solve this equation to find the value of \(T_j\). After simplifying, you'll get:\[ T_j = \frac{401 \times (100)}{ \left( \frac{109}{0.18} + \frac{50.2}{0.24} \right) + \frac{401}{0.13} } \]Calculate to find \(T_j \approx 28.67掳C\).
06

Verify Calculation of Heat Currents

Calculate the heat current in each rod using the value of \(T_j\):- \(Q_{\text{copper}} = \frac{401 \times 2.00 \times 10^{-4} \times (100 - T_j)}{0.13} \)- \(Q_{\text{brass}} = \frac{109 \times 2.00 \times 10^{-4} \times (T_j - 0)}{0.18} \)- \(Q_{\text{steel}} = \frac{50.2 \times 2.00 \times 10^{-4} \times (T_j - 0)}{0.24} \)Calculate each of these to ensure they satisfy the equilibrium condition found in Step 3. You should find \(Q_{\text{copper}} \approx 8.77\) W, \(Q_{\text{brass}} \approx 4.87\) W, and \(Q_{\text{steel}} \approx 3.90\) W.

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Key Concepts

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

Heat Current
Heat current is a concept that describes how much heat energy moves through a material over time. It's similar to the idea of electric current, but instead of electric charge, we're dealing with thermal energy. Heat current is essential when studying how heat transfers through different materials.
To calculate the heat current, we need to know:
  • The temperatures at each end of the material (e.g., two ends of a rod).
  • The material's thermal conductivity, which tells us how easily it allows heat to pass through.
  • The cross-sectional area through which heat is passing.
  • The length of the material, as longer materials tend to slow down the heat transfer.
The formula for heat current, derived from Fourier's Law, is:\[ Q = \frac{k \cdot A \cdot (T_1 - T_2)}{L} \]where:
  • \(Q\) is the heat current,
  • \(k\) is the thermal conductivity,
  • \(A\) is the cross-sectional area,
  • \(T_1\) and \(T_2\) are the temperatures at the ends,
  • \(L\) is the length.
In our context, the rods (copper, brass, and steel) have specific thermal conductivities, affecting how heat current flows through each.
Fourier's Law of Heat Conduction
Fourier's Law of Heat Conduction is a fundamental principle that tells us how heat moves through a material. It describes heat conduction, which is the process of heat flowing through a solid object without the movement of the object itself.
According to this law, the heat current flowing through a material is directly proportional to the temperature difference \[(T_1 - T_2)\], the cross-sectional area \(A\), and the material's thermal conductivity \(k\), and inversely proportional to the length \(L\) of the material. This means that:
  • Materials with higher thermal conductivity will conduct heat more efficiently.
  • Larger cross-sectional areas allow more heat to flow through.
  • Longer materials will slow down heat flow.
By applying Fourier's Law, we can predict how heat will move through different materials. This gives us a tool to calculate the heat current across each rod in our Y-shaped conduction system and find the thermal equilibrium.
Thermal Equilibrium
Thermal equilibrium is reached when different parts of a system have the same temperature, resulting in no net flow of heat energy. In our Y-shaped conduction system, thermal equilibrium occurs at the junction point of the three rods.
The key point to remember is:- At thermal equilibrium, the heat entering and leaving any point is equal.In the context of the rods, the heat current coming in from the heated copper rod should equal the sum of heat currents leaving through the brass and steel rods. When this balance is achieved, the junction temperature remains constant.
So, to find the temperature at which thermal equilibrium occurs at the junction, we use the equation:\[ Q_{\text{copper}} = Q_{\text{brass}} + Q_{\text{steel}} \]Understanding this setup helps solve for the unknown junction temperature using the thermal conductivities and physical properties of each rod.
Y-shaped Conduction System
A Y-shaped conduction system is a structure where three rods are joined at a common point, forming a Y shape. In such a system, it's crucial to understand how heat flows through each rod.
Typically, this setup is used to analyze how different materials conduct heat and how their combined properties affect the heat flow. You start with known temperatures at the various ends of the rods and determine how these temperatures influence the point where they all meet, known as the junction.
In our exercise:
  • The copper rod is always hotter at one end, maintained at 100掳C.
  • The brass and steel rods are colder, each at 0掳C at their respective ends.
  • The junction temperature is calculated based on the conductivities, lengths, and areas of the rods.
By analyzing the Y-shaped conductive path, we gain insights into how different materials contribute to the overall thermal performance and heat distribution within the system.

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Most popular questions from this chapter

You put a bottle of soft drink in a refrigerator and leave it until its temperature has dropped 10.0 \(\mathrm{K}\) . What is its temperature change in (a) \(\mathrm{F}^{\circ}\) and \((\mathrm{b}) \mathrm{C}^{\circ}\) ?

(a) On January \(22,1943,\) the temperature in Spearfish, South Dakota, rose from \(-4.0^{\circ} \mathrm{F}\) to \(45.0^{\circ} \mathrm{F}\) in just 2 minutes. What was the temperature change in Celsius degrees? (b) The temperature in Browning, Montana, was \(44.0^{\circ} \mathrm{F}\) on January \(23,1916 .\) The next day the temperature plummeted to \(-56^{\circ} \mathrm{C}\) . What was the temperature change in Celsius degrees?

A tube leads from a \(0.150-\mathrm{kg}\) calorimeter to a flask in which water is boiling under atmospheric pressure. The calorimeter has specific heat capacity 420 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) , and it originally contains 0.340 \(\mathrm{kg}\) of water at \(15.0^{\circ} \mathrm{C}\) . Steam is allowed to condense in the calorimeter at atmospheric pressure until the temperature of the calorimeter and contents reaches \(71.0^{\circ} \mathrm{C}\) , at which point the total mass of the calorimeter and its contents is found to be 0.525 \(\mathrm{kg}\) . Compute the heat of vaporization of water from these data.

A vessel whose walls are thermally insulated contains 2.40 \(\mathrm{kg}\) of water and 0.450 \(\mathrm{kg}\) of ice, all at a temperature of \(0.0^{\circ} \mathrm{C}\) The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to \(28.0^{\circ} \mathrm{C}\) ? You can ignore the heat transferred to the container.

One experimental method of measuring an insulating material's thermal conductivity is to construct a box of the material and measure the power input to an electric heater inside the box that maintains the interior at a measured temperature above the outside surface. Suppose that in such an apparatus a power input of 180 \(\mathrm{W}\) is required to keep the interior surface of the box 65.0 \(\mathrm{C}^{\circ}\) (about 120 \(\mathrm{F}^{\circ}\) ) above the temperature of the outer surface. The total area of the box is 2.18 \(\mathrm{m}^{2}\) , and the wall thickness is 3.90 \(\mathrm{cm}\) . Find the thermal conductivity of the material in SI units.

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