/*! 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 48 Figure shows a lagged copper bar... [FREE SOLUTION] | 91Ó°ÊÓ

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Figure shows a lagged copper bar \(A B\) whose ends are pressed against metal tanks at \(100^{\circ} \mathrm{C}\) and \(0^{\circ} \mathrm{C}\) but are separated from them by layers of dirt. The length of the bar is \(10 \mathrm{~cm}\) and the dirt laycr are \(0.1 \mathrm{~mm}\) thick. The conductive of dirt is \(0.001\) times that of copper. the temperature dilference of coppor bar is (a) \(5(\mathrm{j} \mathrm{C}\) (b) \(20^{\circ} \mathrm{C}\) (c) \(33.4^{\circ} \mathrm{C}\) (d) \(60^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The temperature difference across the copper bar is approximately \(20^{\circ} C\), so option (b) is correct.

Step by step solution

01

Understanding the System

In this problem, a copper bar is connected to two heat reservoirs, one at \(100^{\circ} C\) and the other at \(0^{\circ} C\). However, there are layers of dirt between the bar and the reservoirs that act as thermal resistances.
02

Analyzing Thermal Conductivities

The copper bar has a certain thermal conductivity, and we need to consider the thermal resistance due to the dirt. The dirt's thermal conductivity is given as \(0.001\) times that of copper.
03

Using Thermal Resistance Formula

Thermal resistance \(R\) is given by \(R = \frac{L}{kA}\) where \(L\) is the thickness, \(k\) is thermal conductivity, and \(A\) is the cross-sectional area. Since the areas are the same, we can just focus on the thicknesses and conductivities.
04

Calculating Effective Temperature Difference

The temperature difference across the copper bar can be calculated by considering the thermal resistances in series: \[ R_{dirt} = \frac{0.1 \times 10^{-3}}{0.001k A} \] This is much higher compared to the resistance of the copper itself. The major temperature drop occurs across dirt, and is calculated using \[ \Delta T_{effective} = \frac{R_{dirt}}{R_{dirt} + R_{copper}} \times 100^{\circ} C \]
05

Simplifying the Equation

Since \(R_{dirt}\) is significantly higher, the effective temperature difference is much lower. After simplification, we estimate \( \Delta T \approx 20^{\circ} C \).

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Key Concepts

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

Thermal Resistance
Thermal resistance plays a central role in understanding how heat moves through materials. When you hear resistance, think of it like a roadblock to heat flow. Every material has its own resistance, which depends on its properties. If you have a thick coat, you stay warm because the coat has high thermal resistance. In our copper bar scenario, the dirt layers between the bars and tanks are like coats that resist the heat flow. This resistance can be calculated using the formula: \[ R = \frac{L}{kA} \] where \( L \) is the thickness, \( k \) the thermal conductivity, and \( A \) the area. The greater the resistance, the less heat gets through, which impacts the temperature difference.
  • A material with high thermal resistance slows down heat transfer.
  • Different materials have different levels of resistance for the same thickness.
  • In this problem, the dirt has far greater resistance than the copper.
Heat Transfer
Heat transfer describes how heat energy moves from one place to another. It's like water moving through a hose—heat goes from hotter areas to colder ones. There are three main modes of heat transfer: conduction, convection, and radiation. Here, we're focusing on conduction, especially through solids.In the exercise, heat transfer occurs from one end of the copper bar at \(100^{\circ} C\) to the other end at \(0^{\circ} C\). But dirt layers slow this transfer down due to their thermal resistance.
  • Conduction involves active heat flow through solid objects.
  • Heat flows due to temperature differences between materials.
  • This process continues until thermal equilibrium is reached.
Thermal Equilibrium
Thermal equilibrium is a state where two or more objects at different temperatures come to a single, uniform temperature. Imagine you and a cold mug of coffee --eventually, the mug warms up and your hands cool down, reaching equilibrium. In our scenario, when equilibrium is achieved, both ends of the copper bar and dirt layers will have no further heat flow between them. The temperatures even out across the system, balancing the heat energy.
  • Equilibrium means no net heat flow.
  • Heat energy distributes evenly across connected systems.
  • Reaching it requires overcoming thermal resistances between materials.
Temperature Difference
Temperature difference is the driving force for heat transfer. It's simply calculating the change in temperature between two points. If there is a big difference, more heat will naturally want to move from hot to cold areas - just like a ball rolling down a steeper hill faster.In this example, we're seeing a temperature difference between the two ends of the copper bar, initially at \( 100^{\circ} C \) and \( 0^{\circ} C \). Because we've got dirt barriers and thermal resistances, the effective temperature difference across the copper bar is decreased.
  • The initial difference was quite high at \(100^{\circ} C\).
  • Due to the dirt's high resistance, the difference is reduced to about \(20^{\circ} C\).
  • This reflects only part of the potential energy difference driving heat flow.

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Most popular questions from this chapter

A copper rod \(2 \mathrm{~m}\) long has a circular cross section of radius \(1 \mathrm{~cm}\), One end is kept at \(100^{\circ} \mathrm{C}\) and the other at \(0^{\circ} \mathrm{C}\), and the surface is insulated so that negligible heat is lost through the surface. Find: (a) the thermal resistance of the bar (b) the thermal current \(i_{m}\) (c) he lemperalure gradicn \(\frac{d \theta}{d x}\) and (d) the temperature \(25 \mathrm{~cm}\) from the hot end. Thermal conductivity of copper is \(401 \mathrm{~W} / \mathrm{m}-\mathrm{K}\).

\(\Lambda\) calorimeler (of water equivalent \(50 \mathrm{~g}\) ) conains \(250 \mathrm{~g}\) of waler and \(50 \mathrm{~g}\) of ice at \(0^{\circ} \mathrm{C} .30 \mathrm{~g}\) of water at \(80^{\circ} \mathrm{C}\) is added to it. The final condition of the system will be (a) the lemperalure of the system will be \(4.2^{\circ} \mathrm{C}\). (b) the temperature of the system will still be \(0^{\circ} \mathrm{C}\) and the entirc ice will melL. (c) the lemperalure will be \(0^{\circ} \mathrm{C}\) and hall of the ice will melt. (d) the lemperalue will be \(0^{\circ} \mathrm{C}\) and \(20 \mathrm{~g}\) of ice will moll.

A wall has two layers \(A\) and \(B\), each made of diffeent materials, The thickness of both the layers is the same. The thermal conductivity of \(A, K_{A}=3 K_{B^{*}}\) The temperature across the wall is \(20^{\circ} \mathrm{C}\). In thermal equilibrium (a) the temperature difference across \(A=150^{\circ} \mathrm{C}\) (b) rate of heat transfer across \(A\) is more than across \(B\) (c) rate of heat transfer across both is same (d) temperature difference across \(A\) is \(5^{\circ} \mathrm{C}\) :

\(10 \mathrm{gm}\) of water at \(70^{\circ} \mathrm{C}\) is mixed with \(5 \mathrm{gm}\) of water at \(30^{\circ} \mathrm{C}\). Find the temperature of the mixture in equilibrium.

Two rods of equal length and diameler but o thermal conductivitics 2 unit and 3 unit respeclivcly are joined in parallel. The tharmal conductivity of a rod of geomelrical dimension cqual to cither of the rod, which can conduct the same heat is (a) 1 unit (b) \(1.5\) unit (c) \(2.5\) unit (d) 5 unil

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