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Chapter 19: Problem 58

(I) One end of a 45 -cm-long copper rod with a diameter of 2.0 \(\mathrm{cm}\) is kept at \(460^{\circ} \mathrm{C},\) and the other is immersed in water at \(22^{\circ} \mathrm{C}\) Calculate the heat conduction rate along the rod.

Short Answer

Expert verified
The heat conduction rate along the rod is approximately 118.2 W.

Step by step solution

01

Understand the Problem

We need to calculate the rate of heat conduction along the rod, given the temperature at two ends, the length of the rod, its diameter, and its material.
02

Formula for Heat Conduction Rate

The formula for the rate of heat conduction is given by Fourier's Law of Heat Conduction: \[ Q = \frac{k A \Delta T}{L} \] where \( Q \) is the rate of heat conduction, \( k \) is the thermal conductivity of the material (copper in this case), \( A \) is the cross-sectional area, \( \Delta T \) is the temperature difference, and \( L \) is the length of the rod.
03

Determine Necessary Values

From tables, the thermal conductivity of copper \( k \) is approximately \( 385 \, \text{W/m掳C} \). The temperature difference \( \Delta T \) is \(460掳C - 22掳C = 438掳C \). The length \( L \) of the rod is \(0.45 \text{ m} \) since 45 cm = 0.45 m.
04

Calculate Cross-Sectional Area

The cross-sectional area \( A \) can be calculated using the formula for the area of a circle: \( A = \pi r^2 \). The radius \( r \) is half of the diameter, so \( r = \frac{2.0 \text{ cm}}{2} = 1.0 \text{ cm} = 0.01 \text{ m} \). Thus, \[ A = \pi (0.01)^2 = 3.14 \times 10^{-4} \text{ m}^2 \].
05

Substitute Values into the Formula

Substitute the known values into the heat conduction formula: \[ Q = \frac{385 \times 3.14 \times 10^{-4} \times 438}{0.45} \].
06

Calculate the Heat Conduction Rate

Perform the calculations: \[ Q = \frac{385 \times 3.14 \times 10^{-4} \times 438}{0.45} \approx 118.2 \text{ W} \].

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a property that measures a material's ability to conduct heat. Materials with high thermal conductivity are good conductors of heat, like metals, while materials with low thermal conductivity, such as wood or air, are good insulators.

- **Definition**: It is defined as the quantity of heat transferred per unit time through a unit area of the material, per unit temperature gradient. - **Units**: The standard unit is watts per meter per degree Celsius (W/m掳C). - **Example**: In this problem, copper is the material with a thermal conductivity of approximately 385 W/m掳C. This high value indicates that copper is an excellent heat conductor.
Thermal conductivity plays a critical role in determining how quickly heat can transfer through a material. Understanding this property is essential when selecting materials for thermal management.
Fourier's Law
Fourier's Law is a fundamental principle that describes how heat is conducted through materials. It forms the basis for calculating the rate of heat transfer in scenarios like our copper rod example.

- **Formula**: The mathematical expression is

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Most popular questions from this chapter

[The Problems in this Section are ranked I, II, or III according to estimated difficulty, with (I) Problems being easiest. Level (III) Problems are meant mainly as a challenge for the best students, for "extra credit." The Problems are arranged by Sections, meaning that the reader should have read up to and including that Section, but this Chapter also has a group of General Problems that are not arranged by Section and not ranked. \(\begin{array}{l}{\text { (I) To what temperature will } 8700 \mathrm{J} \text { of heat raise } 3.0 \mathrm{kg} \text { of }} \\ {\text { water that is initially at } 10.0^{\circ} \mathrm{C} \text { ? }}\end{array}\)

(II) One and one-half moles of an ideal monatomic gas expand adiabatically, performing \(7500 \mathrm{~J}\) of work in the process. What is the change in temperature of the gas during this expansion?

(II) When a diver jumps into the ocean, water leaks into the gap region between the diver's skin and her wetsuit, forming a water layer about \(0.5 \mathrm{~mm}\) thick. Assuming the total surface area of the wetsuit covering the diver is about \(1.0 \mathrm{~m}^{2},\) and that ocean water enters the suit at \(10^{\circ} \mathrm{C}\) and is warmed by the diver to skin temperature of \(35^{\circ} \mathrm{C},\) estimate how much energy (in units of candy bars \(=300 \mathrm{kcal}\) ) is required by this heating process.

(II) The heat capacity, \(C,\) of an object is defined as the amount of heat needed to raise its temperature by 1 \(\mathrm{C}^{\circ} .\) Thus, to raise the temperature by \(\Delta T\) requires heat \(Q\) given by $$Q=C \Delta T$$ (a) Write the heat capacity \(C\) in terms of the specific heat, \(c,\) of the material. \((b)\) What is the heat capacity of 1.0 \(\mathrm{kg}\) of water? \((c)\) Of 35 \(\mathrm{kg}\) of water?

(III) A house thermostat is normally set to \(22^{\circ} \mathrm{C}\), but at night it is turned down to \(12^{\circ} \mathrm{C}\) for \(9.0 \mathrm{~h} .\) Estimate how much more heat would be needed (state as a percentage of daily usage) if the thermostat were not turned down at night. Assume that the outside temperature averages \(0^{\circ} \mathrm{C}\) for the \(9.0 \mathrm{~h}\) at night and \(8^{\circ} \mathrm{C}\) for the remainder of the day, and that the heat loss from the house is proportional to the difference in temperature inside and out. To obtain an estimate from the data, you will have to make other simplifying assumptions; state what these are.
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