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Chapter 11: Problem 41

A steam pipe is covered with \(1.50-\mathrm{cm}\)-thick insulating material of thermal conductivity \(0.200 \mathrm{cal} / \mathrm{cm} \cdot{ }^{\circ} \mathrm{C} \cdot \mathrm{s}\). How much energy is lost every second when the steam is at \(200^{\circ} \mathrm{C}\) and the surrounding air is at \(20.0^{\circ} \mathrm{C}\) ? The pipe has a circumference of \(800 \mathrm{~cm}\) and a length of \(50.0 \mathrm{~m}\). Neglect losses through the ends of the pipe.

Short Answer

Expert verified
The energy lost every second when the steam is at \(200^{\circ}C\) and the surrounding air is at \(20.0^{\circ}C\) is about \(9.6 \times 10^7 \,\mathrm{cal/s}\).

Step by step solution

01

Identify the given information

The question provides the following information: \n\n\(k = 0.200 \, \mathrm{cal/cm} \cdot \, ^{\circ} \mathrm{C} \cdot\, s\) (thermal conductivity), \n\n\(\Delta T = 200^{\circ}C - 20.0^{\circ}C =180^{\circ}C\) (temperature difference), \n\n\(d = 1.5 \,cm\) (thickness of insulating material), \n\nCircumference = 800 cm, \n\nLength = 50.0m = 5000cm (converted metres to centimetres because other measurements are in cm)
02

Calculate the surface area

The surface area (A) of a cylinder, neglecting the ends, is calculated by the formula \(A = 2 \pi r h\), where r is the radius and h is the height. In this case, since the circumference is provided, we can calculate r as \(r = \frac{Circumference} {2\pi} \) . \n\nSo, substituting given values we get, \n\n \(r = \frac{800}{2\pi} \approx 127.32\, \mathrm{cm}\) \n\nSubstitute r and h (height = length of pipe) into area formula: \n\n\(A = 2 *\pi * 127.32 * 5000 \approx 4 \times 10^6 \,\mathrm{cm}^2\)
03

Substitute the values into the formula

The formula to calculate the heat energy lost per second for the given condition is \(Q = \frac{k * A * \Delta T}{d}\). \n\nSubstitute the values into this formula, we get: \n\n\(Q = \frac{0.200 * 4 \times 10^6 * 180}{1.5} \approx 9.6 \times 10^7 \,\mathrm{cal/s}\)
04

Interpret the results

The result indicates the amount of energy lost every second due to heat transfer from the steam pipe to the surrounding air through the insulation. This heat loss is quantified as about \(9.6 \times 10^7 \,\mathrm{cal/s}\), assuming no heat loss from the ends of the pipe

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Key Concepts

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

Understanding Heat Transfer
Heat transfer is the movement of thermal energy from one object or medium to another. This process is driven by a temperature difference between the two, where energy flows from the hotter body to the cooler one until thermal equilibrium is reached. In the context of our exercise, heat transfer occurs from the steam within the pipe, which is at a high temperature of 200鈩�, to the cooler surrounding air at 20鈩�.

There are three main modes of heat transfer: conduction, convection, and radiation. The exercise primarily deals with conduction, which is the heat transfer through a material without any motion of the material itself. It is described by Fourier's law, and the rate of heat transfer by conduction is proportional to the thermal conductivity of the material and the temperature difference across it. The thicker and less conductive the insulation around the steam pipe, the slower the rate of heat loss.
Importance of Thermal Insulation
Thermal insulation is a material or process which significantly reduces the transfer of heat. In the exercise, a 1.5 cm-thick layer of insulating material surrounds the steam pipe to limit energy loss. The effectiveness of insulation is measured by its thermal conductivity, denoted as 'k', which indicates how well a material can conduct heat.

Materials with low thermal conductivity are excellent insulators because they hinder the flow of thermal energy. Examples include fiberglass, wool, and foam. These materials are commonly used in building construction, clothing, and industrial processes to maintain temperatures and improve energy efficiency. The exercise demonstrates how insulation material with a given 'k' value impacts the rate of energy loss through the pipe wall, underlining the role of insulation in controlling heat transfer and conserving energy.
Quantifying Energy Loss
Energy loss in the context of heat transfer refers to the amount of thermal energy that escapes from a warmer area to a cooler one over time. Our exercise calculates the energy loss through the insulated steam pipe using a formula derived from the principles of thermal conductivity.

The formula incorporates various factors including thermal conductivity (k), the surface area (A) through which heat is transferred, the temperature difference (鈭員), and the thickness of the insulating material (d). The result, expressed in calories per second (cal/s), quantifies the rate of energy loss. This calculation is pivotal for engineers and designers who aim to enhance energy efficiency by choosing appropriate insulation materials and thicknesses that minimize unwanted heat loss, reducing energy consumption and costs.

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Most popular questions from this chapter

When you jog, most of the food energy you burn above your basal metabolic rate (BMR) ends up as internal energy that would raise your body temperature if it were not eliminated. The evaporation of perspiration is the primary mechanism for eliminating this energy. Determine the amount of water you lose to evaporation when running for 30 minutes at a rate that uses \(400 \mathrm{kcal} / \mathrm{h}\) above your BMR. (That amount is often considered to be the "maximum fat-burning" energy output.) The metabolism of 1 gram of fat gen- erates approximately \(9.0 \mathrm{kcal}\) of energy and produces approximately 1 gram of water. (The hydrogen atoms in the fat molecule are transferred to oxygen to form water.) What fraction of your need for water will be provided by fat metabolism? (The latent heat of vaporization of water at room temperature is \(2.5 \times 10^{6} \mathrm{~J} / \mathrm{kg}\).)

The apparatus shown in Figure P11.10 was used by Joule to measure the mechanical equivalent of heat. Work is done on the water by a rotating paddle wheel, which is driven by two blocks falling at a constant speed. The temperature of the stirred water increases due to the friction between the water and the paddles. If the energy lost in the bearings and through the walls is neglected, then the loss in potential energy associated with the blocks equals the work done by the paddle wheel on the water. If each block has a mass of \(1.50 \mathrm{~kg}\) and the insulated tank is filled with \(200 \mathrm{~g}\) of water, what is the increase in temperature of the water after the blocks fall through a distance of \(3.00 \mathrm{~m}\) ?

An aluminum calorimeter with a mass of \(100 \mathrm{~g}\) contains \(250 \mathrm{~g}\) of water. The calorimeter and water are in thermal equilibrium at \(10.0^{\circ} \mathrm{C}\). Two metallic blocks are placed into the water. One is a \(50.0-\mathrm{g}\) piece of copper at \(80.0^{\circ} \mathrm{C}\). The other has a mass of \(70.0 \mathrm{~g}\) and is originally at a temperature of \(100^{\circ} \mathrm{C}\). The entire system stabilizes at a final temperature of \(20.0^{\circ} \mathrm{C}\). (a) Determine the spe- cific heat of the unknown sample. (b) Using the data in Tific heat of the unknown sample. (b) Using the data in unknown material? Can you identify a possible material? (c) Explain your answers for part (b).

Measurements on two stars indicate that Star X has a surface temperature of \(5727^{\circ} \mathrm{C}\) and Star Y has a surface temperature of \(11727^{\circ} \mathrm{C}\). If both stars have the same radius, what is the ratio of the luminosity (total power output) of Star Y to the luminosity of Star X? Both stars can be considered to have an emissivity of \(1.0\).

The highest recorded waterfall in the world is found at Angel Falls in Venezuela. Its longest single waterfall has a height of \(807 \mathrm{~m}\). If water at the top of the falls is at \(15.0^{\circ} \mathrm{C}\), what is the maximum temperature of the water at the bottom of the falls? Assume all the kinetic energy of the water as it reaches the bottom goes into raising the water's temperature.
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