/*! 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 83 Two large blocks of different ma... [FREE SOLUTION] | 91Ó°ÊÓ

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Two large blocks of different materials, such as copper and concrete, have been sitting in a room \(\left(23^{\circ} \mathrm{C}\right)\) for a very long time. Which of the two blocks, if either, will feel colder to the touch? Assume the blocks to be semi-infinite solids and your hand to be at a temperature of \(37^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The copper block will feel colder to the touch than the concrete block because it has a higher thermal conductivity (\(k_c = 398 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}}\)) compared to concrete (\(k_{cc} = 1.4 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}}\)). This means that copper transfers heat from the hand more quickly, making it feel colder.

Step by step solution

01

Identify the thermal conductivity values

We need to look up the thermal conductivity values of both copper and concrete. The values are as follows: - Copper: \(k_c = 398 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}}\) - Concrete: \(k_{cc} = 1.4 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}}\)
02

Compare the thermal conductivity values

As mentioned earlier, the material with the higher thermal conductivity will transfer heat from one's hand more quickly, making it feel colder. We can now compare the values of k for both materials: \(k_c > k_{cc}\) This means that copper has a greater thermal conductivity than concrete.
03

Determine the block that feels colder

Since copper has a higher thermal conductivity than concrete, it transfers heat from the hand more quickly. Therefore, the copper block will feel colder to the touch than the concrete block.

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Key Concepts

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

Understanding Heat Transfer
The sensation of cold or warmth when we touch an object is a direct result of heat transfer, a process central to thermodynamics and thermal engineering. Heat naturally flows from a hotter object to a cooler one until both reach thermal equilibrium. The rate of heat transfer can vary significantly depending on several factors, including the materials involved and their respective thermal properties.

When you place your hand on a solid surface, heat transfer occurs as the warmer hand attempts to give off heat to the cooler object. The feeling of coldness doesn't arise from the object's temperature alone but also how quickly it can draw heat away from your hand. Materials with higher thermal conductivity are more efficient at transferring heat, thereby giving a cooler sensation more rapidly.

In educational contexts, understanding heat transfer often involves solving problems and conceptualizing how different materials interact thermally. This can include calculations using thermal conductivity values to predict and explain phenomena, such as why certain materials feel colder to the touch than others even at the same temperature.
The Role of Semi-Infinite Solids in Heat Transfer
Our daily experiences may not include the term 'semi-infinite solids', but this concept plays a pivotal role in simplifying thermal problems. A semi-infinite solid is an idealized material that extends infinitely in one or more dimensions. In practice, this approximation can apply to solids where heat transfer occurs over a distance that is very small compared to the physical size of the solid.

Considering the blocks of copper and concrete in the exercise, assuming they are semi-infinite solids allows the analysis to ignore any complexities arising from the edges or thicknesses of the materials. This assumption is especially useful when we are interested in surface interactions, like touching the block with our hand. Heat transfer at the surface can be treated independently of the rest of the (effectively infinite) material, making calculations and conceptual understanding more straightforward.

Moreover, semi-infinite solids are used in various applications beyond feeling temperatures, such as cooling of large machinery parts or in the study of geothermal energy extraction where the Earth is treated as a semi-infinite solid.
Thermal Conductivity Values and Their Significance
Thermal conductivity, symbolized by the letter 'k', is a measure of a material's ability to conduct heat. Higher values of thermal conductivity indicate that a material can transfer heat more effectively, which is why metals like copper feel much colder than materials such as wood or plastic under the same conditions.

The thermal conductivity values of copper and concrete dramatically differ, as seen in the exercise. Copper's commanding value of around 398 W/m·K, compared to concrete's modest 1.4 W/m·K, explains the pronounced sensation of coldness when touching copper. This is because the high thermal conductivity of copper allows it to rapidly draw heat from your hand, leading to a faster decrease in the temperature at the point of contact.

Recognizing the significance of these values not only aids in solving textbook problems but also enhances our understanding of real-world applications. For instance, knowing the thermal conductivity of materials can guide decisions in construction for thermal insulation or in electronics for dissipating heat from devices. The role of these values in designing efficient systems makes it a critical concept in engineering and physics education.

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Most popular questions from this chapter

Consider the fuel element of Example \(5.11\), which operates at a uniform volumetric generation rate of \(\dot{q}_{1}=10^{7} \mathrm{~W} / \mathrm{m}^{3}\) until the generation rate suddenly changes to \(\dot{q}_{2}=2 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\). Use the finite-element software \(F E H T\) to obtain the following solutions. (a) Calculate the temperature distribution \(1.5 \mathrm{~s}\) after the change in operating power and compare your results with those tabulated in the example. Hint: First determine the steady-state temperature distribution for \(\dot{q}_{1}\), which represents the initial condition for the transient temperature distribution after the step change in power to \(\dot{q}_{2}\). Next, in the Setup menu, click on Transient: in the Specify/Internal Generation box, change the value to \(\dot{q}_{2}\); and in the Run command, click on Continue (not Calculate). See the Run menu in the FEHT Help section for background information on the Continue option. (b) Use your \(F E H T\) model to plot temperature histories at the midplane and surface for \(0 \leq t \leq 400 \mathrm{~s}\). What are the steady-state temperatures, and approximately how long does it take to reach the new equilibrium condition after the step change in operating power?

A constant-property, one-dimensional plane wall of width \(2 L\), at an initial uniform temperature \(T_{i}\), is heated convectively (both surfaces) with an ambient fluid at \(T_{\infty}=T_{\infty, 1}, h=h_{1}\). At a later instant in time, \(t=t_{1}\), heating is curtailed, and convective cooling is initiated. Cooling conditions are characterized by \(T_{\infty}=T_{\infty, 2}=T_{i}, h=h_{2}\) (a) Write the heat equation as well as the initial and boundary conditions in their dimensionless form for the heating phase (Phase 1). Express the equations in terms of the dimensionless quantities \(\theta^{*}, x^{*}, B i_{1}\), and \(F o\), where \(B i_{1}\) is expressed in terms of \(h_{1}\). (b) Write the heat equation as well as the initial and boundary conditions in their dimensionless form for the cooling phase (Phase 2). Express the equations in terms of the dimensionless quantities \(\theta^{*}, x^{*}, B i_{2}\), \(\mathrm{Fo}_{1}\), and \(\mathrm{Fo}\) where \(\mathrm{Fo}_{1}\) is the dimensionless time associated with \(t_{1}\), and \(B i_{2}\) is expressed in terms of \(h_{2}\). To be consistent with part (a), express the dimensionless temperature in terms of \(T_{\infty}=T_{\infty, 1^{*}}\) (c) Consider a case for which \(B i_{1}=10, B i_{2}=1\), and \(F o_{1}=0.1\). Using a finite-difference method with \(\Delta x^{*}=0.1\) and \(\Delta F o=0.001\), determine the transient thermal response of the surface \(\left(x^{*}=1\right)\), midplane \(\left(x^{*}=0\right)\), and quarter-plane \(\left(x^{*}=0.5\right)\) of the slab. Plot these three dimensionless temperatures as a function of dimensionless time over the range \(0 \leq F o \leq 0.5\). (d) Determine the minimum dimensionless temperature at the midplane of the wall, and the dimensionless time at which this minimum temperature is achieved.

A steel strip of thickness \(\delta=12 \mathrm{~mm}\) is annealed by passing it through a large furnace whose walls are maintained at a temperature \(T_{w}\) corresponding to that of combustion gases flowing through the furnace \(\left(T_{w}=T_{\infty}\right)\). The strip, whose density, specific heat, thermal conductivity, and emissivity are \(\rho=7900 \mathrm{~kg} / \mathrm{m}^{3}\), \(c_{p}=640 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\varepsilon=0.7\), respectively, is to be heated from \(300^{\circ} \mathrm{C}\) to \(600^{\circ} \mathrm{C}\). (a) For a uniform convection coefficient of \(h=\) \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{w}=T_{\infty}=700^{\circ} \mathrm{C}\), determine the time required to heat the strip. If the strip is moving at \(0.5 \mathrm{~m} / \mathrm{s}\), how long must the furnace be? (b) The annealing process may be accelerated (the strip speed increased) by increasing the environmental temperatures. For the furnace length obtained in part (a), determine the strip speed for \(T_{w}=T_{\infty}=\) \(850^{\circ} \mathrm{C}\) and \(T_{w}=T_{\infty}=1000^{\circ} \mathrm{C}\). For each set of environmental temperatures \(\left(700,850\right.\), and \(\left.1000^{\circ} \mathrm{C}\right)\), plot the strip temperature as a function of time over the range \(25^{\circ} \mathrm{C} \leq T \leq 600^{\circ} \mathrm{C}\). Over this range, also plot the radiation heat transfer coefficient, \(h_{r}\), as a function of time.

A horizontal structure consists of an \(L_{\mathrm{A}}=10\)-mm-thick layer of copper and an \(L_{\mathrm{B}}=10\)-mm-thick layer of aluminum. The bottom surface of the composite structure receives a heat flux of \(q^{\prime \prime}=100 \mathrm{~kW} / \mathrm{m}^{2}\), while the top surface is exposed to convective conditions characterized by \(h=40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, T_{\infty}=25^{\circ} \mathrm{C}\). The initial temperature of both materials is \(T_{i, \mathrm{~A}}=T_{i, \mathrm{~B}}=25^{\circ} \mathrm{C}\), and a contact resistance of \(R_{t, c}^{\prime \prime}=400 \times 10^{-6} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) exists at the interface between the two materials. (a) Determine the times at which the copper and aluminum each reach a temperature of \(T_{f}=90^{\circ} \mathrm{C}\). The copper layer is on the bottom. (b) Repeat part (a) with the copper layer on the top. Hint: Modify Equation \(5.15\) to include a term associated with heat transfer across the contact resistance. Apply the modified form of Equation \(5.15\) to each of the two slabs. See Comment 3 of Example 5.2.

Circuit boards are treated by heating a stack of them under high pressure, as illustrated in Problem 5.45. The platens at the top and bottom of the stack are maintained at a uniform temperature by a circulating fluid. The purpose of the pressing-heating operation is to cure the epoxy, which bonds the fiberglass sheets, and impart stiffness to the boards. The cure condition is achieved when the epoxy has been maintained at or above \(170^{\circ} \mathrm{C}\) for at least \(5 \mathrm{~min}\). The effective thermophysical properties of the stack or book (boards and metal pressing plates) are \(k=0.613 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\rho c_{p}=2.73 \times 10^{6} \mathrm{~J} / \mathrm{m}^{3} \cdot \mathrm{K}\) (a) If the book is initially at \(15^{\circ} \mathrm{C}\) and, following application of pressure, the platens are suddenly brought to a uniform temperature of \(190^{\circ} \mathrm{C}\), calculate the elapsed time \(t_{e}\) required for the midplane of the book to reach the cure temperature of \(170^{\circ} \mathrm{C}\). (b) If, at this instant of time, \(t=t_{e}\), the platen temperature were reduced suddenly to \(15^{\circ} \mathrm{C}\), how much energy would have to be removed from the book by the coolant circulating in the platen, in order to return the stack to its initial uniform temperature?

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