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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

A spherical vessel used as a reactor for producing pharmaceuticals has a 5 -mm-thick stainless steel wall \((k=17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and an inner diameter of \(D_{i}=1.0 \mathrm{~m}\). During production, the vessel is filled with reactants for which \(\rho=1100 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c=2400 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), while exothermic reactions release energy at a volumetric rate of \(\dot{q}=10^{4} \mathrm{~W} / \mathrm{m}^{3}\). As first approximations, the reactants may be assumed to be well stirred and the thermal capacitance of the vessel may be neglected. (a) The exterior surface of the vessel is exposed to ambient air \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) for which a convection coefficient of \(h=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) may be assumed. If the initial temperature of the reactants is \(25^{\circ} \mathrm{C}\), what is the temperature of the reactants after \(5 \mathrm{~h}\) of process time? What is the corresponding temperature at the outer surface of the vessel? (b) Explore the effect of varying the convection coefficient on transient thermal conditions within the reactor.

As permanent space stations increase in size, there is an attendant increase in the amount of electrical power they dissipate. To keep station compartment temperatures from exceeding prescribed limits, it is necessary to transfer the dissipated heat to space. A novel heat rejection scheme that has been proposed for this purpose is termed a Liquid Droplet Radiator (LDR). The heat is first transferred to a high vacuum oil, which is then injected into outer space as a stream of small droplets. The stream is allowed to traverse a distance \(L\), over which it cools by radiating energy to outer space at absolute zero temperature. The droplets are then collected and routed back to the space station. Consider conditions for which droplets of emissivity \(\varepsilon=0.95\) and diameter \(D=0.5 \mathrm{~mm}\) are injected at a temperature of \(T_{i}=500 \mathrm{~K}\) and a velocity of \(V=0.1 \mathrm{~m} / \mathrm{s}\). Properties of the oil are \(\rho=885 \mathrm{~kg} / \mathrm{m}^{3}, c=1900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=0.145 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming each drop to radiate to deep space at \(T_{\text {sur }}=0 \mathrm{~K}\), determine the distance \(L\) required for the droplets to impact the collector at a final temperature of \(T_{f}=300 \mathrm{~K}\). What is the amount of thermal energy rejected by each droplet?

Plasma spray-coating processes are often used to provide surface protection for materials exposed to hostile environments, which induce degradation through factors such as wear, corrosion, or outright thermal failure. Ceramic coatings are commonly used for this purpose. By injecting ceramic powder through the nozzle (anode) of a plasma torch, the particles are entrained by the plasma jet, within which they are then accelerated and heated. During their time-in-fbht, the ceramic particles must be heated to their melting point and experience complete conversion to the liquid state. The coating is formed as the molten droplets impinge (splat) on the substrate material and experience rapid solidification. Consider conditions for which spherical alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right.\) ) particles of diameter \(D_{p}=50 \mu \mathrm{m}\), density \(\rho_{p}=\) \(3970 \mathrm{~kg} / \mathrm{m}^{3}\), thermal conductivity \(k_{p}=10.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and specific heat \(c_{p}=1560 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) are injected into an arc plasma, which is at \(T_{\infty}=10,000 \mathrm{~K}\) and provides a coefficient of \(h=30,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for convective heating of the particles. The melting point and latent heat of fusion of alumina are \(T_{\text {mp }}=2318 \mathrm{~K}\) and \(h_{s f}=3577 \mathrm{~kJ} / \mathrm{kg}\), respectively. (a) Neglecting radiation, obtain an expression for the time-in-flight, \(t_{i-f}\), required to heat a particle from its initial temperature \(T_{i}\) to its melting point \(T_{\text {mp }}\), and, once at the melting point, for the particle to experience complete melting. Evaluate \(t_{i-f}\) for \(T_{i}=300 \mathrm{~K}\) and the prescribed heating conditions. (b) Assuming alumina to have an emissivity of \(\varepsilon_{p}=0.4\) and the particles to exchange radiation with large surroundings at \(T_{\text {sur }}=300 \mathrm{~K}\), assess the validity of neglecting radiation.

A long rod of \(60-\mathrm{mm}\) diameter and thermophysical properties \(\rho=8000 \mathrm{~kg} / \mathrm{m}^{3}, \quad c=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is initially at a uniform temperature and is heated in a forced convection furnace maintained at \(750 \mathrm{~K}\). The convection coefficient is estimated to be \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What is the centerline temperature of the rod when the surface temperature is \(550 \mathrm{~K}\) ? (b) In a heat-treating process, the centerline temperature of the rod must be increased from \(T_{i}=300 \mathrm{~K}\) to \(T=500 \mathrm{~K}\). Compute and plot the centerline temperature histories for \(h=100,500\), and \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). In each case the calculation may be terminated when \(T=500 \mathrm{~K}\).

A thick steel slab \(\left(\rho=7800 \mathrm{~kg} / \mathrm{m}^{3}, c=480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is initially at \(300^{\circ} \mathrm{C}\) and is cooled by water jets impinging on one of its surfaces. The temperature of the water is \(25^{\circ} \mathrm{C}\), and the jets maintain an extremely large, approximately uniform convection coefficient at the surface. Assuming that the surface is maintained at the temperature of the water throughout the cooling, how long will it take for the temperature to reach \(50^{\circ} \mathrm{C}\) at a distance of \(25 \mathrm{~mm}\) from the surface?

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