For which solid is the lumped system analysis more likely to be applicable: an actual apple or a golden apple of the same size? Why?

Short Answer

Expert verified
Answer: An actual apple is more suitable for a lumped system analysis than a golden apple of the same size because it has a lower thermal conductivity, resulting in a more uniform temperature distribution.

Step by step solution

01

Identify the thermal conductivity values

First, we must identify the thermal conductivities of both an actual apple and a golden apple (gold). The thermal conductivity of an actual apple is approximately k_apple = 0.45 W/m·K, while the thermal conductivity of gold is k_gold = 315 W/m·K.
02

Compare the thermal conductivities

Now that we have identified the thermal conductivities of both an actual apple and a golden apple, we must compare these values to determine which is more suitable for a lumped system analysis. Since a lower thermal conductivity value results in a more uniform temperature distribution, we must compare the values to see which one is lower: k_apple = 0.45 W/m·K or k_gold = 315 W/m·K. It is clear that the thermal conductivity of an actual apple (k_apple = 0.45 W/m·K) is much lower than that of a golden apple (k_gold = 315 W/m·K).
03

Determine which solid is more suitable for lumped system analysis

Based on the comparison in Step 2, we can see that an actual apple has a much lower thermal conductivity value than a golden apple. Since lower thermal conductivity values result in more uniform temperature distribution (which is a key assumption for the lumped system analysis), we can conclude that the lumped system analysis is more likely to be applicable for an actual apple than for a golden apple of the same size.

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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 measure of a material's ability to conduct heat. It represents how quickly heat can pass through a substance when a temperature difference exists. In the context of our exercise, comparing an apple and a golden apple, the concept of thermal conductivity is crucial for determining the applicability of lumped system analysis.
Thermal conductivity is quantitatively represented by the symbol \( k \) and is expressed in units of watts per meter per Kelvin (W/m\( \cdot \)K). A lower \( k \) value means that the material is a poorer conductor of heat and will not transfer heat as quickly as a material with a higher \( k \) value. Therefore, substances with low thermal conductivity are likely to maintain a more uniform temperature distribution because heat diffuses through them more slowly.
Understanding thermal conductivity helps in predicting the suitability for lumped system analysis. With regard to the exercise, the actual apple (\( k_{apple} = 0.45 \) W/m\( \cdot \)K) is more likely to have a uniform temperature distribution than the golden apple (\( k_{gold} = 315 \) W/m\( \cdot \)K), due to its significantly lower thermal conductivity.
Uniform Temperature Distribution
Uniform temperature distribution within a solid is one of the critical assumptions of lumped system analysis. This principle implies that at any given moment, the temperature variation throughout the object is negligible, hence it can be approximated as being uniform. When heat transfer occurs, if the material has a low thermal conductivity and is sufficiently small, it will quickly reach a state where its entire volume settles at one temperature.
For an object to maintain a uniform temperature, several conditions should be met, including low thermal conductivity, small size, and good internal mixing or low Biot number (the ratio of external to internal resistance to heat flow). The property of a material to resist change in temperature when exposed to a uniform heat source is linked to its thermal mass, which is a function of both its specific heat capacity and density.
In our exercise, the actual apple, with its lower thermal conductivity, tends to exhibit a more uniform temperature distribution compared to the golden apple. This characteristic makes the actual apple better suited for lumped system analysis, which simplifies the complex heat transfer problem by considering the temperature of the object as spatially uniform.
Heat Transfer in Solids
Heat transfer in solids occurs mainly through conduction, which involves the transfer of energy from more energetic particles to less energetic ones due to temperature difference, without any actual movement of the particles. This contrasts with heat transfer in fluids, where convection (the movement of fluid itself) can also play a significant role.
In the context of the exercise, understanding heat transfer in solids like an apple or a golden apple helps us apply the lumped system analysis correctly. Solids with lower thermal conductivity, like the actual apple, transfer heat more slowly, which facilitates the approximation of uniform temperature throughout the object. Conversely, heat rapidly disperses through solids with high thermal conductivity, like gold, often leading to significant temperature gradients within the object.
A lumped system analysis is typically used when the rate of heat transfer to and from an object is slow enough that the object's temperature can be assumed fairly uniform. This is more easily justified in solids like the actual apple, which has a lower thermal conductivity and consequently a slower heat transfer rate, than in high conductivity materials like gold.

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Most popular questions from this chapter

Chickens with an average mass of \(1.7 \mathrm{~kg}(k=\) \(0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=0.13 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) initially at a uniform temperature of \(15^{\circ} \mathrm{C}\) are to be chilled in agitated brine at \(-7^{\circ} \mathrm{C}\). The average heat transfer coefficient between the chicken and the brine is determined experimentally to be \(440 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the average density of the chicken to be \(0.95 \mathrm{~g} / \mathrm{cm}^{3}\) and treating the chicken as a spherical lump, determine the center and the surface temperatures of the chicken in \(2 \mathrm{~h}\) and \(45 \mathrm{~min}\). Also, determine if any part of the chicken will freeze during this process. Solve this problem using analytical one-term approximation method (not the Heisler charts).

A 5-mm-thick stainless steel strip \((k=21 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=\) \(8000 \mathrm{~kg} / \mathrm{m}^{3}\), and \(\left.c_{p}=570 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is being heat treated as it moves through a furnace at a speed of \(1 \mathrm{~cm} / \mathrm{s}\). The air temperature in the furnace is maintained at \(900^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the furnace length is \(3 \mathrm{~m}\) and the stainless steel strip enters it at \(20^{\circ} \mathrm{C}\), determine the temperature of the strip as it exits the furnace.

A 6-cm-diameter 13-cm-high canned drink ( \(\rho=\) \(\left.977 \mathrm{~kg} / \mathrm{m}^{3}, k=0.607 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) initially at \(25^{\circ} \mathrm{C}\) is to be cooled to \(5^{\circ} \mathrm{C}\) by dropping it into iced water at \(0^{\circ} \mathrm{C}\). Total surface area and volume of the drink are \(A_{s}=\) \(301.6 \mathrm{~cm}^{2}\) and \(V=367.6 \mathrm{~cm}^{3}\). If the heat transfer coefficient is \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine how long it will take for the drink to \(\operatorname{cool}\) to \(5^{\circ} \mathrm{C}\). Assume the can is agitated in water and thus the temperature of the drink changes uniformly with time. (a) \(1.5 \mathrm{~min}\) (b) \(8.7 \mathrm{~min}\) (c) \(11.1 \mathrm{~min}\) (d) \(26.6 \mathrm{~min}\) (e) \(6.7 \mathrm{~min}\)

In Betty Crocker's Cookbook, it is stated that it takes \(2 \mathrm{~h} \mathrm{} 45 \mathrm{~min}\) to roast a \(3.2-\mathrm{kg}\) rib initially at \(4.5^{\circ} \mathrm{C}\) "rare" in an oven maintained at \(163^{\circ} \mathrm{C}\). It is recommended that a meat thermometer be used to monitor the cooking, and the rib is considered rare done when the thermometer inserted into the center of the thickest part of the meat registers \(60^{\circ} \mathrm{C}\). The rib can be treated as a homogeneous spherical object with the properties \(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4.1 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=\) \(0.91 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\). Determine \((a)\) the heat transfer coefficient at the surface of the rib; \((b)\) the temperature of the outer surface of the rib when it is done; and \((c)\) the amount of heat transferred to the rib. \((d)\) Using the values obtained, predict how long it will take to roast this rib to "medium" level, which occurs when the innermost temperature of the rib reaches \(71^{\circ} \mathrm{C}\). Compare your result to the listed value of \(3 \mathrm{~h} \mathrm{} 20 \mathrm{~min}\). If the roast rib is to be set on the counter for about \(15 \mathrm{~min}\) before it is sliced, it is recommended that the rib be taken out of the oven when the thermometer registers about \(4^{\circ} \mathrm{C}\) below the indicated value because the rib will continue cooking even after it is taken out of the oven. Do you agree with this recommendation? Solve this problem using analytical one-term approximation method (not the Heisler charts).

Chickens with an average mass of \(2.2 \mathrm{~kg}\) and average specific heat of \(3.54 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\) are to be cooled by chilled water that enters a continuous-flow-type immersion chiller at \(0.5^{\circ} \mathrm{C}\). Chickens are dropped into the chiller at a uniform temperature of \(15^{\circ} \mathrm{C}\) at a rate of 500 chickens per hour and are cooled to an average temperature of \(3^{\circ} \mathrm{C}\) before they are taken out. The chiller gains heat from the surroundings at a rate of \(210 \mathrm{~kJ} / \mathrm{min}\). Determine \((a)\) the rate of heat removal from the chicken, in \(\mathrm{kW}\), and \((b)\) the mass flow rate of water, in \(\mathrm{kg} / \mathrm{s}\), if the temperature rise of water is not to exceed \(2^{\circ} \mathrm{C}\).

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