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Chapter 3: Problem 10

Consider two cold canned drinks, one wrapped in a blanket and the other placed on a table in the same room. Which drink will warm up faster?

Short Answer

Expert verified
Answer: The uncovered canned drink warms up faster because it has a higher rate of heat absorption due to direct contact with the air in the room. The blanket on the other canned drink acts as an insulator, slowing down the heat transfer process and causing it to warm up slower.

Step by step solution

01

Determine the conditions for heat transfer

The two drinks are in the same room. Therefore, they are exposed to the same room temperature and conditions. The difference between them is the presence of the blanket, which covers one of the drinks.
02

Explain the role of the blanket

When a cold object is placed in a warmer environment, it will absorb heat from the surroundings, causing its temperature to rise. The blanket acts as an insulating material, which slows down the transfer of heat between the drink and the air in the room.
03

Analyze the impact of insulation on heat transfer

The drink with the blanket wrapped around it will experience a slower rate of heat transfer due to the insulating effect of the blanket. On the other hand, the uncovered drink has direct contact with the air in the room, and therefore will absorb heat faster.
04

Determine which drink warms up faster

The drink that warms up faster is the one with a higher rate of heat absorption, which in this case is the drink without the blanket as it has no additional insulation slowing down the heat transfer.

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Most popular questions from this chapter

Computer memory chips are mounted on a finned metallic mount to protect them from overheating. A \(152 \mathrm{MB}\) memory chip dissipates \(5 \mathrm{~W}\) of heat to air at \(25^{\circ} \mathrm{C}\). If the temperature of this chip is to not exceed \(50^{\circ} \mathrm{C}\), the overall heat transfer coefficient- area product of the finned metal mount must be at least (a) \(0.2 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (b) \(0.3 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (c) \(0.4 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (d) \(0.5 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (e) \(0.6 \mathrm{~W} /{ }^{\circ} \mathrm{C}\)

Consider a house with a flat roof whose outer dimensions are \(12 \mathrm{~m} \times 12 \mathrm{~m}\). The outer walls of the house are \(6 \mathrm{~m}\) high. The walls and the roof of the house are made of \(20-\mathrm{cm}-\) thick concrete \((k=0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The temperatures of the inner and outer surfaces of the house are \(15^{\circ} \mathrm{C}\) and \(3^{\circ} \mathrm{C}\), respectively. Accounting for the effects of the edges of adjoining surfaces, determine the rate of heat loss from the house through its walls and the roof. What is the error involved in ignoring the effects of the edges and corners and treating the roof as a \(12 \mathrm{~m} \times 12 \mathrm{~m}\) surface and the walls as \(6 \mathrm{~m} \times 12 \mathrm{~m}\) surfaces for simplicity?

One wall of a refrigerated warehouse is \(10.0\)-m-high and \(5.0\)-m-wide. The wall is made of three layers: \(1.0\)-cm-thick aluminum \((k=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}), 8.0\)-cm-thick fibreglass \((k=\) \(0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and \(3.0-\mathrm{cm}\) thick gypsum board \((k=\) \(0.48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The warehouse inside and outside temperatures are \(-10^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\), respectively, and the average value of both inside and outside heat transfer coefficients is \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Calculate the rate of heat transfer across the warehouse wall in steady operation. (b) Suppose that 400 metal bolts \((k=43 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), each \(2.0 \mathrm{~cm}\) in diameter and \(12.0 \mathrm{~cm}\) long, are used to fasten (i.e., hold together) the three wall layers. Calculate the rate of heat transfer for the "bolted" wall. (c) What is the percent change in the rate of heat transfer across the wall due to metal bolts?

Heat is lost at a rate of \(275 \mathrm{~W}\) per \(\mathrm{m}^{2}\) area of a \(15-\mathrm{cm}-\) thick wall with a thermal conductivity of \(k=1.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The temperature drop across the wall is (a) \(37.5^{\circ} \mathrm{C}\) (b) \(27.5^{\circ} \mathrm{C}\) (c) \(16.0^{\circ} \mathrm{C}\) (d) \(8.0^{\circ} \mathrm{C}\) (e) \(4.0^{\circ} \mathrm{C}\)

Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially that of the surrounding air, i.e. \(20^{\circ} \mathrm{C}\). Its width is \(5.0 \mathrm{~cm}\); thickness is \(1.0 \mathrm{~mm}\); thermal conductivity is \(200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\); and base temperature is \(40^{\circ} \mathrm{C}\). The heat transfer coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Estimate the fin temperature at a distance of \(5.0 \mathrm{~cm}\) from the base and the rate of heat loss from the entire fin.
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