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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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Key Concepts

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

Thermal Insulation
Thermal insulation is an essential aspect of controlling temperature change in objects. In our exercise, the blanket wrapped around one of the canned drinks serves as a thermal insulator. The primary function of an insulating material is to reduce the rate of heat transfer. Essentially, it acts as a barrier that slows down heat from moving into or out of the drink.

There are various materials known for their insulation properties, such as fiberglass, foam, and even everyday items like blankets. These materials typically have low thermal conductivity, meaning they do not easily absorb or transfer heat. By wrapping a drink in a blanket, we are decreasing the rate at which it gets warmer by preventing heat from the environment from reaching the drink rapidly.

This principle is not only used to keep things cold but also to maintain objects warm. Insulation is a key factor in maintaining the desired temperature and ensuring energy efficiency in various applications.
Heat Absorption
Heat absorption occurs when an object gains heat from its surroundings. This process naturally follows the law of thermodynamics, where heat travels from warmer to cooler areas to reach a thermal equilibrium. In our scenario, the drinks are initially at a lower temperature than the room, prompting heat transfer towards them.

The uncovered drink absorbs heat faster because it has direct contact with the air in the room, lacking any insulation to impede the flow of heat. The rate of heat absorption depends on several factors:

  • Surface area: Larger surface areas allow for more heat transfer.
  • Material: Some materials absorb heat more aggressively due to their thermal properties.
  • Temperature difference: A greater difference between the drink and room temperature increases the heat absorption rate.


Understanding how heat absorption works is crucial for designing systems involving heating or cooling processes, such as in appliances, building design, and thermal engineering.
Temperature Gradient
A temperature gradient refers to the rate of change in temperature over a specific distance. In physics, this concept is pivotal because it determines the direction and speed of heat transfer. In our example, the gradient is between the temperature of the cold drink and the warmer room.

The greater the temperature gradient, the faster the heat will transfer from the room air to the drink. For the uncovered drink, this gradient remains consistent and strong because nothing impedes the flow of heat. As a result, the uncovered drink warms up faster because the temperature difference drives a quick heat exchange.

For the drink wrapped in a blanket, the insulating material creates a buffer that effectively reduces the temperature gradient. This means the heat transfer rate decreases, leading the drink to warm up more slowly.

Temperature gradients are an integral part of understanding how heat energy moves naturally in the environment, whether in climate science, meteorology, or everyday weather conditions.

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Most popular questions from this chapter

Cold conditioned air at \(12^{\circ} \mathrm{C}\) is flowing inside a \(1.5\)-cm- thick square aluminum \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) duct of inner cross section \(22 \mathrm{~cm} \times 22 \mathrm{~cm}\) at a mass flow rate of \(0.8 \mathrm{~kg} / \mathrm{s}\). The duct is exposed to air at \(33^{\circ} \mathrm{C}\) with a combined convection-radiation heat transfer coefficient of \(13 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The convection heat transfer coefficient at the inner surface is \(75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the air temperature in the duct should not increase by more than \(1^{\circ} \mathrm{C}\) determine the maximum length of the duct.

Determine the winter \(R\)-value and the \(U\)-factor of a masonry cavity wall that is built around 4-in-thick concrete blocks made of lightweight aggregate. The outside is finished with 4 -in face brick with \(\frac{1}{2}\)-in cement mortar between the bricks and concrete blocks. The inside finish consists of \(\frac{1}{2}\)-in gypsum wallboard separated from the concrete block by \(\frac{3}{4}\)-in-thick (1-in by 3 -in nominal) vertical furring whose center- to-center distance is 16 in. Neither side of the \(\frac{3}{4}\)-in-thick air space between the concrete block and the gypsum board is coated with any reflective film. When determining the \(R\)-value of the air space, the temperature difference across it can be taken to be \(30^{\circ} \mathrm{F}\) with a mean air temperature of \(50^{\circ} \mathrm{F}\). The air space constitutes 80 percent of the heat transmission area, while the vertical furring and similar structures constitute 20 percent.

A hot plane surface at \(100^{\circ} \mathrm{C}\) is exposed to air at \(25^{\circ} \mathrm{C}\) with a combined heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat loss from the surface is to be reduced by half by covering it with sufficient insulation with a thermal conductivity of \(0.10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming the heat transfer coefficient to remain constant, the required thickness of insulation is (a) \(0.1 \mathrm{~cm}\) (b) \(0.5 \mathrm{~cm}\) (c) \(1.0 \mathrm{~cm}\) (d) \(2.0 \mathrm{~cm}\) (e) \(5 \mathrm{~cm}\)

Circular cooling fins of diameter \(D=1 \mathrm{~mm}\) and length \(L=25.4 \mathrm{~mm}\), made of copper \((k=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), are used to enhance heat transfer from a surface that is maintained at temperature \(T_{s 1}=132^{\circ} \mathrm{C}\). Each rod has one end attached to this surface \((x=0)\), while the opposite end \((x=L)\) is joined to a second surface, which is maintained at \(T_{s 2}=0^{\circ} \mathrm{C}\). The air flowing between the surfaces and the rods is also at \(T_{\infty}=0^{\circ} \mathrm{C}\), and the convection coefficient is \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Express the function \(\theta(x)=T(x)-T_{\infty}\) along a fin, and calculate the temperature at \(x=L / 2\). (b) Determine the rate of heat transferred from the hot surface through each fin and the fin effectiveness. Is the use of fins justified? Why? (c) What is the total rate of heat transfer from a \(10-\mathrm{cm}\) by 10 -cm section of the wall, which has 625 uniformly distributed fins? Assume the same convection coefficient for the fin and for the unfinned wall surface.

Steam at \(320^{\circ} \mathrm{C}\) flows in a stainless steel pipe \((k=\) \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) whose inner and outer diameters are \(5 \mathrm{~cm}\) and \(5.5 \mathrm{~cm}\), respectively. The pipe is covered with \(3-\mathrm{cm}\)-thick glass wool insulation \((k=0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Heat is lost to the surroundings at \(5^{\circ} \mathrm{C}\) by natural convection and radiation, with a combined natural convection and radiation heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the heat transfer coefficient inside the pipe to be \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from the steam per unit length of the pipe. Also determine the temperature drops across the pipe shell and the insulation.

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