Can all three modes of heat transfer occur simultaneously (in parallel) in a medium?

Short Answer

Expert verified
Answer: Yes, all three modes of heat transfer - conduction, convection, and radiation - can occur simultaneously in a medium, provided that the medium is in contact with another body, in motion, and experiencing a temperature difference. An example of this is a pot of water being heated on a stove.

Step by step solution

01

Understand the three modes of heat transfer

Conduction is the transfer of heat within a solid or between two solids in direct contact. Convection is the transfer of heat in liquids and gases due to the movement of molecules. Radiation is the transfer of heat through empty space or even through a medium without a change in the temperature of the medium itself.
02

Determine if conduction, convection, and radiation can occur simultaneously in a medium

Yes, all three modes of heat transfer can occur simultaneously in a medium. The key condition for such a situation is that the medium must be in contact with another body, be in motion, and be in the presence of a temperature difference. In that case, heat can be transferred by conduction, as the temperature difference drives the transfer of energy. Convection can also occur, as the moving fluid can transmit heat in regions where temperature varies. Lastly, radiation can happen as heat can be absorbed or emitted through electromagnetic waves in any situation.
03

Example scenario

An example of this would be a pot of water being heated on a stove. Conduction occurs between the pot and the water due to direct contact, convection is happening within the water as the heated water at the bottom rises, and radiation occurs since the stove emits infrared radiation which is absorbed by both the pot and the water.

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Key Concepts

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

Conduction
When we place our hand on a hot mug of tea, we feel the heat because of a process called conduction. This is the direct transfer of heat from the mug into our hand due to the physical contact. The thermal energy moves from the hotter area (the mug) to the cooler one (our hand), flowing along a temperature gradient until both materials reach thermal equilibrium, meaning they are the same temperature.

Metals like copper or aluminum are excellent conductors and quickly transfer heat, which is why they're commonly used in cookware. However, materials like wood or plastic are poor conductors and are often used as insulators, preventing heat from escaping or entering, such as in cooler boxes or thermal drink containers.
Convection
Ever noticed how a spoon in a cup of hot soup eventually becomes warm, even if it's not touching the bottom of the cup? This is because of convection, the transfer of heat through the movement of fluids (which include liquids and gases). Warm parts of a fluid rise because they are less dense, and cooler, denser parts sink, creating a circulation pattern known as a convection current.

A home heating system is a typical example of convection. Warm air from a heater rises and circulates throughout a room, while cooler air sinks to be warmed by the heater, creating a continuous flow of heat distribution.
Radiation
Imagine feeling the warmth of the sun on your skin, even though it's millions of kilometers away. This is possible because of radiation, a method of heat transfer that does not rely on any contact or medium to occur. Instead, heat is emitted through electromagnetic waves, primarily in the infrared spectrum.

These waves can travel through the vacuum of space to reach Earth, providing warmth. Similarly, things like toasters and radiators emit infrared waves, heating nearby objects without needing to physically touch them.
Heat Transfer in Mediums
Different materials can transfer heat in different ways. This is called heat transfer in mediums. For instance, solids typically conduct heat well, while gases and liquids are better at convection. The material's properties, such as its density, specific heat, and thermal conductivity, affect how easily heat can travel through it.

Materials that trap air, like foam or wool, are good insulators because air is a poor conductor of heat. These materials prevent the easy flow of heat and are used in applications designed to maintain a specific temperature, like thermal clothing or building insulation.
Temperature Difference
The driving force behind all modes of heat transfer is the temperature difference. Heat naturally flows from regions of higher temperature to regions of lower temperature. This flow will continue until a state of thermal equilibrium is reached where no temperature difference exists.

The larger the temperature difference between two bodies, the faster the rate of heat transfer. This principle is the foundation of thermodynamics and is crucial in applications ranging from industrial heat exchangers to household refrigeration.

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Most popular questions from this chapter

How do \((a)\) draft and \((b)\) cold floor surfaces cause discomfort for a room's occupants?

It is well-known that at the same outdoor air temperature a person is cooled at a faster rate under windy conditions than under calm conditions due to the higher convection heat transfer coefficients associated with windy air. The phrase wind chill is used to relate the rate of heat loss from people under windy conditions to an equivalent air temperature for calm conditions (considered to be a wind or walking speed of \(3 \mathrm{mph}\) or \(5 \mathrm{~km} / \mathrm{h})\). The hypothetical wind chill temperature (WCT), called the wind chill temperature index (WCTI), is an equivalent air temperature equal to the air temperature needed to produce the same cooling effect under calm conditions. A 2003 report on wind chill temperature by the U.S. National Weather Service gives the WCTI in metric units as WCTI \(\left({ }^{\circ} \mathrm{C}\right)=13.12+0.6215 T-11.37 V^{0.16}+0.3965 T V^{0.16}\) where \(T\) is the air temperature in \({ }^{\circ} \mathrm{C}\) and \(V\) the wind speed in \(\mathrm{km} / \mathrm{h}\) at \(10 \mathrm{~m}\) elevation. Show that this relation can be expressed in English units as WCTI \(\left({ }^{\circ} \mathrm{F}\right)=35.74+0.6215 T-35.75 V^{0.16}+0.4275 T V^{0.16}\) where \(T\) is the air temperature in \({ }^{\circ} \mathrm{F}\) and \(V\) the wind speed in \(\mathrm{mph}\) at \(33 \mathrm{ft}\) elevation. Also, prepare a table for WCTI for air temperatures ranging from 10 to \(-60^{\circ} \mathrm{C}\) and wind speeds ranging from 10 to \(80 \mathrm{~km} / \mathrm{h}\). Comment on the magnitude of the cooling effect of the wind and the danger of frostbite.

An engineer who is working on the heat transfer analysis of a house in English units needs the convection heat transfer coefficient on the outer surface of the house. But the only value he can find from his handbooks is \(22 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), which is in SI units. The engineer does not have a direct conversion factor between the two unit systems for the convection heat transfer coefficient. Using the conversion factors between \(\mathrm{W}\) and \(\mathrm{Btu} / \mathrm{h}, \mathrm{m}\) and \(\mathrm{ft}\), and \({ }^{\circ} \mathrm{C}\) and \({ }^{\circ} \mathrm{F}\), express the given convection heat transfer coefficient in Btu/ \(\mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\). Answer: \(3.87 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\)

Water enters a pipe at \(20^{\circ} \mathrm{C}\) at a rate of \(0.50 \mathrm{~kg} / \mathrm{s}\) and is heated to \(60^{\circ} \mathrm{C}\). The rate of heat transfer to the water is (a) \(20 \mathrm{~kW}\) (b) \(42 \mathrm{~kW}\) (c) \(84 \mathrm{~kW}\) (d) \(126 \mathrm{~kW}\) (e) \(334 \mathrm{~kW}\)

The roof of a house consists of a 22-cm-thick (st) concrete slab \((k=2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that is \(15 \mathrm{~m}\) wide and \(20 \mathrm{~m}\) long. The emissivity of the outer surface of the roof is \(0.9\), and the convection heat transfer coefficient on that surface is estimated to be \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The inner surface of the roof is maintained at \(15^{\circ} \mathrm{C}\). On a clear winter night, the ambient air is reported to be at \(10^{\circ} \mathrm{C}\) while the night sky temperature for radiation heat transfer is \(255 \mathrm{~K}\). Considering both radiation and convection heat transfer, determine the outer surface temperature and the rate of heat transfer through the roof. If the house is heated by a furnace burning natural gas with an efficiency of 85 percent, and the unit cost of natural gas is \(\$ 1.20\) / therm ( 1 therm \(=105,500 \mathrm{~kJ}\) of energy content), determine the money lost through the roof that night during a 14-hour period.

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