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How would the rate of heat transfer between a thermal reservoir at a higher temperature and one at a lower temperature differ if the reservoirs were in contact with a 10 -cm-long glass rod instead of a 10 -m-long aluminum rod having an identical cross-sectional area?

Short Answer

Expert verified
Question: Compare the heat transfer rate between two thermal reservoirs when they are connected by a glass rod and an aluminum rod, both having the same length and cross-sectional area. Answer: The heat transfer rate of the glass rod is only about 0.39% of the aluminum rod. Therefore, the rate of heat transfer is significantly lower when using a glass rod compared to an aluminum rod, with the same cross-sectional area and length.

Step by step solution

01

Understand the Formula for Heat Transfer via Conduction

We need to use the formula for heat transfer through conduction. The formula is given by: Q = (k * A * (Th - Tc) * t) / L Where Q is the heat transfer, k is the thermal conductivity of the material, A is the cross-sectional area, Th is the higher temperature, Tc is the lower temperature, t is the time period for which the heat transfer occurs, and L is the length of the rod. In this exercise, we are concerned with comparing heat transfer rates (Q/t). Therefore, we can simplify the formula as follows: Q/t = (k * A * (Th - Tc)) / L
02

Find the Thermal Conductivities of Glass and Aluminum

We need to know the thermal conductivities of glass and aluminum to calculate the heat transfer rates. The general values for the thermal conductivities are: k_glass ≈ 0.8 W/(m·K) k_aluminum ≈ 205 W/(m·K)
03

Compare Heat Transfer Rates for Glass and Aluminum Rods

Let's denote the heat transfer rates of glass and aluminum rods as Q/t_glass and Q/t_aluminum, respectively. To compare the heat transfer rates of the two rods, we can put the values from Step 2 into the simplified formula from Step 1: Q/t_glass = (k_glass * A * (Th - Tc)) / L_glass Q/t_aluminum = (k_aluminum * A * (Th - Tc)) / L_aluminum Given that the cross-sectional area (A), temperature difference (Th - Tc), and lengths of the rods (L_glass = L_aluminum = 10 cm = 0.1m) are identical, we can find the ratio of the heat transfer rates: (Q/t_glass) / (Q/t_aluminum) = k_glass / k_aluminum Plug in the values for the thermal conductivities: (Q/t_glass) / (Q/t_aluminum) = 0.8 / 205 = 0.0039 Therefore, the heat transfer rate of the glass rod is only about 0.39% (0.0039 * 100) of the aluminum rod. In conclusion, the rate of heat transfer between the thermal reservoirs would be significantly lower when using a glass rod compared to an aluminum rod, with the same cross-sectional area and length.

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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 property that measures a material's ability to conduct heat. It is denoted by the symbol \( k \) and is expressed in units of watts per meter-Kelvin \( W/(m \cdot K) \). This property is crucial in understanding how efficiently different materials can transfer heat when there is a temperature difference.
Materials with high thermal conductivity, like aluminum, transfer heat more efficiently compared to those with low thermal conductivity, like glass. This means that for the same conditions, aluminum will transfer heat faster than glass.
  • High thermal conductive materials: Suitable for applications that require quick heat dissipation.
  • Low thermal conductive materials: Used as insulators to reduce heat loss.
Knowing the thermal conductivity values helps in making informed decisions in engineering and everyday applications involving heat transfer.
Conduction
Conduction is the process of heat transfer through a material without the material itself moving. It occurs when there is a temperature difference across a medium, and energy is transferred from high to low temperature regions.
The rate of heat transfer by conduction is calculated using Fourier's law, expressed as: \[Q = \frac{k \cdot A \cdot (T_h - T_c) \cdot t}{L}\] Where:
  • \(Q\) is the quantity of heat transferred.
  • \(k\) is the thermal conductivity.
  • \(A\) is the area through which heat flows.
  • \(T_h\) and \(T_c\) are the high and low temperatures, respectively.
  • \(t\) is the time.
  • \(L\) is the length over which the heat flows.
Conduction is one of the primary methods of heat transfer and is significant in processes where heat has to be conducted through solid materials.
Thermal Reservoir
A thermal reservoir is a system that can absorb or supply an unlimited amount of heat without undergoing any change in temperature. It's often idealized in physics and engineering as a constant temperature source or sink.
Thermal reservoirs are key in understanding heat engines and refrigerators, as they provide the heat source or sink necessary for these systems to operate. In this context, the reservoir at a higher temperature provides heat (the source), and the one at a lower temperature absorbs heat (the sink).
  • The high-temperature reservoir is frequently referred to as a heat source.
  • The low-temperature reservoir acts as a heat sink.
These concepts help in analyzing systems like power plants, where efficient management of heat transfer between reservoirs is crucial for optimal performance.
Heat Transfer Rate
The heat transfer rate is the amount of heat energy transferred per unit time, and it is often measured in watts (\(W\)). It tells us how quickly heat is being transferred from one place to another.
In the formula for heat conduction, the rate of heat transfer \[Q/t = \frac{k \cdot A \cdot (T_h - T_c)}{L}\] is of particular interest as it reveals how different parameters affect the speed of heat transfer.
Factors influencing the heat transfer rate include:
  • Thermal conductivity (\(k\)): Higher values mean faster heat transfer.
  • Area (\(A\)): Larger areas result in more heat being transferred.
  • Temperature difference \((T_h - T_c)\)): Greater differences drive faster transfer.
  • Length \((L)\): Shorter lengths enable faster heat flow.
Understanding the rate of heat transfer is essential in designing efficient thermal systems and determining the suitability of materials for conducting or insulating heat.

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Most popular questions from this chapter

Water is an excellent coolant as a result of its very high heat capacity. Calculate the amount of heat that is required to change the temperature of \(10.0 \mathrm{~kg}\) of water by \(10.0 \mathrm{~K}\). Now calculate the kinetic energy of a car with \(m=1.00 \cdot 10^{3} \mathrm{~kg}\) moving at a speed of \(27.0 \mathrm{~m} / \mathrm{s}(60.0 \mathrm{mph}) .\) Compare the two quantities.

Determine the ratio of the heat flow into a six-pack of aluminum soda cans to the heat flow into a 2.00 - \(\mathrm{L}\) plastic bottle of soda when both are taken out of the same refrigerator, that is, have the same initial temperature difference with the air in the room. Assume that each soda can has a diameter of \(6.00 \mathrm{~cm}\), a height of \(12.0 \mathrm{~cm}\), and a thickness of \(0.100 \mathrm{~cm}\). Use \(205 \mathrm{~W} /(\mathrm{m} \mathrm{K})\) as the thermal conductivity of aluminum. Assume that the 2.00 - \(\mathrm{L}\) bottle of soda has a diameter of \(10.0 \mathrm{~cm}\), a height of \(25.0 \mathrm{~cm}\), and a thickness of \(0.100 \mathrm{~cm} .\) Use \(0.100 \mathrm{~W} /(\mathrm{mK})\) as the thermal conductivity of plastic.

Assuming the severity of a burn increases as the amount of energy put into the skin increases, which of the following would cause the most severe burn (assume equal masses)? a) water at \(90^{\circ} \mathrm{C}\) b) copper at \(110^{\circ} \mathrm{C}\) c) steam at \(180^{\circ} \mathrm{C}\) d) aluminum at \(100^{\circ} \mathrm{C}\) e) lead at \(100^{\circ} \mathrm{C}\)

Knife blades are often made of hardened carbon steel. The hardening process is a heat treatment in which the blade is first heated to a temperature of \(1346^{\circ} \mathrm{F}\) and then cooled down rapidly by immersing it in a bath of water. To achieve the desired hardness, after heating to \(1346^{\circ} \mathrm{F}\), a blade needs to be brought to a temperature below \(5.00 \cdot 10^{2}{ }^{\circ} \mathrm{F}\). If the blade has a mass of \(0.500 \mathrm{~kg}\) and the water is in an open copper container of mass \(2.000 \mathrm{~kg}\) and sufficiently large volume, what is the minimum quantity of water that needs to be in the container for this hardening process to be successful? Assume the blade is not in direct mechanical (and thus thermal) contact with the container, and neglect cooling through radiation into the air. Assume no water boils but reaches \(100^{\circ} \mathrm{C} .\) The heat capacity of copper around room temperature is \(c_{\text {copper }}=386 \mathrm{~J} /(\mathrm{kg} \mathrm{K}) .\) Use the data in the table below for the heat capacity of carbon steel

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