/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 In the conduction equation \(Q=(... [FREE SOLUTION] | 91Ó°ÊÓ

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In the conduction equation \(Q=(k A \Delta T) t / L,\) the combination of terms \(k A / L\) is called the conductance. The human body has the ability to vary the conductance of the tissue beneath the skin by means of vasoconstriction and vasodilation, in which the flow of blood to the veins and capillaries is decreased and increased, respectively. The conductance can be adjusted over a range such that the tissue beneath the skin is equivalent to a thickness of \(0.080 \mathrm{mm}\) of Styrofoam or \(3.5 \mathrm{mm}\) of air. By what factor (high/low) can the body adjust the conductance?

Short Answer

Expert verified
The conductance can be adjusted by a factor of 0.0183.

Step by step solution

01

Understand the Conductance Formula

The conductance is defined as \( \frac{k A}{L} \), where \( k \) is the thermal conductivity, \( A \) is the area, and \( L \) is the thickness of the material. The conductance determines the rate at which heat is transferred through an object.
02

Identify the Materials and Their Properties

The problem states that the tissue can be equivalent to 0.080 mm of Styrofoam and 3.5 mm of air. We need the thermal conductivities of Styrofoam and air: \( k_{\text{Styrofoam}} \approx 0.03 \frac{W}{m \cdot K} \) and \( k_{\text{Air}} \approx 0.024 \frac{W}{m \cdot K} \).
03

Calculate the Conductance for Styrofoam

For Styrofoam, \( L = 0.080 \times 10^{-3} \) m, \( k = 0.03 \frac{W}{m \cdot K} \). The conductance is given by \( \frac{k A}{L} = \frac{0.03 \cdot A}{0.080 \times 10^{-3}} \).
04

Calculate the Conductance for Air

For air, \( L = 3.5 \times 10^{-3} \) m, \( k = 0.024 \frac{W}{m \cdot K} \). The conductance is \( \frac{k A}{L} = \frac{0.024 \cdot A}{3.5 \times 10^{-3}} \).
05

Determine the Conductance Factor

The factor by which conductance can be adjusted is the ratio of the conductance for air to the conductance for Styrofoam. This is calculated as: \[\text{Factor} = \frac{\frac{0.024 \cdot A}{3.5 \times 10^{-3}}}{\frac{0.03 \cdot A}{0.080 \times 10^{-3}}} \] The area \( A \) cancels out, simplifying to:\[\text{Factor} = \frac{0.024 \cdot 0.080}{3.5 \times 0.03}\approx \frac{0.00192}{0.105}\approx 0.0183\] So, the conductance can be adjusted by a factor of about 0.0183.

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Key Concepts

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

Conduction Equation
The conduction equation is a vital part of understanding how heat transfer works in different materials. It is expressed as \( Q = \frac{(k A \Delta T) t}{L} \), where the entire equation helps us calculate the heat transferred over time. Here, \( Q \) is the total heat transferred, \( k \) represents the thermal conductivity of the material, \( A \) indicates the area through which the heat is being conducted, \( \Delta T \) is the temperature difference across the material, \( t \) is the time period, and \( L \) denotes the thickness of the material.

One of the key components in this equation is "conductance," represented by \( \frac{k A}{L} \). Conductance essentially tells us how good a specific material is at allowing heat to move through it quickly. The higher the conductance, the more efficient the material is in heat transfer.
Vasoconstriction and Vasodilation
Vasoconstriction and vasodilation are physiological processes that control the flow of blood in the human body and directly impact thermal conductance. They play a crucial role in regulating body temperature.

- **Vasoconstriction:** This process involves the tightening or narrowing of blood vessels. When blood vessels constrict, there is less blood flow to the skin and outer layers of the body, which reduces heat loss.
- **Vasodilation:** In contrast, this process involves the widening of blood vessels. When blood vessels dilate, more blood flows to the surface of the skin, which helps to release extra body heat into the environment.

In response to temperature changes, the body can increase or decrease conductance by changing the thickness of equivalent tissue, simulating materials like Styrofoam or air, as described in the exercise.
Thermal Conductivity of Materials
Understanding thermal conductivity is fundamental when discussing heat transfer through materials. Thermal conductivity, denoted by \( k \), is a property that indicates how well a material can conduct heat. Measured in watts per meter-kelvin (\( \frac{W}{m \cdot K} \)), it shows the rate of heat transfer through a unit area of the material for a temperature difference of one Kelvin per unit thickness.

- **Styrofoam:** Usually has a low thermal conductivity around \( 0.03 \frac{W}{m \cdot K} \). It's an excellent insulator, which means it doesn’t allow heat to pass through easily.
- **Air:** Has an even lower thermal conductivity of approximately \( 0.024 \frac{W}{m \cdot K} \), making it a poor conductor of heat.

These values help in calculating the conductance of materials, as shown in the exercise, to assess how the body can simulate the thermal resistance of different materials by adjusting blood flow.
Heat Transfer
Heat transfer is a process central to thermal dynamics, encompassing how heat moves from a hotter object to a cooler one. It can occur in three main ways: conduction, convection, and radiation. This article focuses primarily on conduction, the direct transfer of heat through materials.

- **Conduction:** Takes place when heat passes through a solid or stationary fluid by means of molecular vibrations and energy transfer. This method of heat transfer depends on the thermal conductivity of the material.
- **Impact on the Human Body:** The body uses blood flow adjustments to influence conductance by altering tissue thickness akin to inorganic insulators. This helps in maintaining a stable body temperature across various environmental conditions.

By understanding these concepts, we grasp how physics and biology work together intricately to manage heat transfer, directly affecting comfort and survival in different climates.

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Most popular questions from this chapter

A wall in a house contains a single window. The window consists of a single pane of glass whose area is \(0.16 \mathrm{m}^{2}\) and whose thickness is \(2.0 \mathrm{mm}\). Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are \(18 \mathrm{m}^{2}\) and \(0.10 \mathrm{m},\) respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window?

A solar collector is placed in direct sunlight where it absorbs energy at the rate of \(880 \mathrm{J} / \mathrm{s}\) for each square meter of its surface. The emissivity of the solar collector is \(e=0.75 .\) What equilibrium temperature does the collector reach? Assume that the only energy loss is due to the emission of radiation.

A solid sphere has a temperature of \(773 \mathrm{K}\). The sphere is melted down and recast into a cube that has the same emissivity and emits the same radiant power as the sphere. What is the cube's temperature?

The amount of radiant power produced by the sun is approximately \(3.9 \times 10^{26}\) W. Assuming the sun to be a perfect blackbody sphere with a radius of \(6.96 \times 10^{8} \mathrm{m}\), find its surface temperature (in kelvins).

A baking dish is removed from a hot oven and placed on a cooling rack. As the dish cools down to \(35^{\circ} \mathrm{C}\) from \(175^{\circ} \mathrm{C}\), its net radiant power decreases to \(12.0 \mathrm{W}\). What was the net radiant power of the baking dish when it was first removed from the oven? Assume that the temperature in the kitchen remains at \(22^{\circ} \mathrm{C}\) as the dish cools down.

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