/*! 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 85 Heat Transport in the Human Body... [FREE SOLUTION] | 91Ó°ÊÓ

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Heat Transport in the Human Body The core temperature of the human body is \(37.0^{\circ} \mathrm{C}\), and the skin, with a surface area of \(1.40 \mathrm{m}^{2},\) has a temperature of \(34.0^{\circ} \mathrm{C}\). (a) Find the rate of heat transfer out of the body under the following as sumptions: (i) The average thickness of tissue between the core and the skin is \(1.20 \mathrm{cm} ;\) (ii) the thermal conductivity of the tissue is that of water. (b) Without repeating the calculation of part (a), what rate of heat transfer would you expect if the skin temperature were to fall to \(31.0^{\circ} \mathrm{C}\) ? Explain.

Short Answer

Expert verified
(a) 210 W, (b) 420 W, doubling due to doubled temperature difference.

Step by step solution

01

Understanding Heat Transfer

We are asked to find the rate of heat transfer through the tissue. The heat transfer equation for conduction is \[ Q = \frac{kA(T_2 - T_1)}{d} \]where \(k\) is the thermal conductivity, \(A\) is the surface area, \(T_2 - T_1\) is the temperature difference, and \(d\) is the thickness of the material.
02

Calculating Heat Transfer for Part (a)

For part (a), the core temperature \(T_2\) is \(37.0^{\circ} \mathrm{C}\), and the skin temperature \(T_1\) is \(34.0^{\circ} \mathrm{C}\). The thermal conductivity \(k\) of water is approximately \(0.6 \mathrm{W/m\cdot K}\). The surface area \(A\) is \(1.40 \mathrm{m}^2\), and the thickness \(d\) is \(1.20 \mathrm{cm} = 0.012 \mathrm{m}\).Substitute these values into the formula:\[ Q = \frac{0.6 \times 1.40 \times (37.0 - 34.0)}{0.012} \]Solve this to find the rate \(Q\).
03

Solving Part (a)

Evaluating the expression from Step 2:\[ Q = \frac{0.6 \times 1.40 \times 3.0}{0.012} = \frac{2.52}{0.012} = 210 \mathrm{W} \]The rate of heat transfer out of the body is \(210 \mathrm{W}\).
04

Analyzing Changes in Part (b)

For part (b), the new skin temperature \(T_1\) is \(31.0^{\circ} \mathrm{C}\) while \(T_2\) remains the same at \(37.0^{\circ} \mathrm{C}\). The new temperature difference \(T_2 - T_1\) is \(37.0 - 31.0 = 6.0^{\circ} \mathrm{C}\). Since the only change is in the temperature difference, the rate of heat transfer will double because it is directly proportional to the temperature difference.
05

Calculating Rate for Part (b)

Given the doubled temperature difference, substitute back into the heat transfer formula:\[ Q' = \frac{0.6 \times 1.40 \times 6.0}{0.012} = \frac{5.04}{0.012} = 420 \mathrm{W} \]The rate of heat transfer for part (b) is \(420 \mathrm{W}\).

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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 measure of how well a material can conduct heat. It is represented by the symbol \( k \) and typically measured in units of watts per meter per kelvin (\( \mathrm{W/m \cdot K} \)). This property is intrinsic to materials, meaning that each material has a specific thermal conductivity value. For example, metals usually have high thermal conductivity, which allows them to transfer heat quickly, whereas insulators like rubber have low thermal conductivity.
In the context of human body heat transfer, thermal conductivity plays a crucial role in determining how effectively heat can be transferred from the body’s core to its skin. In this exercise, the thermal conductivity of the tissue is assumed to be like that of water, which is approximately \( 0.6 \mathrm{W/m \cdot K} \). This means that the tissue can conduct heat moderately well, facilitating the calculation of heat transfer rates.
  • Higher \( k \): More efficient heat transfer
  • Lower \( k \): Less efficient heat transfer
Understanding thermal conductivity allows us to predict how fast or slow heat will travel through different layers of material, which is essential in biomedical applications and thermal planning in engineering.
Conduction Equation
The conduction equation is an essential formula used to calculate the rate of heat transfer through a material. It is expressed as:
\[ Q = \frac{kA(T_2 - T_1)}{d} \]
where:
  • \( Q \) is the rate of heat transfer
  • \( k \) is the thermal conductivity of the material
  • \( A \) is the surface area through which heat is being transferred
  • \( T_2 - T_1 \) is the temperature difference across the material
  • \( d \) is the thickness of the material
This equation plays a pivotal role in numerous applications, including designing heat exchangers, creating comfortable environments in buildings, and, as seen in the problem, understanding heat transfer within the human body. By inserting the relevant parameters into the formula, we can determine how much heat energy is passing through a specific thickness of material over time.
In this exercise, by calculating \( Q \), students can monitor how changes in parameters like temperature difference or material properties affect the body's heat dispersion. This understanding is crucial in both academic research and practical fields like physiology and thermal engineering.
Temperature Gradient
The temperature gradient is an essential concept in heat transfer, referring to the rate at which temperature changes over a specific distance. It is generally expressed as temperature difference per unit length. In the conduction equation, the temperature gradient is represented by \( \frac{T_2 - T_1}{d} \), where \( T_2 \) and \( T_1 \) are the temperatures at two different positions, and \( d \) is the distance between those positions.
In the human body example, the temperature gradient is the difference between the core temperature \( 37.0^{\circ} \mathrm{C} \) and the skin temperature. The larger the temperature gradient, the faster heat will transfer from the hotter region to the cooler region because heat energy naturally moves from high to low temperature areas.
  • High temperature gradient: Faster heat transfer
  • Low temperature gradient: Slower heat transfer
A strong understanding of temperature gradients is vital in predicting how quickly heat moves through a material. Bodies adapt naturally by adjusting blood flow to manage temperature gradients, illustrating one of the many ways living organisms utilize these principles to maintain homeostasis.

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Most popular questions from this chapter

Two objects are made of the same material but have different temperatures. Object 1 has a mass \(m\) and object 2 has a mass \(2 m\). If the objects are brought into thermal contact, (a) is the temperature change of object 1 greater than, less than, or equal to the temperature change of object \(2 ?\) (b) Choose the best explanation from among the following: I. The larger object gives up more heat, and therefore its temperature change is greatest. II. The heat given up by one object is taken up by the other object. since the objects have the same heat capacity, the temperature changes are the same. III. One object loses heat of magnitude \(Q\), the other gains heat of magnitude Q. With the same magnitude of heat involved, the smaller object has the greater temperature change.

Brain Power As you read this problem, your brain is consuming about 22 W of power. (a) How many steps with a height of \(21 \mathrm{cm}\) must you climb to expend a mechanical energy equivalent to one hour of brain operation? (b) A typical human brain, which is \(77 \%\) water, has a mass of \(1.4 \mathrm{kg}\). Assuming that the \(22 \mathrm{W}\) of brain power is converted to heat, what temperature rise would you estimate for the brain in one hour of operation? Ignore the significant heat transfer that occurs between a human head and its surroundings, as well as the \(23 \%\) of the brain that is not water.

A pendulum consists of a large weight suspended by a steel wire that is \(0.9500 \mathrm{m}\) long. (a) If the temperature increases, does the period of the pendulum increase, decrease, or stay the same? Explain. (b) Calculate the change in length of the pendulum if the temperature increase is \(150.0 \mathrm{C}^{\circ}\) (c) Calculate the period of the pendulum before and after the temperature increase. (Assume that the coefficient of linear expansion for the wire is \(12.00 \times 10^{-6} \mathrm{K}^{-1}\) and that \(g=9.810 \mathrm{m} / \mathrm{s}^{2}\) at the location of the pendulum.)

Suppose you could convert the 525 Calories in the cheeseburger you ate for lunch into mechanical energy with \(100 \%\) efficiency. (a) How high could you throw a \(0.145-\mathrm{kg}\) baseball with the energy contained in the cheeseburger? (b) How fast would the ball be moving at the moment of release?

If heat is transferred to 150 g of water at a constant rate for \(2.5 \mathrm{min},\) its temperature increases by \(13 \mathrm{C}^{\circ} .\) When heat is transferred at the same rate for the same amount of time to a \(150-g\) object of unknown material, its temperature increases by \(61 \mathrm{C}^{\circ}\). (a) From what material is the object made? (b) What is the heating rate?

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