/*! 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 98 Animals in cold climates often d... [FREE SOLUTION] | 91Ó°ÊÓ

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Animals in cold climates often depend on \(t w o\) layers of insulation: a layer of body fat (of thermal conductivity \(0.20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) surrounded by a layer of air trapped inside fur or down. We can model a black bear (Ursus americanus) as a sphere \(1.5 \mathrm{~m}\) in diameter having a layer of fat \(4.0 \mathrm{~cm}\) thick. (Actually, the thickness varies with the season, but we are interested in hibernation, when the fat layer is thickest.) In studies of bear hibernation, it was found that the outer surface layer of the fur is at \(2.7^{\circ} \mathrm{C}\) during hibernation, while the inner surface of the fat layer is at \(31.0^{\circ} \mathrm{C}\). (a) What is the temperature at the fat-inner fur boundary so that the bear loses heat at a rate of \(50.0 \mathrm{~W} ?\) (b) How thick should the air layer (contained within the fur) be?

Short Answer

Expert verified
The temperature at the fat-inner fur boundary is approximately \(14.6^{\circ}C\), and the thickness of the air layer should be approximately \(14.1cm\).

Step by step solution

01

- Determine the surface area of the bear

The bear can be modeled as a sphere of diameter 1.5 m. The surface area of a sphere is given by \(A = 4 \pi r^2\), where \(r\) is the radius of the sphere. Hence, the surface area can be calculated as \(A = 4 \pi (0.75)^2 = 7.07 m^2\).
02

- Determine the temperature at the fat-inner fur boundary

Given that the outer surface of the fur layer is at temperature \(T_{2}=2.7^{\circ}C\), the heat rate is \(Q=50.0W\), the thermal conductivity of fat is \(k=0.20W/mK\), and the thickness of the fat layer is \(d=0.04m\), and using the formula for heat transfer mentioned earlier, we can solve for \(T_{1}\) which is the temperature at the fat-inner fur boundary. Rearranging the formula for \(T_{1}\), we get:\(T_{1} = T_{2} + \frac{Qd}{kA}\)So, substituting in the values, we get \(T_{1} = 2.7 + \frac{(50)(0.04)}{(0.20)(7.07)}\)Upon performing the calculation, we get: \(T_{1} = 14.6^{\circ}C\)
03

- Determine the thickness of the air layer

Given that the inner surface of the fat layer is at temperature \(T_{3}= 31.0^{\circ}C\), the temperature at the fat-inner fur boundary is \(T_{1} = 14.6^{\circ}C\), the heat rate is \(Q = 50.0W\), and the thermal conductivity of air is \(k = 0.026W/mK\), we can solve for \(d\) which is the thickness of the air layer using the formula for heat transfer mentioned earlier. Rearranging the formula for \(d\), we get:\(d = \frac{Q(T_{3}-T_{1})}{kA}\)So, substituting the values, we get:\(d = \frac{(50)(31.0 - 14.6)}{(0.026)(7.07)}\)Upon performing the calculation, we get:\(d = 0.141m\) or \(14.1cm\).

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Key Concepts

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

Thermal Conductivity in Animals
Thermal conductivity refers to a material's ability to conduct heat. In biological contexts, particularly for animals in cold climates, it has significant implications for survival. Animals have developed insulating layers, such as fat and fur, that exhibit varying degrees of thermal conductivity to regulate temperature efficiently.

For instance, when we consider a black bear during hibernation, the bear's body fat has a relatively low thermal conductivity of 0.20 W/mK, implying that it is a good insulator. Low thermal conductivity materials slow the rate of heat transfer, which is critical for conserving energy and maintaining body heat in cold conditions.

The calculation of the temperature at the fat-inner fur boundary involves considering the thermal properties of the insulating layers, the surface area through which the heat is lost, and the constant heat loss rate. This balance helps animals maintain a safe and sustainable internal temperature even when external temperatures are extremely low.
Hibernation Thermoregulation
During hibernation, animals such as bears undergo physiological changes that allow them to conserve energy. The core principle of hibernation thermoregulation is to minimize metabolic rate and hence heat production to conserve energy for the long period of dormancy. While the metabolic rate drops, maintaining a stable internal temperature becomes essential for the animal's survival.

Insulation plays a crucial role in this process. As we see from our example, the bear has a thick layer of fat and a layer of air trapped in its fur, both of which are excellent insulators. By manipulating the thickness of these layers and the material properties such as thermal conductivity, the bear achieves a delicate balance where it loses heat at a steady, sustainable rate of 50.0W during hibernation.
Heat Loss in Cold Climates
Heat loss in cold climates is a critical challenge for warm-blooded animals. To minimize it, these animals rely on behavioral and physiological adaptations. The properties of fat, fur, and trapped air layers are all leveraged to reduce the rate of heat escape to the cold environment.

In our model, the thickness of the air layer within the fur can be determined by relating the heat loss rate, the temperature gradient between the internal and external environments, and the thermal conductivity of air. The calculated thickness serves as an adaptation feature. Thicker layers of trapped air within fur can provide better insulation due to the static nature of air, which is a poor heat conductor. This physical adaptation, combined with the natural behavior of seeking shelter or curling up to reduce exposed surface area, helps animals survive freezing temperatures.

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Most popular questions from this chapter

Size of a Light-Bulb Filament. The operating temperature of a tungsten filament in an incandescent light bulb is \(2450 \mathrm{~K},\) and its emissivity is \(0.350 .\) Find the surface area of the filament of a \(150 \mathrm{~W}\) bulb if all the electrical energy consumed by the bulb is radiated by the filament as electromagnetic waves. (Only a fraction of the radiation appears as visible light.)

Evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. (a) What mass of water must evaporate from the skin of a \(70.0 \mathrm{~kg}\) man to cool his body \(1.00 \mathrm{C}^{\circ}\) ? The heat of vaporization of water at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) The specific heat of a typical human body is \(3480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) (see Exercise 17.27 ) . (b) What volume of water must the man drink to replenish the evaporated water? Compare to the volume of a soft-drink can \(\left(355 \mathrm{~cm}^{3}\right)\).

A U.S. penny has a diameter of \(1.9000 \mathrm{~cm}\) at \(20.0^{\circ} \mathrm{C}\). The coin is made of a metal alloy (mostly zinc) for which the coefficient of linear expansion is \(2.6 \times 10^{-5} \mathrm{~K}^{-1}\). What would its diameter be on a hot day in Death Valley \(\left(48.0^{\circ} \mathrm{C}\right) ?\) On a cold night in the mountains of Greenland \(\left(-53^{\circ} \mathrm{C}\right) ?\)

Derive an equation that gives \(T_{\mathrm{K}}\) as a function of \(T_{\mathrm{F}}\) to the nearest hundredth of a degree. Solve the equation and thereby obtain an equation for \(T_{\mathrm{F}}\) as a function of \(T_{\mathrm{K}}\).

To measure the specific heat in the liquid phase of a newly developed cryoprotectant, you place a sample of the new cryoprotectant in contact with a cold plate until the solution's temperature drops from room temperature to its freezing point. Then you measure the heat transferred to the cold plate. If the system isn't sufficiently isolated from its roomtemperature surroundings, what will be the effect on the measurement of the specific heat? (a) The measured specific heat will be greater than the actual specific heat; (b) the measured specific heat will be less than the actual specific heat; (c) there will be no effect because the thermal conductivity of the cryoprotectant is so low; (d) there will be no effect on the specific heat, but the temperature of the freezing point will change.

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