/*! 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

(a) Normal body temperature. The average normal body temperature measured in the mouth is \(310 \mathrm{~K}\). What would Celsius and Fahrenheit thermometers read for this temperature? (b) Elevated body temperature. During very vigorous exercise, the body's temperature can go as high as \(40^{\circ} \mathrm{C}\). What would Kelvin and Fahrenheit thermometers read for this temperature? (c) Temperature difference in the body. The surface temperature of the body is normally about \(7 \mathrm{C}^{\circ}\) lower than the internal temperature. Express this temperature difference in kelvins and in Fahrenheit degrees. (d) Blood storage. Blood stored at \(4.0^{\circ} \mathrm{C}\) lasts safely for about 3 weeks, whereas blood stored at \(-160^{\circ} \mathrm{C}\) lasts for 5 years. Express both temperatures on the Fahrenheit and Kelvin scales. (e) Heat stroke. If the body's temperature is above \(105^{\circ} \mathrm{F}\) for a prolonged period, heat stroke can result. Express this temperature on the Celsius and Kelvin scales.

An electric kitchen range has a total wall area of \(1.40 \mathrm{~m}^{2}\) and is insulated with a layer of fiberglass \(4.00 \mathrm{~cm}\) thick. The inside surface of the fiberglass has a temperature of \(175^{\circ} \mathrm{C},\) and its outside surface is at \(35.0^{\circ} \mathrm{C}\). The fiberglass has a thermal conductivity of \(0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) What is the heat current through the insulation, assuming it may be treated as a flat slab with an area of \(1.40 \mathrm{~m}^{2}\) ? (b) What electric-power input to the heating element is required to maintain this temperature?

A carpenter builds a solid wood door with dimensions \(2.00 \mathrm{~m} \times 0.95 \mathrm{~m} \times 5.0 \mathrm{~cm} .\) Its thermal conductivity is \(k=0.120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The air films on the inner and outer surfaces of the door have the same combined thermal resistance as an additional \(1.8 \mathrm{~cm}\) thickness of solid wood. The inside air temperature is \(20.0^{\circ} \mathrm{C},\) and the outside air temperature is \(-8.0^{\circ} \mathrm{C}\). (a) What is the rate of heat flow through the door? (b) By what factor is the heat flow increased if a window \(0.500 \mathrm{~m}\) on a side is inserted in the door? The glass is \(0.450 \mathrm{~cm}\) thick, and the glass has a thermal conductivity of \(0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The air films on the two sides of the glass have a total thermal resistance that is the same as an additional \(12.0 \mathrm{~cm}\) of glass.

You are making pesto for your pasta and have a cylindrical measuring cup \(10.0 \mathrm{~cm}\) high made of ordinary glass \(\left[\beta=2.7 \times 10^{-5}\left(\mathrm{C}^{\circ}\right)^{-1}\right]\) that is filled with olive oil \(\left[\beta=6.8 \times 10^{-4}\left(\mathrm{C}^{\circ}\right)^{-1}\right]\) to a height of \(3.00 \mathrm{~mm}\) below the top of the cup. Initially, the cup and oil are at room temperature \(\left(22.0^{\circ} \mathrm{C}\right)\). You get a phone call and forget about the olive oil, which you inadvertently leave on the hot stove. The cup and oil heat up slowly and have a common temperature. At what temperature will the olive oil start to spill out of the cup?

A \(\mathrm 500.0 \mathrm{~g}\) chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating Styrofoam beaker containing \(1.00 \mathrm{~kg}\) of water at room temperature \(\left(20.0^{\circ} \mathrm{C}\right) .\) After waiting and gently stirring for 5.00 minutes, you observe that the water's temperature has reached a constant value of \(22.0^{\circ} \mathrm{C}\). (a) Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specific heat of the metal? (b) Which is more useful for storing thermal energy: this metal or an equal weight of water? Explain. (c) If the heat absorbed by the Styrofoam actually is not negligible, how would the specific heat you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain.

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