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Chapter 17: Problem 47

\(Camels require very little }}\) water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to \(34.0^{\circ} \mathrm{C}\) overnight and rise to \(40.0^{\circ} \mathrm{C}\) during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a \(400 \mathrm{~kg}\) camel would have to drink if it attempted to keep its body temperature at a constant \(34.0^{\circ} \mathrm{C}\) by evaporation of sweat during the day ( 12 hours) instead of letting it rise to \(40.0^{\circ} \mathrm{C}\). (Note: The specific heat of a camel or other mammal is about the same as that of a typical human, \(3480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). The heat of vaporization of water at \(34^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) )

Short Answer

Expert verified
We use the calculation from Step 3 to provide a numerical answer in terms of liters of water.

Step by step solution

01

Calculate the total change in energy

Firstly, we calculate the total heat change in the body of the camel using the formula q = mc(delta)T, where 'q' is the total heat, 'm' is the mass of the camel, 'c' is the specific heat capacity and '(delta)T' is the change in temperature. Given mass of camel is 400 kg, c = 3480 J/kg.K and temperature change (delta)T = 40.0掳C - 34.0掳C = 6.0掳C or 6.0 K, plugging these values into our formula gives, q = 400 kg * 3480 J/kg.K * 6.0 K.
02

Calculate the amount of water needed

The camel can keep the body temperature constant by evaporating water from its body (sweating). The heat q absorbed by water during evaporation is given by the formula q = mL, where 'm' is the mass of water and 'L' is the heat of evaporation. To find out the amount of water needed we rearrange the equation to m = q/L. Substituting L with the given heat of vaporization of water at 34掳C, 2.42 x 10^6 J/kg, and q with the result from Step 1.
03

Convert mass of water to liters

The calculated mass of water in Step 2 is in kg, to convert it to liters we must remember that the density of water is approximately 1 kg/L. So, a mass of water in kg will be approximately the same volume in liters.

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Key Concepts

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

Body Temperature Regulation
Animals have various strategies for maintaining their body temperature within a certain range. This ability is crucial since enzymatic and metabolic processes in animals can be highly sensitive to temperature changes. In endotherms, such as mammals and birds, body temperature is regulated through metabolic heat production and various cooling mechanisms.

  • Endotherms often maintain a stable body temperature, even when the temperature outside varies. Humans, for example, keep their core temperature around 37掳C.
  • Mechanisms such as shivering increase heat production, while sweating and panting facilitate cooling.
However, animals like camels are able to tolerate wider fluctuations in their body temperature with minimal physiological stress. This flexibility allows camels to conserve water, especially in arid environments.
Specific Heat Capacity
Specific heat capacity is a property that defines how much heat energy is required to change the temperature of a substance by a certain amount. Essentially, it tells us how resistant a material is to temperature changes. It is expressed in Joules per kilogram per Kelvin (J/kg路K).

In biological terms, the specific heat capacity of an organism affects how quickly it can change its temperature in response to environmental shifts.
  • For the camel, the specific heat capacity is similar to that of humans, about 3480 J/kg路K.
  • This value indicates the amount of heat needed to raise one kilogram of the body's mass by one degree.
Understanding this property is essential when calculating the total heat loss or gain in an animal鈥檚 body, such as when determining how much energy a camel uses or saves when its body temperature changes.
Heat of Vaporization
The heat of vaporization refers to the amount of energy required to convert a substance from a liquid to a vapor without changing its temperature. This property is particularly important in the context of body temperature regulation through evaporation, such as sweating in mammals.

  • For water at 34掳C, the heat of vaporization is 2.42 x 106 J/kg.
  • This indicates how much energy per kilogram is absorbed when water vaporizes and carries away heat from the surface.
In practice, animals use this endothermic process to dissipate heat. To maintain a consistent and cooler body temperature, an organism like the camel might rely on the evaporation of sweat. This process works efficiently in hot climates, leading to significant water conservation when temperature fluctuations are tolerated.
Camel Physiology
Camels are incredible animals adapted to survive in harsh, arid environments. Their physiological adaptations allow them to handle extreme fluctuations in temperature and conserve water efficiently.
  • Camels can tolerate a body temperature range from 34掳C at night to 40掳C during the day.
  • This ability minimizes the need for constant water intake to cool down through sweating.
During the day, as temperatures rise, camels allow their body temperature to increase, which reduces water loss from sweating compared to maintaining a constant lower temperature.

These adaptations are part of a broader strategy that includes conserving water by having a highly efficient kidney and maintaining a thick fur coat that insulates against high temperatures and radiation. Such physiological traits make camels well-suited to their desert homeland, illustrating the intricate balance between water conservation and temperature regulation in animal physiology.

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Most popular questions from this chapter

One end of an insulated metal rod is maintained at \(100.0^{\circ} \mathrm{C}\), and the other end is maintained at \(0.00^{\circ} \mathrm{C}\) by an ice-water mixture. The rod is \(60.0 \mathrm{~cm}\) long and has a cross-sectional area of \(1.25 \mathrm{~cm}^{2}\). The heat conducted by the rod melts \(8.50 \mathrm{~g}\) of ice in \(10.0 \mathrm{~min} .\) Find the thermal conductivity \(k\) of the metal.

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 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.

One suggested treatment for a person who has suffered a stroke is immersion in an ice-water bath at \(0^{\circ} \mathrm{C}\) to lower the body temperature, which prevents damage to the brain. In one set of tests, patients were cooled until their internal temperature reached \(32.0^{\circ} \mathrm{C}\). To treat a \(70.0 \mathrm{~kg}\) patient, what is the minimum amount of ice (at \(0^{\circ} \mathrm{C}\) ) you need in the bath so that its temperature remains at \(0^{\circ} \mathrm{C} ?\) The specific heat of the human body is \(3480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{C}^{\circ},\) and recall that normal body temperature is \(37.0^{\circ} \mathrm{C}\).

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?
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