/*! 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 53 "The Ship of the Desert." Camels... [FREE SOLUTION] | 91Ó°ÊÓ

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"The Ship of the Desert." 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
The camel would need to drink approximately 3.45 liters of water.

Step by step solution

01

Understand the Problem

The camel allows its body temperature to vary from \(34.0^{\circ} \mathrm{C}\) to \(40.0^{\circ} \mathrm{C}\). If we want to keep it constantly at \(34.0^{\circ} \mathrm{C}\), the camel would need to evaporate water (sweat) to dissipate the excess heat. The task is to calculate how much water would be required for this process.
02

Calculate the Heat Required to Keep Constant Temperature

First, find the amount of thermal energy that the camel needs to dissipate if its temperature rises by \(6.0^{\circ} \mathrm{C}\) (from \(34.0^{\circ} \mathrm{C}\) to \(40.0^{\circ} \mathrm{C}\)). Use the formula: \[ Q = mc\Delta T \]where \( m = 400 \text{ kg} \) is the mass of the camel, \( c = 3480 \text{ J/kg} \cdot \text{K} \) is the specific heat capacity, and \( \Delta T = 6.0 \text{ }^{\circ} \mathrm{C} \). Substitute these values:\[ Q = 400 \times 3480 \times 6 = 8,352,000 \text{ J} \]
03

Calculate the Mass of Water Needed

Now, use the heat of vaporization for water to find the mass of water that needs to be evaporated to dissipate \(8,352,000 \text{ J}\) of heat. The formula is:\[ Q = m_v L_v \]where \( m_v \) is the mass of the evaporated water, and \( L_v = 2.42 \times 10^6 \text{ J/kg} \) is the heat of vaporization of water.Rearranging for \( m_v \):\[ m_v = \frac{Q}{L_v} = \frac{8,352,000}{2.42 \times 10^6} \approx 3.45 \text{ kg} \]
04

Convert the Mass of Water to Liters

Since 1 liter of water has a mass of 1 kg, the camel would need approximately 3.45 kg of water, which is 3.45 liters.

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Key Concepts

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

specific heat capacity
The specific heat capacity is an essential concept when understanding how substances respond to temperature changes. Specific heat capacity is the amount of heat energy required to raise the temperature of 1 kilogram of a substance by 1 degree Celsius (\(1 \text{ kg} \cdot \text{K}\)). It is a measure of a substance's thermal inertia or resistance to temperature change.

In the context of a camel, which has a similar specific heat capacity to humans at 3480 \(\text{J/km} \cdot \text{K}\), it measures how well camels can store and manage heat within their bodies.
  • Higher specific heat capacity means the animal can absorb a lot of heat without a significant increase in temperature.
  • This ability helps camels in hot environments to avoid overheating as their body can store more heat throughout the day without sweating.
This capability is vital in arid environments, as it greatly reduces water loss through sweat, which is crucial for survival in water-scarce regions.

Understanding specific heat capacity can offer insights into how animals regulate their internal temperatures, particularly in extreme environments, by efficiently managing heat energy without excessive water loss. This is critical in understanding thermoregulation strategies in different species.
evaporation and cooling
Evaporation is a key process in thermoregulation for many animals, including camels. When water changes from liquid to vapor, it takes in energy, which results in a cooling effect on the surface from which it evaporates.

This process is known as evaporative cooling and is essential in maintaining a stable internal temperature.
  • For camels, if maintaining constant body temperature through sweating, the heat (\(Q\)) dissipated through evaporation equals the energy needed to increase the body temperature.
  • The heat of vaporization of water at the camel's body temperature (\(34^{\circ} \mathrm{C}\)) is 2.42 × 10^6 \(\text{J/kg}\), representing the energy for each kilogram of water that evaporates to cool the body.
Through this method, camels could use significant amounts of water to shed excess heat if they did not allow their temperature to fluctuate.

The reliance on evaporative cooling demonstrates an elegant balance between temperature regulation and water conservation, a delicate trade-off managed by camels in desert climates.
camel physiology
Camel physiology is fascinating and well-adapted for survival in the harsh, arid desert environment. One remarkable feature is their ability to tolerate significant fluctuations in body temperature.

Camels can allow their body temperature to drop to \(34^{\circ} \mathrm{C}\) and rise to \(40^{\circ} \mathrm{C}\) in response to extremes in environmental temperature. This temperature range helps conserve water by reducing the need for evaporation through sweating.
  • This physiological adaptation means camels don’t have to sweat excessively, conserving vital water resources.
  • Camels also have other adaptations, such as thick fur insulating against heat and sandstorms while reducing water loss.
Their thick coats reflect sunlight and provide insulation, and their unique fat storage in humps also plays a role in water conservation, as metabolizing fat yields water.

By allowing body temperature to vary, camels maximize their scarce resources and adapt to environments where other animals might not survive. This efficient system goes beyond simple cooling; it is a holistic approach to thriving in one of the world’s most challenging climates.

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Most popular questions from this chapter

Hot Air in a Physics Lecture. (a) A typical student listening attentively to a physics lecture has a heat output of 100 \(\mathrm{W}\) . How much heat energy does a class of 90 physics students release into a lecture hall over the course of a 50 -min lecture? (b) Assume that all the heat energy in part (a) is transferred to the 3200 \(\mathrm{m}^{3}\) of air in the room. The air has specific heat 1020 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) and density 1.20 \(\mathrm{kg} / \mathrm{m}^{3} .\) If none of the heat escapes and the air conditioning system is off, how much will the temperature of the air in the room rise during the 50 -min lecture? (c) If the class is taking an exam, the heat output per student rises to 280 \(\mathrm{W} .\) What is the temperature rise during 50 min in this case?

A nail driven into a board increases in temperature. If we assume that 60\(\%\) of the kinetic energy delivered by a 1.80-kg hammer with a speed of 7.80 \(\mathrm{m} / \mathrm{s}\) is transformed into heat that flows into the nail and does not flow out, what is the temperature increase of an \(8.00-\mathrm{g}\) aluminum nail after it is struck ten times?

In a container of negligible mass, 0.200 \(\mathrm{kg}\) of ice at an initial temperature of \(-40.0^{\circ} \mathrm{C}\) is mixed with a mass \(m\) of water that has an initial temperature of \(80.0^{\circ} \mathrm{C}\) . No heat is lost to the surroundings. If the final temperature of the system is \(20.0^{\circ} \mathrm{C},\) what is the mass \(m\) of the water that was initially at \(80.0^{\circ} \mathrm{C} ?\)

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.

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-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 17.31\() .\) (b) What volume of water must the man drink to replenish the evaporated water? Compare to the volume of a soft-drink can \(\left(355 \mathrm{cm}^{3}\right)\).

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