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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 \(\left(12 \text { hours) instead of letting it rise to } 40.0^{\circ} \mathrm{C} \text { . (Note: The }\right.\) 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 \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\) )

Step by step solution

01

Understand the Problem

To solve this problem, we must calculate the amount of heat absorbed when the camel's temperature rises naturally versus trying to maintain a constant body temperature by evaporating sweat.

02

Calculate Heat Absorbed Without Evaporation

First, calculate the heat absorbed due to a temperature rise of the camel's body from \(34^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\). Use the specific heat formula: \(Q = mc\Delta T\), where \(m = 400 \, \mathrm{kg}\), \(c = 3480 \, \mathrm{J/kg \cdot K}\), and \(\Delta T = 6^{\circ} \mathrm{C}\).

03

Plug Values Into the Equation

Substitute the values into the equation: \[ Q = 400 \, \mathrm{kg} \times 3480 \, \mathrm{J/kg \cdot K} \times 6 \, \mathrm{K} = 8352000 \, \mathrm{J}. \]

04

Calculate Water Needed for Evaporation

To calculate how much water is needed to absorb this amount of heat through evaporation, use the formula for heat of vaporization: \(Q = mL\), where \(L = 2.42 \times 10^6 \, \mathrm{J/kg}\). Rearrange to find \(m\): \[ m = \frac{Q}{L} = \frac{8352000 \, \mathrm{J}}{2.42 \times 10^6 \, \mathrm{J/kg}} \approx 3.45 \, \mathrm{kg}. \]

05

Convert Mass to Volume

Since the density of water is approximately 1 \(\mathrm{kg/L}\), 3.45 kg of water is equivalent to 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 a fundamental concept in thermoregulation of animals, including camels. It describes the amount of heat required to raise the temperature of 1 kilogram of a substance by 1 degree Celsius. For camels, as well as humans, this value is around 3480 J/kg·K. Understanding this concept helps explain why camels can regulate their body temperature with minimal water loss. The specific heat capacity determines the amount of energy needed to change the camel's temperature. When the camel's body temperature changes by a few degrees, a significant amount of heat energy is involved due to their high specific heat capacity. In practice, this means camels absorb and release a lot of heat as their temperature fluctuates from 34°C to 40°C, without requiring large quantities of water to stabilize their temperature. This is crucial for their survival in arid environments where water is scarce.

Heat of Vaporization

Heat of vaporization is the energy required to convert a unit mass of a liquid into vapor without changing its temperature. For water, this amount is quite substantial, specifically, 2.42 x 10^6 J/kg at 34°C. This property is essential for understanding how camels could potentially regulate their temperature through evaporation. When water evaporates from the camel's skin, it absorbs a large amount of heat from the body, effectively cooling it down. This process utilizes the heat of vaporization. In the arithmetic of thermoregulation, by allowing their temperature to naturally rise, camels save this significant energy amount they would otherwise need for evaporating water to cool down. Hence, the heat of vaporization highlights the energy efficiency in a camel's strategy to conserve water.

Evaporation and Temperature Regulation

Evaporation is a key mechanism for temperature regulation, especially for animals like camels living in harsh, dry climates. When water evaporates from a surface, it requires heat energy, which is drawn from the surrounding area, thereby producing a cooling effect. This is why sweating is a common cooling method among many animals, including humans. For camels, regulating temperature by allowing their body temperature to fluctuate reduces the need for sweating. If camels kept a stable temperature of 34°C, they would need to evaporate enough sweat to account for a rise in internal heat of 8352000 J over a single day. Allowing their body temperature to vary reduces the need for such significant evaporative cooling. This adaptation frees camels from the necessity to find water regularly, emphasizing how controlling body temperature through selective evaporation aids in water conservation.