91Ó°ÊÓ

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\(^\circ\)C overnight and rise to 40.0\(^\circ\)C during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a 400-kg camel would have to drink if it attempted to keep its body temperature at a constant 34.0\(^\circ\)C by evaporation of sweat during the day (12 hours) instead of letting it rise to 40.0\(^\circ\)C. (Note: The specific heat of a camel or other mammal is about the same as that of a typical human, 3480 J/kg \(\cdot\) K. The heat of vaporization of water at 34\(^\circ\)C is \(2.42 \times 10{^6} J/kg\).)

Step by step solution

01

Understand the Problem

We need to calculate the amount of water a camel would have to drink to maintain its body temperature at 34°C instead of allowing it to rise to 40°C. This involves calculating the heat energy that must be dissipated (by evaporation of sweat) to prevent this temperature rise.

02

Calculate the Temperature Change

The camel's body temperature changes from 34°C to 40°C. This is a temperature change of \[\Delta T = 40.0 - 34.0 = 6.0 \text{ °C} \].

03

Calculate the Heat to be Removed

The heat energy, \( Q \), that needs to be removed to maintain the temperature is given by the formula: \[ Q = mc\Delta T \]where \( m \) is the mass of the camel (400 kg),\( c \) is the specific heat capacity (3480 J/kg \cdot \text{K}),and \( \Delta T \) is the temperature change (6.0°C or 6.0 K).Thus, \[ Q = 400 \times 3480 \times 6 = 8,352,000 \text{ J} \].

04

Calculate the Mass of Water Needed

To find the mass of water required for evaporation, we use the formula:\[ Q = m_w \cdot L \] where \( m_w \) is the mass of water to be evaporated and \( L \) is the latent heat of vaporization (\(2.42 \times 10^6 J/kg\)). Rearrange to find \( m_w \): \[ m_w = \frac{Q}{L} \] \[ m_w = \frac{8,352,000}{2.42 \times 10^6} \approx 3.45 \text{ kg} \].

05

Convert Mass of Water to Liters

Since the density of water is approximately 1 kg/L, the volume in liters is equal to the mass in kilograms. Therefore, \[ V = 3.45 \text{ L} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Specific Heat

Specific heat is a concept in thermodynamics that describes how much heat energy a specific substance can absorb before its temperature rises. In this exercise, it's all about understanding how well the camel regulates its body temperature. The specific heat of the camel is given as 3480 J/kg \( \cdot \) K, similar to humans. This means for every kilogram of the camel's body, it requires 3480 Joules of energy to increase the temperature by 1 Kelvin (or 1 degree Celsius).
This property is crucial when understanding heat management because it dictates how much energy the camel can absorb or release to maintain or alter its body temperature.

For example:

• Specific heat allows us to calculate how much energy is needed to change the camel's body temperature by a certain amount. In our case, it's the rise from 34°C to 40°C.
• Knowing the camel's mass (400 kg), we can calculate the energy required for this temperature change using the formula: \[ Q = mc \Delta T \]
• So, it's about how the camel absorbs and dissipates heat during temperature fluctuations throughout the day and night.

Latent Heat of Vaporization

Latent heat of vaporization is the energy required for a substance to change from liquid to gas without temperature change. In our exercise, this concept is key for understanding how camels use sweat evaporation for cooling.
The amount needed given in the problem, which is \(2.42 \times 10^6\) J/kg, tells us how much energy is needed to completely evaporate 1 kg of water at 34°C. Evaporation absorbs enormous energy, hence cooling the camel's body efficiently.
This is how it works step-by-step:

• When the temperature rises, the camel sweats, and the sweat evaporates, taking away heat.
• This heat removal is quantified by the latent heat of vaporization. The energy calculation \[ Q = m_w \cdot L \] tells us how much energy is used during the evaporation of a calculated mass \( m_w \) of water.
• Therefore, knowing this value helps us estimate how much water is turned into vapor to achieve the desired cooling effect.

Without this efficient cooling process, camels would require significantly more water to control their body temperature during hot days.

Temperature Regulation

In the world of thermoregulation in animals, specifically heat and latent heat of vaporization work together to allow camels to survive harsh conditions.
Temperature regulation is not just about keeping a constant temperature; it's about strategically letting the body temperature vary to minimize water loss and using stored heat effectively. Camels have adapted by allowing their temperature to climb during the day and drop at night, reducing the need for constant water consumption.
This strategy involves three main processes:

• Heat absorption: Due to high specific heat, camels can store more heat, delaying reaching peak temperatures.
• Variable tolerance: They let their body temperature naturally fluctuate, reducing the need to sweat excessively.
• Cooling: When needed, through evaporation using the latent heat of vaporization, they effectively dissipate stored heat.

This smart adaptation helps conserve water, which is vital for survival in desert environments, and explains why they thrive where other animals can't.