Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 \(2.42 \times 10{^6} J/kg\).)

Short Answer

Expert verified
The camel needs to drink about 3.45 liters of water.

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 Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Convert the following Celsius temperatures to Fahrenheit: (a) -62.8\(^\circ\)C, the lowest temperature ever recorded in North America (February 3, 1947, Snag, Yukon); (b) 56.7\(^\circ\)C, the highest temperature ever recorded in the United States (July 10, 1913, Death Valley, California); (c) 31.1\(^\circ\)C, the world’s highest average annual temperature (Lugh Ferrandi, Somalia).

Two rods, one made of brass and the other made of copper, are joined end to end. The length of the brass section is 0.300 m and the length of the copper section is 0.800 m. Each segment has cross-sectional area 0.00500 m\(^2\). The free end of the brass segment is in boiling water and the free end of the copper segment is in an ice–water mixture, in both cases under normal atmospheric pressure. The sides of the rods are insulated so there is no heat loss to the surroundings. (a) What is the temperature of the point where the brass and copper segments are joined? (b) What mass of ice is melted in 5.00 min by the heat conducted by the composite rod?

Convert the following Kelvin temperatures to the Celsius and Fahrenheit scales: (a) the midday temperature at the surface of the moon (400 K); (b) the temperature at the tops of the clouds in the atmosphere of Saturn (95 K); (c) the temperature at the center of the sun \((1.55 \times 10{^7} K)\).

A machinist bores a hole of diameter 1.35 cm in a steel plate that is at 25.0\(^\circ\)C. What is the cross-sectional area of the hole (a) at 25.0\(^\circ\)C and (b) when the temperature of the plate is increased to 175\(^\circ\)C? Assume that the coefficient of linear expansion remains constant over this temperature range.

Styrofoam bucket of negligible mass contains 1.75 kg of water and 0.450 kg of ice. More ice, from a refrigerator at -15.0\(^\circ\)C, is added to the mixture in the bucket, and when thermal equilibrium has been reached, the total mass of ice in the bucket is 0.884 kg. Assuming no heat exchange with the surroundings, what mass of ice was added?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free