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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 C\(^\circ\)? The heat of vaporization of water at body temperature (37\(^\circ\)C) is \(2.42 \times 10{^6} J/kg\). The specific heat of a typical human body is 3480 J/kg \(\cdot\) K (see Exercise 17.25). (b) What volume of water must the man drink to replenish the evaporated water? Compare to the volume of a soft-drink can (355 cm\(^3\)).

Short Answer

Expert verified
To cool by 1°C, evaporate 0.1007 kg of water (100.7 cm³), less than a can's volume.

Step by step solution

01

Identify Heat Loss

First, calculate the amount of heat that needs to be removed from the man's body to reduce the temperature by 1.00°C. Use the formula for heat transfer: \[ Q = mc\Delta T \] where \( m \) is the mass of the man (70.0 kg), \( c \) is the specific heat capacity (3480 J/kg·K), and \( \Delta T \) is the temperature change (1.00°C or 1.00 K).
02

Calculate Heat to be Removed

Substitute the values into the formula: \[Q = (70.0\, \text{kg}) \times (3480\, \text{J/kg·K}) \times (1.00\, \text{K})\]Calculate \( Q \):\[ Q = 243,600\, \text{J} \] This is the heat that must be removed from the body.
03

Determine Mass of Water Evaporated

Find the mass of water that must evaporate using the heat of vaporization equation: \[ Q = m_{\text{water}} \cdot L_{\text{vapor}} \]where \( m_{\text{water}} \) is the mass of water evaporated and \( L_{\text{vapor}} = 2.42 \times 10^6 \) J/kg. Solve for \( m_{\text{water}} \):\[ m_{\text{water}} = \frac{Q}{L_{\text{vapor}}} = \frac{243,600\, \text{J}}{2.42 \times 10^6\, \text{J/kg}} \]Calculate \( m_{\text{water}} \):\[ m_{\text{water}} \approx 0.1007\, \text{kg} \]
04

Volume of Water Evaporated

Convert the mass of evaporated water to volume. Knowing the density of water is approximately 1 kg/L, \[ V = m_{\text{water}} \times 1\, \text{L/kg} = 0.1007\, \text{L} \] Convert liters to cubic centimeters:\[ V = 0.1007\, \text{L} \times 1000\, \text{cm}^3/\text{L} = 100.7\, \text{cm}^3 \]
05

Compare to Soft-Drink Can

Compare the calculated volume of evaporated water to the volume of a soft-drink can (355 cm³). The evaporated volume of 100.7 cm³ is much less than 355 cm³.

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Key Concepts

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

Evaporation
Evaporation is a fascinating natural process through which a liquid changes into a gas. It involves the transformation of molecules from the surface of a liquid to the gas phase. This transformation occurs when molecules in a liquid gain enough energy to break free from the surface and enter the atmosphere as vapor.
When it comes to temperature regulation in living organisms, evaporation plays an especially crucial role. For instance, many mammals use evaporation to cool down, especially in hot climates. Sweat evaporating from the skin removes excess heat, helping maintain a stable body temperature.
In terms of physics, when evaporation happens, energy is absorbed from the surroundings. This energy absorption results in cooling, making evaporation an efficient way for heat dissipation. In warm-blooded animals, like humans, this cooling mechanism is vital to prevent overheating and maintain homeostasis.
Specific Heat Capacity
The concept of specific heat capacity is critical in understanding how different substances respond to heat. It is defined as the quantity of heat required to change the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). The formula for calculating heat using specific heat capacity is given by:
\[ Q = mc riangle T \]
where \( Q \) is the heat added or removed, \( m \) is the mass of the substance, \( c \) is the specific heat capacity, and \( riangle T \) is the temperature change.
For example, the specific heat capacity of water is relatively high, meaning it can absorb a lot of heat with a small temperature increase. This property is why bodies of water and human bodies, which are largely composed of water, are effective at maintaining temperature stability.
Knowing the specific heat capacity of substances allows scientists and engineers to design systems for heating and cooling that take advantage of these properties. It is especially useful for calculating how much energy is required to heat or cool objects or for understanding natural processes such as climate dynamics.
Heat of Vaporization
The heat of vaporization refers to the amount of energy required to convert a unit mass of liquid into a gas at a constant temperature. This property is vital in understanding how evaporation functions in cooling processes.
Heat of vaporization is often expressed in units like joules per kilogram \((J/kg)\). When sufficient energy is applied to a liquid, molecules overcome intermolecular forces and transition into a gaseous state. For water at body temperature, the heat of vaporization is \( 2.42 \times 10^6 \text{ J/kg} \).
In practical terms, this means that evaporating water requires a significant amount of energy, which is extracted from surroundings, such as the human body during sweating. As a warm-blooded organism uses this process to cool down, understanding the heat of vaporization becomes essential for evaluating how much water and energy are involved in bodily cooling mechanisms. Understanding these principles is key in designing artificial cooling systems that mimic natural processes.
Temperature Regulation
Temperature regulation is a crucial physiological process for maintaining a consistent internal environment in living organisms. This results in optimal functioning of metabolic processes. Mechanisms involved encompass a range of physiological and behavioral responses.
One of the primary methods of temperature control in humans is through thermoregulation. This involves sweating and subsequent evaporation to dissipate excess body heat.
  • Physiological thermoregulation utilizes the skin, blood flow, and sweating.
  • Behaviorally, organisms may seek shade or alter activity levels.
Cooler blood from the skin is circulated back to the core, helping lower overall body temperature.
In engineering and design, principles of temperature regulation are utilized to develop systems like HVAC for controlled environments. By understanding how natural systems maintain homeostasis, better artificial systems can be engineered, leveraging the efficiency observed in biological organisms.

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Most popular questions from this chapter

A surveyor's 30.0-m steel tape is correct at 20.0\(^\circ\)C. The distance between two points, as measured by this tape on a day when its temperature is 5.00\(^\circ\)C, is 25.970 m. What is the true distance between the points?

One experimental method of measuring an insulating material's thermal conductivity is to construct a box of the material and measure the power input to an electric heater inside the box that maintains the interior at a measured temperature above the outside surface. Suppose that in such an apparatus a power input of 180 W is required to keep the interior surface of the box 65.0 C\(^\circ\) (about 120 F\(^\circ\)) above the temperature of the outer surface. The total area of the box is 2.18 m\(^2\), and the wall thickness is 3.90 cm. Find the thermal conductivity of the material in SI units.

CP While painting the top of an antenna 225 m in height, a worker accidentally lets a 1.00-L water bottle fall from his lunchbox. The bottle lands in some bushes at ground level and does not break. If a quantity of heat equal to the magnitude of the change in mechanical energy of the water goes into the water, what is its increase in temperature?

A laboratory technician drops a 0.0850-kg sample of unknown solid material, at 100.0\(^\circ\)C, into a calorimeter. The calorimeter can, initially at 19.0\(^\circ\)C, is made of 0.150 kg of copper and contains 0.200 kg of water. The final temperature of the calorimeter can and contents is 26.1\(^\circ\)C. Compute the specific heat of the sample.

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