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Chapter 17: Problem 54

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 \(\left(355 \mathrm{cm}^{3}\right)\).

Short Answer

Expert verified
Around 0.101 kg of water must evaporate; this is about 101 cm鲁, or 1/3 of a soft-drink can volume.

Step by step solution

01

Determine Heat Needed to Cool the Body

To calculate the heat needed to cool the body, we use the equation for heat transfer: \( Q = mc\Delta T \), where \( m \) is the mass, \( c \) is the specific heat, and \( \Delta T \) is the temperature change. Here, \( m = 70 \text{ kg} \), \( c = 3480 \text{ J/(kg}\cdot\text{K)} \), and \( \Delta T = 1.00 \text{ } ^{\circ} \text{C} \). Thus, \( Q = 70 \times 3480 \times 1 = 243,600 \text{ J} \).
02

Calculate Mass of Water Needed to Evaporate

Since vaporization requires energy, we use the equation \( Q = m_v L_v \), where \( m_v \) is the mass of water that must evaporate, and \( L_v \) is the latent heat of vaporization. From part (a), \( Q = 243,600 \text{ J} \) and \( L_v = 2.42 \times 10^6 \text{ J/kg} \). Solving for \( m_v \), we find \( m_v = \frac{243,600}{2.42 \times 10^6} \approx 0.101 \text{ kg} \).
03

Determine Volume of Water Evaporated

To find the volume of the evaporated water, use the density of water \( \rho = 1000 \text{ kg/m}^3 \). The volume \( V \) is given by \( V = \frac{m_v}{\rho} \). So, \( V = \frac{0.101 \text{ kg}}{1000 \text{ kg/m}^3} = 0.000101 \text{ m}^3 \).
04

Convert Volume to Compare with a Soft-Drink Can

Convert \( 0.000101 \text{ m}^3 \) to \( \text{cm}^3 \), knowing that \( 1 \text{ m}^3 = 10^6 \text{ cm}^3 \). Thus, \( 0.000101 \text{ m}^3 = 101 \text{ cm}^3 \). Now compare this to a typical soft-drink can volume of \( 355 \text{ cm}^3 \). It takes roughly 1/3 of a soft drink can to replenish the evaporated water.

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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 where a liquid turns into a gas. This phenomenon occurs when molecules at the surface of a liquid gain enough energy to break free from the liquid and enter the air. Think of a puddle on a sunny day slowly disappearing as water molecules transition to water vapor in the air.
Evaporation plays a crucial role in thermoregulation in warm-blooded animals. When we sweat, the water from our skin evaporates into the air, taking heat with it. This process helps cool our bodies.
  • Evaporation occurs at the surface of the liquid.
  • Molecules need energy to break intermolecular bonds.
  • Evaporation can happen at any temperature.
This passive cooling mechanism is what keeps us comfortable on hot days.
Latent Heat of Vaporization
The latent heat of vaporization is the energy required to transform a unit mass of a liquid into a gas at a constant temperature. For water, this is a considerable energy amount, which is why boiling a pot of water takes some time.
At body temperature, the latent heat of vaporization for water is given as \(2.42 \times 10^6 \text{ J/kg}\). This value hints at how much energy is required to turn water from liquid sweat into vapor to cool our bodies.
  • Latent heat "hides" energy during phase changes without raising temperature.
  • High latent heat means effective cooling鈥攅xcellent for thermoregulation.
In the human body, evaporating sweat utilizes this large energy transfer to effectively remove heat.
Specific Heat Capacity
Specific heat capacity refers to the amount of heat energy required to raise the temperature of one kilogram of a substance by one degree Celsius (or one Kelvin).
For the human body, specific heat capacity is estimated at \(3480 \text{ J/(kg} \cdot \text{K)}\). This indicates how much heat is needed to change body temperature, which tends to resist quick changes.
  • High specific heat means more energy is needed for temperature change.
  • Helps maintain stable internal body temperatures.
  • Ensures body warming and cooling are controlled processes.
This concept is essential for understanding temperature regulation in biological entities.
Heat Transfer
Heat transfer is the movement of thermal energy from one object or substance to another. It can happen in several ways, including conduction, convection, and radiation.
In our exercise problem, heat transfer is crucial for calculating how much sweat must evaporate to cool the body. The formula \(Q = mc\Delta T\) helps us figure out the heat energy transferred as the temperature changes.
  • Conduction transfers heat through direct contact.
  • Convection involves heat moving through fluids (liquids and gases).
  • Radiation transmits heat through electromagnetic waves.
Understanding heat transfer is vital for predicting temperature changes and strategizing ways to maintain or alter them efficiently.

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Most popular questions from this chapter

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

A \(25,000-\mathrm{kg}\) subway train initially traveling at 15.5 \(\mathrm{m} / \mathrm{s}\) slows to a stop in a station and then stays there long enough for its brakes to cool. The station's dimensions are 65.0 \(\mathrm{m}\) long by 20.0 \(\mathrm{m}\) wide by 12.0 \(\mathrm{m}\) high. Assuming all the work done by the brakes in stopping the train is transferred as heat uniformly to all the air in the station, by how much does the air temperature in the station rise? Take the density of the air to be 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and its specific heat to be 1020 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\).

Why Do the Seasons Lag? In the northern hemisphere, June 21 (the summer solstice) is both the longest day of the year and the day on which the sun's rays strike the earth most vertically, hence delivering the greatest amount of heat to the surface. Yet the hottest summer weather usually occurs about a month or so later. Let us see why this is the case. Because of the large specific heat of water, the oceans are slower to warm up than the land (and also slower to cool off in winter). In addition to perusing pertinent information in the tables included in this book, it is useful to know that approximately two-thirds of the earth's surface is ocean composed of salt water having a specific heat of 3890 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) and that the oceans, on the average, are 4000 m deep. Typically, an average of 1050 \(\mathrm{W} / \mathrm{m}^{2}\) of solar energy falls on the earth's surface, and the oceans absorb essentially all of the light that strikes them. However, most of that light is absorbed in the upper 100 \(\mathrm{m}\) of the surface. Depths below that do not change temperature seasonally. Assume that the sunlight falls on the surface for only 12 hours per day and that the ocean retains all the heat it absorbs. What will be the rise in temperature of the upper 100 \(\mathrm{m}\) of the oceans during the month following the summer solstice? Does this seem to be large enough to be perceptible?

The emissivity of tungsten is \(0.350 .\) A tungsten sphere with radius 1.50 \(\mathrm{cm}\) is suspended within a large evacuated enclosure whose walls are at 290.0 \(\mathrm{K}\) . What power input is required to maintain the sphere at a temperature of 3000.0 \(\mathrm{K}\) if heat conduction along the supports is neglected?

What is the rate of energy radiation per unit area of a blackbody at a temperature of \((\) a ) 273 \(\mathrm{K}\) and (b) 2730 \(\mathrm{K} ?\)
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