Problem 36 - College Physics

Question

The 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 \(70.0 \mathrm{~kg}\) 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 \(3480 \mathrm{~J} /(\mathrm{kg} \cdot \mathrm{K})\) (b) What volume of water must the man drink to replenish the evaporated water? Compare this result with the volume of a soft-drink can, which is \(355 \mathrm{~cm}^{3}\).

Step-by-Step Solution

Answer

Approximately 0.1006 kg of water must evaporate, equating to about 100.6 cm³, less than a soft-drink can's volume.

Step 1: Calculate Heat Energy Lost

To find the heat energy lost due to the body cooling down by 1°C, use the formula \( Q = mc\Delta T \), where \( m = 70.0 \, \text{kg} \), \( c = 3480 \, \text{J/kg} \cdot \text{K} \), and \( \Delta T = 1.00 \, \text{K} \). Thus,\[Q = 70.0 \, \text{kg} \times 3480 \, \text{J/kg} \cdot \text{K} \times 1.00 \, \text{K} = 243600 \, \text{J}.\]

Step 2: Determine Mass of Water Evaporated

To find the mass of water that must evaporate, use the formula \( Q = m_v L_v \), where \( L_v = 2.42 \times 10^6 \, \text{J/kg} \). Rearrange to find \( m_v \):\[m_v = \frac{Q}{L_v} = \frac{243600 \, \text{J}}{2.42 \times 10^6 \, \text{J/kg}} \approx 0.1006 \, \text{kg}.\]The mass of water that must evaporate is approximately \( 0.1006 \, \text{kg} \).

Step 3: Convert Mass to Volume

Assuming the density of water is \( 1000 \, \text{kg/m}^3 \), convert the mass of water evaporated to volume using \( \text{Volume} = \frac{\text{mass}}{\text{density}} \):\[\text{Volume} = \frac{0.1006 \, \text{kg}}{1000 \, \text{kg/m}^3} = 0.0001006 \, \text{m}^3.\]Convert \( \text{m}^3 \) to \( \text{cm}^3 \) (1 \( \text{m}^3 = 10^6 \, \text{cm}^3 \)):\[0.0001006 \, \text{m}^3 = 100.6 \, \text{cm}^3.\]

Step 4: Compare with Soft-drink Can

Compare this volume of water (100.6 cm³) to the volume of a typical soft-drink can, which is 355 cm³.

Key Concepts

Evaporation

Evaporation is a process where liquid water turns into vapor, a gas. This occurs when water molecules gain enough energy to break free from the surface. In living organisms, evaporation is crucial for thermoregulation. Sweat evaporates from the skin, taking away excess heat, which cools the body. This is especially important for maintaining a stable temperature in warm-blooded animals, like humans.
Understanding how much water needs to evaporate from the skin is vital. It helps determine how much heat can be removed from the body, ensuring homeostasis, or balance, in body temperature. This evaporative cooling effect also shows how the body's natural processes work to keep us comfortable, even in warm climates.

• This process is effective because it utilizes water's physical properties to absorb heat as it transitions to vapor.
• The rate of evaporation increases with temperature, humidity, and airflow over the skin.

Specific Heat

Specific heat is the amount of heat energy required to raise the temperature of one kilogram of a substance by 1°C. For the human body, the specific heat is approximately 3480 J/(kg·K). This means it takes 3480 joules to raise the temperature of 1 kg of body mass by 1 degree Celsius.
The specific heat of the human body is relatively high, signifying that a lot of energy is required to change its temperature. This property helps maintain stable internal conditions, preventing rapid temperature changes due to environmental factors. It is an essential factor when calculating how much heat is needed to cool or warm the body.

• Specific heat helps in determining how efficiently the body can dissipate heat through processes like sweating.
• A higher specific heat enables the body to store and release heat in a controlled manner, ensuring temperature regulation.

Heat of Vaporization

The heat of vaporization refers to the amount of energy required to convert 1 kilogram of a liquid into a gas without changing its temperature. For water at body temperature (37°C), this value is 2.42 x 10^6 J/kg. This high value signifies why water is such an effective coolant.
When sweat evaporates from the body, it absorbs a large amount of heat, precisely through this high heat of vaporization. This energy absorption cools the skin and, consequently, the blood circulating near the body's surface. The heat of vaporization plays a crucial role in thermoregulation by efficiently dissipating excess body heat.

• This high energy requirement means water can carry away a significant amount of heat when it vaporizes, making it effective for cooling.
• It explains why the body relies on sweating as a primary mechanism for heat release.

Density of Water

Density is a measure of how much mass is contained in a given volume. For water, the density is 1000 kg/m³. This means that for every cubic meter, there are 1000 kilograms of water. Density is a critical parameter when converting between mass and volume of water.
When discussing sweat evaporation and replenishing lost fluids, density allows us to compute volume from mass. For instance, knowing the mass of water needed to cool the body, we can easily find how much water volume needs to evaporate or be consumed to achieve the same effect. Understanding this fundamental property of water aids in practical conversions and comparisons, like evaluating water consumption relative to common measures such as a soft drink can.

• Density allows easy conversion between mass and volume, essential for practical real-world calculations in thermoregulation.
• It ensures accuracy in measuring and comparing water loss and intake in various scenarios.