Problem 36 - College Physics

Question

\(\cdot\) Evaporative cooling. The evaporation of sweat is an important mechanism for temperature control in some warmblooded 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 capacity 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

A 70 kg man must evaporate 0.1008 kg of water, which is equivalent to 100.8 cm³, to cool by 1°C. This is about 28.4% of a soft-drink can's volume.

Step 1: Calculate the Heat Lost

To cool the man's body by 1°C, we first need to determine the total heat loss needed. This can be calculated using the formula: \[ Q = m c \Delta T \] where \( m = 70.0 \text{ kg} \) (mass of the body), \( c = 3480 \text{ J/(kg⋅K)} \) (specific heat capacity), and \( \Delta T = 1.0 \text{ °C} \) (change in temperature). Substituting these values, we have: \[ Q = 70.0 \times 3480 \times 1.0 = 243,600 \text{ J} \] Therefore, the total heat lost is 243,600 J.

Step 2: Calculate the Mass of Water to Evaporate

Next, we find out how much water needs to evaporate to lose this amount of heat. The heat required to vaporize water is given by: \[ Q = m_{\text{water}} L \] where \( Q = 243,600 \text{ J} \) as found previously, and \( L = 2.42 \times 10^6 \text{ J/kg} \) (heat of vaporization). Rearranging for the mass of water: \[ m_{\text{water}} = \frac{Q}{L} = \frac{243,600}{2.42 \times 10^6} \approx 0.1008 \text{ kg} \]So, about 0.1008 kg of water must evaporate.

Step 3: Convert Mass to Volume

To find out how much water this corresponds to in volume, consider that the density of water is approximately \( 1000 \text{ kg/m}^3 \). Therefore, the volume of water needed is: \[ V = \frac{m_{\text{water}}}{\text{density}} = \frac{0.1008}{1000} = 0.0001008 \text{ m}^3 \] Converting meters cubed to centimeters cubed (since \( 1 \text{ m}^3 = 10^6 \text{ cm}^3 \)), we find: \[ V = 0.0001008 \times 10^6 = 100.8 \text{ cm}^3 \] Thus, the volume of water to replenish is 100.8 cm³.

Step 4: Compare with Soft Drink Can Volume

A typical soft-drink can has a volume of 355 cm³. Comparing this to the volume of water that needs to be replaced: \[ \frac{100.8 \text{ cm}^3}{355 \text{ cm}^3} \approx 0.284 \] This indicates that about 28.4% of a soft-drink can is needed to replenish the evaporated water.

Key Concepts

Understanding the Heat of Vaporization

When it comes to evaporative cooling, the heat of vaporization plays a crucial role. This is the amount of energy required for turning liquid water into vapor, without any temperature change. At a body temperature of 37°C, the heat of vaporization for water is a substantial 2.42 x 10^6 J/kg.

This high energy requirement means the body can efficiently use sweat evaporation for cooling, as a significant amount of heat can be dissipated with a relatively small amount of water. The heat energy from the body's surface is absorbed by the water, allowing it to change states from liquid to gas.

This supports the regulation of body temperature, by ensuring excess heat is lost through the transformation and escape of water molecules.

Exploring Specific Heat Capacity

The specific heat capacity is a measure of the amount of heat energy required to change the temperature of a unit mass by one degree. For the human body, this number is approximately 3480 J/(kg·K).

Since the human body has a relatively high specific heat capacity, it means it can absorb or release a significant amount of heat with little change in temperature. This property contributes to the body's ability to maintain stable internal temperatures, even when experiencing external temperature fluctuations.

During evaporative cooling, the specific heat capacity ensures that the heat energy required to raise or lower body temperature is accounted for, allowing precise regulation through heat energy exchange with evaporating sweat.

Mechanisms of Temperature Control

Temperature control in warm-blooded animals like humans is vital for maintaining homeostasis. One efficient method that our bodies use to control temperature is through evaporative cooling.

When body temperature rises, sweat glands in the skin produce sweat. As the sweat evaporates, it absorbs heat from the body, lowering the body temperature. This is essential for preventing overheating and maintaining optimal physiological function.

The process involves a balance between heat gain and heat loss. Factors such as ambient temperature, humidity levels, and physical activity influence the rate and efficiency of this cooling method. The body also uses blood vessels near the skin to help regulate temperature, expanding them to release heat and contracting to conserve heat.

Fundamentals of Thermodynamics in Evaporative Cooling

Thermodynamics, the branch of physics concerning heat and temperature, provides an essential framework for understanding how evaporative cooling works. In this system, energy transformations follow physical rules such as the conservation of energy—the first law of thermodynamics.

During evaporative cooling, the human body releases energy in the form of heat when converting sweat to vapor. This aligns with the principle that energy cannot be created or destroyed, only transformed. The energy absorbed by the water molecules allows them to escape into the air, thus cooling the body.

The second law of thermodynamics also plays a role, which involves entropy or disorder. Evaporation increases entropy by dispersing water molecules into the air, contributing to an even temperature distribution. Through these principles, thermodynamics provides insight into the effectiveness and necessity of this biological cooling system.