Problem 78 - Physics

Question

The latent heat of vaporization of \(\mathrm{H}_{2} \mathrm{O}\) at body temperature \(\left(37.0^{\circ} \mathrm{C}\right)\) is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) To cool the body of a \(75-\mathrm{kg}\) jogger [average specific heat capacity \(\left.=3500 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\right]\) by \(1.5 \mathrm{C}^{\circ},\) how many kilograms of water in the form of sweat have to be evaporated?

Step-by-Step Solution

Answer

Approximately 0.1627 kg of water must evaporate.

Step 1: Understand the Problem

We need to find how many kilograms of water must evaporate to cool the jogger's body by \(1.5\, ^{\circ}\text{C}\) by using the latent heat of vaporization. The jogger's mass is 75 kg, and the specific heat capacity is \(3500\, \text{J/(kg} \cdot \text{C}^{\circ})\). The latent heat of vaporization for water is \(2.42 \times 10^6\, \text{J/kg}\).

Step 2: Calculate the Heat to Remove

Calculate the total heat that must be removed from the jogger's body to lower the temperature by \(1.5\, ^{\circ}\text{C}\). This can be calculated using the formula for heat: \[Q = mc\Delta T\]where \(m = 75\, \text{kg}\), \(c = 3500\, \text{J/(kg} \cdot \text{C}^{\circ})\), and \(\Delta T = 1.5\, ^{\circ}\text{C}\).

Step 3: Compute Heat Removal Value

Substitute the given values into the equation:\[Q = 75\, \text{kg} \times 3500\, \text{J/(kg} \cdot \text{C}^{\circ}) \times 1.5\, ^{\circ}\text{C} = 393750\, \text{J}\]This is the amount of heat that needs to be removed through evaporation.

Step 4: Relate to Latent Heat of Vaporization

We know that the heat removed by evaporating sweat is given by\[Q = m_wL\]where \(m_w\) is the mass of water evaporated and \(L = 2.42 \times 10^6 \text{J/kg}\) is the latent heat of vaporization.

Step 5: Solve for Evaporated Water Mass

Rearrange the equation to solve for \(m_w\):\[m_w = \frac{Q}{L} = \frac{393750\, \text{J}}{2.42 \times 10^6\, \text{J/kg}} \approx 0.1627\, \text{kg}\]

Step 6: Conclusion

The jogger must evaporate approximately 0.1627 kilograms of water to cool down by \(1.5\, ^{\circ}\text{C}.\)

Key Concepts

Specific Heat Capacity

Specific heat capacity is a concept that helps us understand how much energy is required to change the temperature of an object by a certain amount. It's an intrinsic property of materials that tells us how well a substance can absorb heat. To put it simply, it is the amount of energy needed to raise 1 kilogram of a substance by 1 degree Celsius (°C). The unit of specific heat capacity is Joules per kilogram per degree Celsius (\(\text{J/(kg} \cdot \text{C}^{\circ})\)).In our exercise, the jogger's body has a specific heat capacity of 3500 \(\text{J/(kg} \cdot \text{C}^{\circ})\). This means that to increase 1 kg of their body mass by 1°C, 3500 Joules of energy are required. If we want to cool the body by 1.5°C, we will remove this energy, multiplied by the body's mass. Hence, specific heat capacity plays a crucial role in determining how much heat needs to be added or removed to achieve the desired temperature change.

Heat Transfer

Heat transfer is the process by which heat energy moves from one place to another. This can happen in several ways: conduction, convection, and radiation. In our context, heat will transfer from the jogger's body to the environment via sweat evaporation.For the jogger, the body needs to lose a specific amount of heat energy to drop in temperature. This is calculated using the heat transfer formula:\[Q = mc\Delta T\]where:

• \(Q\) is the heat energy transferred, in Joules (J)
• \(m\) is the mass, in kilograms (kg)
• \(c\) is the specific heat capacity, in Joules per kilogram per degree Celsius (\(\text{J/(kg} \cdot \text{C}^{\circ})\))
• \(\Delta T\) is the change in temperature, in Celsius (°C)

In the exercise, we calculated that 393750 Joules of heat must be removed from the jogger's body to lower their temperature by 1.5°C. Effective heat transfer to the environment ensures the jogger cools down efficiently.

Evaporation

Evaporation is a type of heat transfer that involves the phase change of a substance from liquid to gas. In the context of sweating, evaporation is a cooling process. When sweat on the skin's surface evaporates, it absorbs a large amount of heat from the body, effectively lowering its temperature.The energy required for this phase change is known as the latent heat of vaporization. Specifically, it is the amount of energy needed to transform 1 kilogram of liquid into vapor without changing its temperature. For water, at the body's temperature (37°C), this value is \(2.42 \times 10^6 \text{J/kg}\).In the exercise problem, we used the formula:\[Q = m_wL\]where:

• \(Q\) is the heat energy removed, in Joules (J)
• \(m_w\) is the mass of water evaporated, in kilograms (kg)
• \(L\) is the latent heat of vaporization, in Joules per kilogram (J/kg)

By knowing the heat to be removed and the latent heat of vaporization, we calculated the mass of water (sweat) that must evaporate to achieve the required cooling effect on the jogger's body. It's this latent heat absorbed by the evaporating sweat that allows the jogger to lose that specific heat and cool down.