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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?

Short Answer

Expert verified
Approximately 0.163 kg of water must evaporate.

Step by step solution

01

Calculate Heat Required to Cool the Body

First, we calculate the amount of heat that needs to be removed from the jogger's body to lower the temperature by 1.5°C. This can be calculated using the formula: \( Q = mc\Delta T \), where:- \( m \) is the mass of the jogger (75 kg),- \( c \) is the specific heat capacity (3500 J/kg°C), and- \( \Delta T \) is the change in temperature (1.5°C).Calculate: \( Q = 75 \times 3500 \times 1.5 \).
02

Calculate Heat Required for Evaporation of Water

The heat required to evaporate a certain mass of water is given by \( Q = mL \), where:- \( m \) is the mass of water to be evaporated,- \( L \) is the latent heat of vaporization (\(2.42 \times 10^6\) J/kg).We need the same amount of heat calculated in Step 1 to be removed by evaporation.
03

Solve for Mass of Water Evaporated

We equate the heat removed by evaporation to the heat required to cool the jogger's body to find the mass of water:\( mcL = Q \).Rearrange to solve for \( m \): \( m = \frac{Q}{L} \).Substitute the values computed in Steps 1 and 2 to solve for \( m \).
04

Perform Calculations

First, compute the heat that needs to be removed: \( Q = 75 \times 3500 \times 1.5 = 393750 \) J.Next, solve the equation \( m = \frac{393750}{2.42 \times 10^6} \) kg to find the mass of water that needs to evaporate.

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Key Concepts

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

Latent Heat
Latent heat is the amount of heat energy that is required or released when a substance changes its state without changing temperature. For example, converting water from liquid to vapor requires an input of heat even though the temperature remains constant during the transition. This hidden heat energy is known as the latent heat of vaporization for processes involving the change from liquid to gas.

A practical example of latent heat can be found in the process of sweating. When we sweat, water from our skin evaporates, absorbing latent heat from our body. This helps cool us down. The latent heat of vaporization of water is particularly high, at approximately 2.42 million joules per kilogram at body temperature, which means it can absorb a significant amount of heat during evaporation, making it an effective cooling method.

When calculating tasks involving latent heat, such as evaporating a certain mass of sweat, use the formula:
  • \( Q = mL \)
where \( Q \) is the heat absorbed or released, \( m \) is the mass, and \( L \) is the latent heat of vaporization. This concept plays a crucial role in thermodynamic problems involving phase changes.
Specific Heat Capacity
Specific heat capacity is a property of materials that describes the amount of heat required to change the temperature of one kilogram of the material by one degree Celsius. High specific heat capacity means the material can store more heat per unit of temperature change.

In the scenario described in the exercise, the jogger's body can be modeled as a system that absorbs or releases heat. The specific heat capacity of the human body can be assumed to be around 3500 J/(kg°C). This means every kilogram of body mass requires 3500 joules to change the temperature by one degree Celsius.

Using the formula for heat capacity:
  • \( Q = mc\Delta T \)

Where:
  • \( Q \) is the heat absorbed or released,
  • \( m \) is the mass,
  • \( c \) is the specific heat capacity,
  • \( \Delta T \) is the change in temperature.

This formula allows us to calculate how much heat needs to be removed or added to a body, such as a jogger needing to cool down after running. By understanding this property, we can predict and control changes in temperature effectively.
Vaporization
Vaporization is the process in which a liquid is transformed into a gas. This can occur through evaporation or boiling. During vaporization, molecules gain enough kinetic energy to overcome intermolecular forces, allowing them to enter the gas phase.

From a thermodynamic perspective, vaporization is an endothermic process, absorbing heat from the surroundings. This makes it a useful mechanism for cooling, as the absorbed heat can significantly lower temperatures nearby. For this reason, our body relies on the vaporization of sweat to remove excess heat and maintain temperature.
  • Evaporation occurs at the surface of a liquid, below boiling point.
  • Boiling happens throughout the liquid at its boiling point.

Vaporization can be utilized in numerous applications, from cooling systems to industrial processes, wherever heat transfer and fluid dynamics are involved. The energy required or released through vaporization is quantitatively described by the latent heat of vaporization, combining both scientific curiosity and practical application in everyday life.

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Most popular questions from this chapter

The Eiffel Tower is a steel structure whose height increases by \(19.4 \mathrm{cm}\) when the temperature changes from -9 to \(+41^{\circ} \mathrm{C}\) . What is the approximate height (in meters) at the lower temperature?

A thick, vertical iron pipe has an inner diameter of \(0.065 \mathrm{m}\). A thin aluminum disk, heated to a temperature of \(85^{\circ} \mathrm{C},\) has a diameter that is \(3.9 \times 10^{-5} \mathrm{m}\) greater than the pipe's inner diameter. The disk is laid on top of the open upper end of the pipe, perfectly centered on it, and allowed to cool. What is the temperature of the aluminum disk when the disk falls into the pipe? Ignore the temperature change of the pipe.

(a) Objects A and B have the same mass of 3.0 kg. They melt when \(3.0 \times 10^{4} \mathrm{J}\) of heat is added to object \(\mathrm{A}\) and when \(9.0 \times 10^{4} \mathrm{J}\) is added to object B. Determine the latent heat of fusion for the substance from which each object is made. (b) Find the heat required to melt object A when its mass is \(6.0 \mathrm{kg}\).

If a nonhuman civilization were to develop on Saturn's largest moon, Titan, its scientists might well devise a temperature scale based on the properties of methane, which is much more abundant on the surface than water is. Methane freezes at \(-182.6^{\circ} \mathrm{C}\) on Titan, and boils at \(-155.2^{\circ} \mathrm{C} .\) Taking the boiling point of methane as \(100.0^{\circ} \mathrm{M}\) (degrees Methane) and its freezing point as \(0^{\circ} \mathrm{M},\) what temperature on the Methane scale corresponds to the absolute zero point of the Kelvin scale?

A 0.35-kg coffee mug is made from a material that has a specific heat capacity of \(920 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) and contains \(0.25 \mathrm{kg}\) of water. The cup and water are at \(15^{\circ} \mathrm{C} .\) To make a cup of coffee, a small electric heater is immersed in the water and brings it to a boil in three minutes. Assume that the cup and water always have the same temperature and determine the minimum power rating of this heater.

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