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What is latent heat? How is the latent heat loss from the human body affected by \((a)\) skin wettedness and \((b)\) relative humidity of the environment? How is the rate of evaporation from the body related to the rate of latent heat loss?

Short Answer

Expert verified
Answer: Skin wettedness and relative humidity both play important roles in the latent heat loss process of the human body. An increase in skin wettedness typically leads to an increase in latent heat loss, as more sweat is available to evaporate and transfer heat away from the body. However, latent heat loss decreases as relative humidity increases because the air's capacity to absorb additional moisture through evaporation is reduced. As a result, evaporative cooling is more effective at lower humidity levels, allowing the body to lose more heat through latent heat loss.

Step by step solution

01

Definition of latent heat

Latent heat is the amount of heat absorbed or released by a substance during a phase change without changing its temperature. In the context of the human body, latent heat is primarily involved in the process of sweating, where the body releases heat to the environment through the evaporation of sweat on the skin.
02

(a) Skin wettedness and latent heat loss

Skin wettedness refers to the fraction of the body's surface area covered by sweat. An increase in skin wettedness typically increases latent heat loss, as more sweat is available to evaporate and transfer heat away from the body. The higher the skin wettedness, the more heat is transferred through evaporation, which leads to more efficient cooling of the body.
03

(b) Relative humidity and latent heat loss

Relative humidity is the amount of moisture present in the air as a percentage of the maximum moisture the air can hold at a given temperature. When the relative humidity is high, the air contains more water vapor, which reduces its capacity to absorb additional moisture (like sweat) through evaporation. As a result, latent heat loss from the human body decreases as relative humidity increases. When humidity is low, evaporative cooling is more effective, and the body can lose more heat through latent heat loss.
04

Rate of evaporation and latent heat loss

The rate of evaporation is the amount of liquid (i.e., sweat) turning into vapor per unit time. The rate of latent heat loss is directly proportional to the rate of evaporation. When the evaporation rate increases, more sweat is turned into vapor, which requires more heat energy to be taken away from the body in the form of latent heat. Conversely, when the evaporation rate decreases, less heat is lost through latent processes, making the cooling effect less efficient.

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Key Concepts

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

Evaporation and Cooling
When we talk about evaporation and cooling, we're exploring a natural process that impacts everything from the weather to our own comfort on a hot day. Imagine a hot summer afternoon, where you instinctively reach for a cold drink and perhaps sit near a fan. What's actually happening to your body is an intricate dance of heat exchange. Evaporation is the process where liquid water turns into water vapor, which is a gas. It requires heat to do this—a concept known as 'latent heat of vaporization'. This heat is taken from your body or the surroundings.

For us humans, we rely on sweating as a portable air conditioning system. Our sweat glands produce moisture on the surface of our skin, and as this sweat evaporates, it absorbs heat from our body, leading to a decrease in body temperature. This is an incredibly effective way to cool down, but it does depend on other factors, such as humidity, wind, and temperature.
Skin Wettedness
Moving to the idea of skin wettedness, this is essentially how much sweat covers your skin. It's not just about the amount you sweat, but also the spread over your skin's surface area. Think of it as the body's version of spreading out a wet towel to dry—it covers more area, so it dries, or in our case, evaporates, more efficiently.

The more your skin is wetted with sweat, the greater the area for evaporation, and consequently, the more heat can be lost through this process. However, it's important to reach a balance, as too much sweat that doesn't evaporate effectively feels uncomfortable and doesn't contribute to additional cooling. To maximize evaporation, the body must release sweat at a rate that matches the evaporation capacity of the surrounding environment.
Relative Humidity
Now, let's turn to relative humidity. This measures the current amount of water vapor in the air relative to the maximum it can hold at that temperature. So, when the weather anchor says 'It's a humid day', they're really saying that the air is holding a lot of moisture already.

In conditions of high relative humidity, the air is almost saturated and has a lower capacity to take on more water vapor from sweat evaporation. This means less sweat can evaporate from the skin, which limits the cooling effect and reduces the latent heat loss from the body. Conversely, in low humidity, the air is dry and can absorb more sweat, enhancing the evaporation and cooling processes.
Phase Change Thermodynamics
Diving deeper into phase change thermodynamics involves understanding the energy exchanges that occur when a substance changes from one state to another—in our case, from liquid to gas during sweating. Latent heat is the heat energy that must be absorbed or released during a phase change without causing a temperature change. This energy breaks the bonds between water molecules in sweat to allow them to escape as vapor.

In this context, phase change is a cooling process. When sweat evaporates, it absorbs the necessary latent heat directly from the body’s surface—a process that's essential for regulating body temperature. The thermodynamic principles governing this process are unchanging, but the rate at which evaporation occurs and thus, the efficiency of cooling, can be influenced by ambient conditions such as humidity, temperature, and airflow.

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Most popular questions from this chapter

Consider two concentric spheres forming an enclosure with diameters of \(12 \mathrm{~cm}\) and \(18 \mathrm{~cm}\) and surface temperatures \(300 \mathrm{~K}\) and \(500 \mathrm{~K}\), respectively. Assuming that the surfaces are black, the net radiation exchange between the two spheres is (a) \(21 \mathrm{~W}\) (b) \(140 \mathrm{~W}\) (c) \(160 \mathrm{~W}\) (d) \(1275 \mathrm{~W}\) (e) \(3084 \mathrm{~W}\)

This experiment is conducted to determine the emissivity of a certain material. A long cylindrical rod of diameter \(D_{1}=0.01 \mathrm{~m}\) is coated with this new material and is placed in an evacuated long cylindrical enclosure of diameter \(D_{2}=0.1 \mathrm{~m}\) and emissivity \(\varepsilon_{2}=0.95\), which is cooled externally and maintained at a temperature of \(200 \mathrm{~K}\) at all times. The rod is heated by passing electric current through it. When steady operating conditions are reached, it is observed that the rod is dissipating electric power at a rate of \(8 \mathrm{~W}\) per unit of its length and its surface temperature is \(500 \mathrm{~K}\). Based on these measurements, determine the emissivity of the coating on the rod.

Two concentric spheres are maintained at uniform temperatures \(T_{1}=45^{\circ} \mathrm{C}\) and \(T_{2}=280^{\circ} \mathrm{C}\) and have emissivities \(\varepsilon_{1}=0.25\) and \(\varepsilon_{2}=0.7\), respectively. If the ratio of the diameters is \(D_{1} / D_{2}=0.30\), the net rate of radiation heat transfer between the two spheres per unit surface area of the inner sphere is (a) \(86 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(1169 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(1181 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(2510 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(3306 \mathrm{~W} / \mathrm{m}^{2}\)

Consider a cylindrical enclosure with \(A_{1}, A_{2}\), and \(A_{3}\) representing the internal base, top, and side surfaces, respectively. Using the length to diameter ratio, \(K=L / D\), determine (a) the expression for the view factor from the side surface to itself \(F_{33}\) in terms of \(K\) and \((b)\) the value of the view factor \(F_{33}\) for \(L=D\).

Two very large parallel plates are maintained at uniform temperatures of \(T_{1}=1000 \mathrm{~K}\) and \(T_{2}=800 \mathrm{~K}\) and have emissivities of \(\varepsilon_{1}=\varepsilon_{2}=0.5\), respectively. It is desired to reduce the net rate of radiation heat transfer between the two plates to one-fifth by placing thin aluminum sheets with an emissivity of \(0.1\) on both sides between the plates. Determine the number of sheets that need to be inserted.

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