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On a certain day the temperature is \(20^{\circ} \mathrm{C}\) and the dew point is \(5.0\) \({ }^{\circ} \mathrm{C}\). What is the relative humidity? Saturated air at \(20{ }^{\circ} \mathrm{C}\) and \(5.0\) \({ }^{\circ} \mathrm{C}\) contains \(17.12\) and \(6.80 \mathrm{~g} / \mathrm{m}^{3}\) of water, respectively.

Short Answer

Expert verified
The relative humidity is approximately 39.8%.

Step by step solution

01

Understand the Problem

To find the relative humidity, we need to determine the ratio of the actual water vapor content in the air to the maximum possible water vapor content at the same temperature, expressed as a percentage. The dew point tells us the temperature at which air becomes saturated with moisture. We have the temperature at which the air is saturated and the amount of water it can contain at those temperatures.
02

Use the Formula of Relative Humidity

Relative humidity is calculated using the formula: \( RH = \frac{e_a}{e_s} \times 100 \% \), where \( e_a \) is the actual vapor density (at the dew point), and \( e_s \) is the saturated vapor density (at the current temperature).
03

Identify the Values

From the problem, we know that the actual vapor density, \( e_a \), at the dew point of \(5^{\circ}C\) is \(6.80\, g/m^3\). The saturated vapor density, \( e_s \), at the current temperature of \(20^{\circ}C\) is \(17.12\, g/m^3\).
04

Calculate Relative Humidity

Substitute the known values into the formula: \( RH = \frac{6.80}{17.12} \times 100 \% \). Perform the division to find \( RH = 0.39766187 \times 100 \% \), which equals approximately \( 39.77\% \).
05

Round to a Practical Answer

Round the calculated relative humidity to a reasonable number of decimal places, as appropriate for practical use. In this case, \( RH \approx 39.8\%\).

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Key Concepts

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

Dew Point
The dew point is a critical concept in meteorology and everyday life because it helps us understand how humid the air feels. Think of it as the temperature at which air becomes so saturated with moisture that water starts to condense into dew. If the environmental temperature drops to the dew point, the air cannot hold any more water vapor, leading to condensation. This is why on cool mornings you might see dew on the grass.

For example, in our exercise, the dew point is given as \(5^{\circ} \mathrm{C}\). This means that if the air cools down to \(5^{\circ} \mathrm{C}\), it will reach saturation point and moisture will start forming as dew. Understanding the dew point can give us insights into potential weather conditions, like fog or heavy dew.

To determine the dew point practically, we need to measure the actual vapor content and assess how much more or less muggy it might feel as the temperature changes.
Saturated Vapor Density
Saturated vapor density is the maximum amount of water vapor that air can hold at a given temperature. It varies with temperature – warmer air can hold more moisture. When the air is saturated, it is at the dew point, meaning it contains the maximum possible amount of water vapor without condensing into liquid.

For instance, in our exercise, the saturated vapor density at \(20^{\circ} \mathrm{C}\) is \(17.12 \, \mathrm{g/m}^3\). And at \(5^{\circ} \mathrm{C}\) it is \(6.80 \, \mathrm{g/m}^3\). These values indicate how much moisture the air can hold when saturated at these respective temperatures.

Understanding saturated vapor density helps us calculate the relative humidity, as it serves as the denominator in the fraction representing the ratio of actual water vapor content to maximum capacity.
Vapor Density Calculation
In calculating vapor density, we mainly focus on two values: the actual vapor density and the saturated vapor density at the current temperature. The actual vapor density is the amount of water vapor present in the air, which typically is measured by the dew point.

Using the formula for relative humidity:\[RH = \frac{e_a}{e_s} \times 100\% \],where:
  • \(e_a\) is the actual vapor density.
  • \(e_s\) is the saturated vapor density.
This formula allows us to compare the current amount of moisture in the air with what could be the maximum amount under those temperature conditions, giving a good sense of how humid or dry it actually feels.

By substituting the given values (like in our problem \(6.80 \, \mathrm{g/m}^3\) for actual vapor density and \(17.12 \, \mathrm{g/m}^3\) for the saturated vapor density at \(20^{\circ} \mathrm{C}\)), we calculate the relative humidity and get tangible measures of air moisture conditions.
Temperature and Humidity
Temperature and humidity go hand in hand in determining the comfort level of our environment. As temperature varies, it impacts how much moisture the air can hold. Higher temperatures allow the air to hold more water vapor, making it feel hotter and stickier. Conversely, lower temperatures may mean the air is drier, since it can hold less moisture.

By understanding how temperature affects humidity, we can better design our living environments for comfort and health. For instance, at \(20^{\circ} \mathrm{C}\) with a relative humidity of approximately \(39.8\%\), the air may feel comfortable and neither too dry nor excessively moist.

Heating and cooling systems in buildings also adjust both temperature and humidity to maintain a level that is pleasant for humans, reducing discomfort and promoting well-being. Recognizing the interplay between these two factors can help students and professionals forecast weather conditions or manage indoor climates more effectively.

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Most popular questions from this chapter

Two identical metal plates (mass \(=m\), specific heat \(=c\) ) have different temperatures; one is at \(20^{\circ} \mathrm{C}\), and the other is at \(90^{\circ} \mathrm{C}\). They are placed in good thermal contact. What is their final temperature? Because the plates are identical, we would guess the final temperature to be midway between \(20^{\circ} \mathrm{C}\) and \(90^{\circ} \mathrm{C}\), namely 55 \({ }^{\circ} \mathrm{C}\). This is correct, but let us show it mathematically. From the law of conservation of energy, the heat lost by one plate must equal the heat gained by the other. Thus, the total heat change of the system is zero. In equation form, (Heat change of hot plate) \(+\) (Heat change of cold plate) =0 $$ m c(\Delta T)_{\text {hox }}+m c(\Delta T)_{\text {cold }}=0 $$ which is short-hand for \(m_{\text {hot }} c_{\text {hot }} \Delta T_{\text {hot }}+m_{\text {cold }} C_{\text {cold }} \Delta T_{\text {cold }}=0\). Be careful about \(\Delta T\) : It is the final temperature (which we denote by \(T_{f}\) in this case) minus the initial temperature. The above equation thus becomes $$ m c\left(T_{f}-90^{\circ} \mathrm{C}\right)+m c\left(T_{f}-20^{\circ} \mathrm{C}\right)=0 $$ After canceling \(m c\) from each term, solve the equation and find \(T_{f}\) \(=55^{\circ} \mathrm{C}\), the expected answer. Alternative Method .Notice that this analysis is identical to starting with Eq. (18.3); namely, \(Q_{\text {in }}=-Q_{\text {out }}\) inasmuch as \(Q_{\text {in }}+Q_{\text {out }}=0\).

Calculate the heat of fusion of ice from the following data for ice at \(0{ }^{\circ} \mathrm{C}\) added to water: \(\begin{array}{lr}\text { Mass of calorimeter } & 60 \mathrm{~g} \\ \text { Mass of calorimeter plus water } & 460 \mathrm{~g} \\ \text { Mass of calorimeter plus water and ice } & 618 \mathrm{~g} \\ \text { Initial temperature of water } & 38.0^{\circ} \mathrm{C} \\ \text { Final temperature of mixture } & 5.0^{\circ} \mathrm{C} \\ \text { Specific heat of calorimeter } & 0.10 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\end{array}\)

Suppose a person who eats 2500 Cal of food each day loses the heat equivalent of the food through evaporation of water from the body. How much water must evaporate each day? At body temperature, \(L_{v}\) for water is about \(600 \mathrm{cal} / \mathrm{g}\).

An electric heater that produces \(900 \mathrm{~W}\) of power is used to vaporize water. How much water at \(100{ }^{\circ} \mathrm{C}\) can be changed to steam at \(100^{\circ} \mathrm{C}\) in \(3.00\) min by the heater? (For water at \(100^{\circ} \mathrm{C}\), \(\left.L_{0}=2.26 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\right)\) The heater produces \(900 \mathrm{~J}\) of heat energy per second. So the heat produced in \(3.00 \mathrm{~min}\) is $$ \Delta Q=(900 \mathrm{~J} / \mathrm{s})(180 \mathrm{~s})=162 \mathrm{~kJ} $$ The heat required to vaporize a mass \(m\) of water is $$ \Delta Q=m L_{v}=m\left(2.26 \times 10^{6} \mathrm{~J} / \mathrm{kg}\right) $$ Equating these two expressions for \(\Delta Q\) and solving for \(m\) gives \(m\) \(=0.0717 \mathrm{~kg}=71.7 \mathrm{~g}\) as the mass of water vaporized.

A \(0.250\) -hp drill causes a dull \(50.0\) -g steel bit to heat up rather than to deepen a hole in a block of hard wood. Assuming that \(75.0\) percent of the friction-loss energy causes heating of the bit, by what amount will its temperature change in \(20.0\) s? For steel, \(c\) \(=450 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\)

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