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Chapter 18: Problem 17

On a day when the temperature is \(28^{\circ} \mathrm{C}\), moisture forms on the outside of a glass of cold drink if the glass is at a temperature of \(16{ }^{\circ} \mathrm{C}\) or lower. What is the R.H. on that day? Saturated air at 28 \({ }^{\circ} \mathrm{C}\) contains \(26.93 \mathrm{~g} / \mathrm{m}^{3}\) of water, while, at \(16{ }^{\circ} \mathrm{C}\), it contains \(13.50 \mathrm{~g} / \mathrm{m}^{3}\) Dew forms at a temperature of \(16^{\circ} \mathrm{C}\) or lower, so the dew point is \(16{ }^{\circ} \mathrm{C}\). The air is saturated at that temperature and therefore contains \(13.50 \mathrm{~g} / \mathrm{m}^{3}\). Then $$ \text { R.H. }=\frac{\text { Mass present } / \mathrm{m}^{3}}{\text { Mass } / \mathrm{m}^{3} \text { in saturated air }}=\frac{13.50}{26.93}=0.50=50 \% $$

Short Answer

Expert verified
The relative humidity is 50%.

Step by step solution

01

Understand the Problem

The task is to find the relative humidity (R.H.) on a day when the temperature is \(28^{\circ} \mathrm{C}\). We know that condensation occurs at the dew point, which is given at \(16^{\circ} \mathrm{C}\). The relative humidity is the ratio of the actual water vapor density to the water vapor density of saturated air at the same temperature.
02

Determine Dew Point Details

As specified, dew forms at a temperature of \(16{ }^{\circ} \mathrm{C}\). At this point, the air becomes saturated with moisture, so it contains \(13.50 \mathrm{~g} / \mathrm{m}^{3}\) of water vapor.
03

Saturated Air Comparison

To find the relative humidity, we need the maximum amount of water vapor that the air can hold at \(28^{\circ} \mathrm{C}\), which is given as \(26.93 \mathrm{~g} / \mathrm{m}^{3}\).
04

Calculate Relative Humidity

The formula for relative humidity is:\[\text{R.H.} = \frac{\text{Mass of water vapor present per m}^3}{\text{Mass of water vapor per m}^3\text{ in saturated air}}\]Substitute the known values:\[\text{R.H.} = \frac{13.50}{26.93} \approx 0.50\]Convert 0.50 to a percentage: \(0.50 \times 100\% = 50\%\).
05

Conclusion

Therefore, the relative humidity on that day, when the temperature is \(28^{\circ} \mathrm{C}\), is 50\%.

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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 the temperature at which the air becomes fully saturated with moisture, leading to condensation. This occurs because the air can no longer hold all of the water vapor present, causing the excess to form liquid water droplets. On the day described, dew forms at a temperature of \(16^{\circ} \mathrm{C}\), marking this as the dew point. This means that if the air is cooled to \(16^{\circ} \mathrm{C}\) or below, it reaches its capacity to hold water vapor, and any additional moisture results in dew formation.
Understanding dew point is crucial in predicting weather patterns and understanding humidity levels. It tells us how much cooling is needed to bring air to saturation, which is essential for various applications like agriculture and meteorology.
Saturated Air
Saturated air is the state where the air contains the maximum amount of water vapor it can hold at a particular temperature. Beyond this point, the air cannot hold more water vapor, which leads to condensation. In this exercise, at \(28^{\circ} \mathrm{C}\), the air can hold up to \(26.93 \mathrm{~g} / \mathrm{m}^{3}\) of water vapor, whereas at the dew point temperature of \(16^{\circ} \mathrm{C}\), it can hold only \(13.50 \mathrm{~g} / \mathrm{m}^{3}\).
When air is saturated, relative humidity reaches 100%, and the chances of precipitation increase. Saturated air at different temperatures can hold varying amounts of water vapor, which directly affects weather conditions. Knowing the saturation levels helps in understanding climate dynamics and preparing for weather changes, such as fog or rain.
Water Vapor Density
Water vapor density refers to the amount of water vapor present in the air, measured in grams per cubic meter \(\mathrm{g}/\mathrm{m}^3\). It is directly related to humidity, as higher water vapor density indicates higher moisture levels in the air. In this context, at the dew point of \(16^{\circ} \mathrm{C}\), the water vapor density is \(13.50 \mathrm{~g} / \mathrm{m}^{3}\).
Water vapor density impacts daily weather phenomena and human comfort levels. High water vapor density can make the air feel muggy, while lower densities can make it feel dry. By calculating water vapor density, you can determine relative humidity, which is crucial for understanding and predicting thermal comfort, agricultural needs, and even HVAC system requirements.
  • It helps gauge total atmospheric moisture.
  • Influences weather predictions and climate modeling.
Monitoring and adjusting to the water vapor density can optimize conditions in controlled environments, such as greenhouses and factories, ensuring proper growth and operation efficiency.

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Most popular questions from this chapter

A thermometer in a \(10 \mathrm{~m} \times 8.0 \mathrm{~m} \times 4.0 \mathrm{~m}\) room reads \(22{ }^{\circ} \mathrm{C}\) and a humidistat reads the R.H. to be 35 percent. What mass of water vapor is in the room? Saturated air at \(22^{\circ} \mathrm{C}\) contains \(19.33 \mathrm{~g}\) \(\mathrm{H}_{2} \mathrm{O} / \mathrm{m}^{3}\) $$ \begin{aligned} \% \mathrm{R} . \mathrm{H} . &=\frac{\text { Mass of water } / \mathrm{m}^{3}}{\text { Mass of water } / \mathrm{m}^{3} \text { of saturated air }} \times 100 \\ 35 &=\frac{\mathrm{Mass} / \mathrm{m}^{3}}{0.01933 \mathrm{~kg} / \mathrm{m}^{3}} \times 100 \end{aligned} $$ from which mass \(/ \mathrm{m}^{3}=6.77 \times 10^{-3} \mathrm{~kg} / \mathrm{m}^{3}\). But the room in question has a volume of \(10 \mathrm{~m} \times 8.0 \mathrm{~m} \times 4.0 \mathrm{~m}=320 \mathrm{~m}^{3}\). Therefore, the total mass of water in it is $$ \left(320 \mathrm{~m}^{3}\right)\left(6.77 \times 10^{-3} \mathrm{~kg} / \mathrm{m}^{3}\right)=2.2 \mathrm{~kg} $$

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 \(3.00-\mathrm{g}\) bullet \(\left(c=0.0305 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}=128 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\) moving at 180 \(\mathrm{m} / \mathrm{s}\) enters a bag of sand and stops. By what amount does the temperature of the bullet change if all its KE becomes thermal energy that is added to the bullet? The bullet loses KE in the amount $$ \mathrm{KE}=\frac{1}{2} m v^{2}=\frac{1}{2}\left(3.00 \times 10^{-3} \mathrm{~kg}\right)(180 \mathrm{~m} / \mathrm{s})^{2}=48.6 \mathrm{~J} $$ This results in the addition of \(\Delta Q=48.6 \mathrm{~J}\) of thermal energy to the bullet. Then, since \(\Delta Q=m c \Delta T\), we can find \(\Delta T\) for the bullet: $$ \Delta T=\frac{\Delta Q}{m c}=\frac{48.6 \mathrm{~J}}{\left(3.00 \times 10^{-3} \mathrm{~kg}\right)\left(128 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)}=127^{\circ} \mathrm{C} $$ Notice that we had to use \(c\) in \(\mathrm{J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and not in \(\mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\).

(a) How much heat is required to raise the temperature of \(250 \mathrm{~mL}\) of water from \(20.0{ }^{\circ} \mathrm{C}\) to \(35.0{ }^{\circ} \mathrm{C} ?(b)\) How much heat is lost by the water as it cools back down to \(20.0^{\circ} \mathrm{C}\) ? Since \(250 \mathrm{~mL}\) of water has a mass of \(250 \mathrm{~g}\), and since \(c=1.00\) \(\mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) for water, we have (a) \(\Delta Q=m c \Delta T=(250 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(15.0{ }^{\circ} \mathrm{C}\right)=3.75 \times 10^{3} \mathrm{cal}\) \(=15.7 \mathrm{~kJ}\) (b) \(\Delta Q=m\) c \(\Delta T=(250 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(-15.0{ }^{\circ} \mathrm{C}\right)=-3.75 \times 10^{3}\) \(\mathrm{cal}=-15.7 \mathrm{~kJ}\) Notice that heat-in (i.e., the heat that enters an object) is taken to be positive, whereas heat-out (i.e., the heat that leaves an object) is taken to be negative. Alternative Method Let's redo \((a)\) in SI units: \(250 \mathrm{~mL}=250 \mathrm{~cm}^{3}=250 \times 10^{-6} \mathrm{~m}^{3}\), and \(c\) for water is \(4.186 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\); hence \(\Delta Q=m c \Delta T=(0.250 \mathrm{~kg})(4.186 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(15.0 \mathrm{~K})=15.7 \mathrm{~kJ}\)

How much heat is given up when \(20 \mathrm{~g}\) of steam at \(100{ }^{\circ} \mathrm{C}\) is condensed and cooled to \(20^{\circ} \mathrm{C}\) ? The steam loses an amount of heat to condense into water at 100 \({ }^{\circ} \mathrm{C}\) and more for the water at \(100{ }^{\circ} \mathrm{C}\) to drop in temperature to 20 \({ }^{\circ} \mathrm{C}\). From Table \(18-2, L_{v}=2259 \mathrm{~kJ} / \mathrm{kg}\). Thus the heat that must be removed is $$ Q=m L_{v}+c_{\text {wacer }} \Delta T_{\text {uateter }} $$ \(Q=-(0.020 \mathrm{~kg})(2259 \mathrm{~kJ} / \mathrm{kg})+(4.186 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(0.020 \mathrm{~kg})(-80 \mathrm{~K})\) $$ Q=-45.18 \mathrm{~kJ}-6.6976 \mathrm{~kJ} $$ and \(Q=-51.9 \mathrm{~kJ}\), or to two figures \(-52 \mathrm{~kJ}\). Alternative Method conter
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