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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 (R.H.) on that day is 50\%.

Step by step solution

01

Understanding the Problem

We are tasked to find the relative humidity (R.H.) on a day at a temperature of \(28{ }^{\circ} \mathrm{C}\), given certain moisture conditions related to dew point and saturation levels at \(28{ }^{\circ} \mathrm{C}\) and \(16^{\circ} \mathrm{C}\).
02

Identifying Dew Point

Dew forms on the glass when the temperature is \(16^{\circ} \mathrm{C}\) or lower, meaning the air is saturated at this temperature. Therefore, the dew point is \(16^{\circ} \mathrm{C}\).
03

Mass of Water Vapor at Dew Point

Since the dew point temperature is \(16^{\circ} \mathrm{C}\), the air contains \(13.50 \, \mathrm{g/m^{3}}\) of water vapor, based on given saturation mass data for \(16^{\circ} \mathrm{C}\).
04

Calculating Relative Humidity (R.H.)

Use the formula for relative humidity: \(\text{R.H.} = \frac{\text{mass of water vapor present}}{\text{mass of water vapor at saturation}}\). Here, it is \(\frac{13.50}{26.93}\).
05

Final Calculation for R.H.

Perform the division: \(\frac{13.50}{26.93} = 0.50\), which is expressed as a percentage, giving R.H. = 50\%.

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Key Concepts

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

Dew Point Calculation
The dew point is a crucial concept when dealing with humidity and weather conditions. It's the temperature at which air becomes saturated with water vapor, meaning the air holds as much moisture as it can before dew starts to form. In simpler terms, it鈥檚 the point when the air gets so cool it cannot hold all its moisture, leading it to condense.
Understanding the dew point is essential for different practical applications. For example, it helps in predicting fog, frost, and dew formation. In our exercise, dew forms at and below the temperature of 16掳C, marking 16掳C as the dew point for that specific condition.
To identify the dew point:
  • Recognize when visible moisture or condensation is starting to form.
  • Look at temperature changes and conditions when air seems saturated, as in our glass example.
  • Consult saturation point data for different temperatures to confirm the dew point accurately.
Once we determine the dew point as 16掳C, we know that at this temperature, the air holds exactly 13.50 g/m鲁 of water vapor, according to given data for saturation levels.
Water Vapor Saturation
Water vapor saturation refers to the maximum amount of water vapor that air can hold at a given temperature. When air reaches this point, it cannot hold any additional moisture and, if cooled further, starts to release water as dew, rain, or other forms of precipitation.
In the context of our problem, saturation is evaluated by comparing the mass of water vapor present to the mass needed to reach saturation at specific temperatures, namely 28掳C and 16掳C.
To understand saturation:
  • Remember, warmer air can hold more water vapor. At 28掳C, the air holds 26.93 g/m鲁 of water vapor when saturated.
  • As temperature drops, the capacity decreases. At 16掳C, it drops to 13.50 g/m鲁.
  • This shift informs us of the moisture content relative to temperature and how likely dew or precipitation is to occur.
Accurate saturation data enables us to functionally calculate relative humidity and determine comfort levels and weather predictions.
Temperature Effect on Humidity
Temperature significantly affects humidity, determining how much water vapor the air can hold. Higher temperatures mean more capacity for moisture, while lower temperatures mean less.
Consider our scenario: At 28掳C, air can potentially hold a lot of moisture, up to 26.93 g/m鲁. But as the temperature cools to 16掳C, the air's capacity drops significantly to just 13.50 g/m鲁. This difference highlights how the temperature controls saturation points and, thus, relative humidity.
Key points about temperature's impact on humidity:
  • Warmer temperatures increase air's moisture capacity, often leading to lower relative humidity unless more moisture is added.
  • As air cools, its moisture holding capacity decreases, raising relative humidity even with the same amount of moisture present.
  • This concept is also why dew forms on cooler surfaces, like windows or glasses, when air reaches its saturation point on those cooler surfaces.
Understanding these principles allows us to better grasp environmental changes and their resultant impacts on atmospheric moisture and comfort levels.

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Most popular questions from this chapter

How many calories are required to heat each of the following from \(15^{\circ} \mathrm{C}\) to \(65^{\circ} \mathrm{C} ?(a) 3.0 \mathrm{~g}\) of aluminum, (b) \(5.0 \mathrm{~g}\) of Pyrex glass, \((c) 20 \mathrm{~g}\) of platinum. The specific heats, in cal \(/ \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), for aluminum, Pyrex, and platinum are \(0.21,0.20\), and \(0.032\), respectively.

Outside air at \(5{ }^{\circ} \mathrm{C}\) and 20 percent relative humidity is introduced into a heating and air-conditioning plant where it is heated to \(20^{\circ} \mathrm{C}\) and its relative humidity is increased to a comfortable 50 percent. How many grams of water must be evaporated into a cubic meter of outside air to accomplish this? Saturated air at \(5^{\circ} \mathrm{C}\) contains \(6.8 \mathrm{~g} / \mathrm{m}^{3}\) of water, and at \(20^{\circ} \mathrm{C}\) it contains \(17.3 \mathrm{~g} / \mathrm{m}^{3}\). $$ \begin{array}{l} \text { Mass } / \mathrm{m}^{3} \text { of water vapor in air at } 5^{\circ} \mathrm{C}=0.20 \times 6.8 \mathrm{~g} / \mathrm{m}^{3}=1.36 \mathrm{~g} / \mathrm{m}^{3}\\\ \text { Comfortable mass } / \mathrm{m}^{3} \text { at } 20^{\circ} \mathrm{C}=0.50 \times 17.3 \mathrm{~g} / \mathrm{m}^{3}=8.65 \mathrm{~g} / \mathrm{m}^{3}\\\ 1 \mathrm{~m}^{3} \text { of air at } 5^{\circ} \mathrm{C} \text { expands to }(293 / 278) \mathrm{m}^{3}=1.054 \mathrm{~m}^{3} \text { at } 20^{\circ} \mathrm{C}\\\ \text { Mass of water vapor in } 1.054 \mathrm{~m}^{3} \text { at } 20^{\circ} \mathrm{C}=1.054 \mathrm{~m}^{3} \times 8.65 \mathrm{~g} / \mathrm{m}^{3}=9.12 \mathrm{~g}\\\ \text { Mass of water to be added to each } m^{3} \text { of air at } 5^{\circ} \mathrm{C}=(9.12-1.36) \mathrm{g}=7.8 \mathrm{~g} \end{array} $$

In a calorimeter can (which behaves thermally as if it were equivalent to \(40 \mathrm{~g}\) of water) are \(200 \mathrm{~g}\) of water and \(50 \mathrm{~g}\) of ice, all at exactly \(0{ }^{\circ} \mathrm{C}\). Into this is poured \(30 \mathrm{~g}\) of water at \(90{ }^{\circ} \mathrm{C}\). What will be the final condition of the system? Let us start by assuming (perhaps incorrectly) that the final temperature is \(T_{f}>0{ }^{\circ} \mathrm{C}\). Then $$ \begin{array}{c} \left(\begin{array}{l} \text { Heat change of } \\ \text { hot water } \end{array}\right)+\left(\begin{array}{l} \text { Heat to } \\ \text { melt ice } \end{array}\right)+\left(\begin{array}{l} \text { Heat to warm } \\ 250 \mathrm{~g} \text { of water } \end{array}\right)+\left(\begin{array}{l} \text { Heat to warm } \\ \text { calorimeter } \end{array}\right)=0 \\ (30 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(T_{f}-90{ }^{\circ} \mathrm{C}\right)+(50 \mathrm{~g})(80 \mathrm{cal} / \mathrm{g})+(250 \mathrm{~g})\left(1 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(T_{f}-0{ }^{\circ} \mathrm{C}\right) \\ +(40 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(T_{f}-0{ }^{\circ} \mathrm{C}\right)=0 \end{array} $$ Solving gives \(T_{f}=-4.1^{\circ} \mathrm{C}\), contrary to our assumption that the final temperature is above \(0{ }^{\circ} \mathrm{C}\). Apparently, not all the ice melts. Therefore, \(T_{f}=0{ }^{\circ} \mathrm{C}\). To find how much ice melts, we write Heat lost by hot water \(=\) Heat gained by melting ice $$ (30 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(90{ }^{\circ} \mathrm{C}\right)=(80 \mathrm{cal} / \mathrm{g}) m $$ where \(m\) is the mass of ice that melts. Solving this equation yields \(m=34 \mathrm{~g}\). The final system has \(50 \mathrm{~g}-34 \mathrm{~g}=16 \mathrm{~g}\) of ice not melted.

A thermos bottle contains \(250 \mathrm{~g}\) of coffee at \(90{ }^{\circ} \mathrm{C}\). To this is added \(20 \mathrm{~g}\) of milk at \(5{ }^{\circ} \mathrm{C}\). After equilibrium is established, what is the temperature of the liquid? Assume no heat loss to the thermos bottle. Water, coffee, and milk all have the same value of \(c, 1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). The law of energy conservation allows us to write (Heat change of coffee) \(+\) (Heat change of milk) \(=0\) $$ (\mathrm{cm} \Delta T)_{\text {coffee }}+(\mathrm{cm} \Delta T)_{\text {milk }}=0 $$ In other words, the heat lost by the coffee equals the heat gained by the milk. If the final temperature of the liquid is \(T_{f}\), then $$ \Delta T_{\text {coffee }}=T_{f}=90^{\circ} \mathrm{C} \quad \Delta T_{\text {milk }}=T_{f}-5^{\circ} \mathrm{C} $$ Substituting and canceling \(c\) yields $$ (250 \mathrm{~g})\left(T_{f}-90^{\circ} \mathrm{C}\right)+(20 \mathrm{~g})\left(T_{f}-5{ }^{\circ} \mathrm{C}\right)=0 $$ Solving this equation leads to \(T_{f}=84{ }^{\circ} \mathrm{C}\).

Furnace oil has a heat of combustion of \(44 \mathrm{MJ} / \mathrm{kg}\). Assuming that 70 percent of the heat is useful, how many kilograms of oil are required to raise the temperature of \(2000 \mathrm{~kg}\) of water from \(20{ }^{\circ} \mathrm{C}\) to \(99{ }^{\circ} \mathrm{C}\) ?
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