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Weather & water: Earth Science Regents Practice

Practice Weather & water for the Earth Science Regents with real-style questions and full worked solutions. Miss one? Read the solution — the exam reuses the same ideas every year.

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Question 8 · 1 pt

Which instrument is used to measure air pressure?

Worked solution
$\text{air pressure} \Rightarrow \text{barometer}$ Match the instrument to the variable.
Answer: \text{barometer}

A barometer measures atmospheric pressure; falling pressure often signals approaching stormy weather.

Question 49 · 2 pts

A student uses a sling psychrometer and records a dry-bulb temperature of $20^\circ\text{C}$ and a wet-bulb temperature of $14^\circ\text{C}$. Using the ESRT, what is the relative humidity of the air?

Worked solution
$\Delta T = T_{dry} - T_{wet} = 20^\circ\text{C} - 14^\circ\text{C} = 6^\circ\text{C}$ Find the wet-bulb depression.
$\text{Relative Humidity table: dry-bulb} = 20^\circ\text{C},\ \Delta T = 6^\circ\text{C}$ Locate the row and column on the ESRT.
$\text{RH} = 51\%$ Read the value at the intersection.
Answer: 51%

A psychrometer compares dry-bulb and wet-bulb temperatures; a larger wet-bulb depression means drier air and a lower relative humidity.

Question 54 · 2 pts

A surface air temperature is measured as $50^\circ\text{C}$. Using the ESRT temperature scale, what is this temperature on the Kelvin scale?

Worked solution
$K = {}^\circ\text{C} + 273$
$K = 50 + 273 = 323\ \text{K}$
Answer: 323 K

Kelvin temperature equals Celsius temperature plus 273, so the two scales differ only by an offset, not by a scaling factor.

Question 61 · 2 pts

A maritime tropical (mT) air mass moves into New York State during the summer. This air mass is best described as

Worked solution
$m = \text{maritime, formed over ocean} \Rightarrow \text{moist}$ The first letter gives the moisture and source surface.
$T = \text{tropical, formed at low latitude} \Rightarrow \text{warm}$
$mT \Rightarrow \text{warm and humid}$
Answer: warm and humid

Air masses are named for their source region: maritime means moist (formed over water) and tropical means warm (formed at low latitudes), so mT air is warm and humid.

Question 64 · 2 pts

Through the evening at a weather station, the air temperature and the dewpoint temperature steadily move closer together. This change indicates that the

Worked solution
$\text{air temp} - \text{dewpoint} \to 0 \Rightarrow \text{air approaches saturation}$ The spread reaches zero at 100% relative humidity.
$\text{smaller spread} \Rightarrow \text{higher relative humidity}$
Answer: relative humidity is increasing

As air temperature and dewpoint converge, the air nears saturation, so relative humidity rises and clouds, fog, or precipitation become more likely.

Question 7 · 2 pts

A sling psychrometer is used to measure air temperature. As the difference between the dry-bulb and wet-bulb temperature readings (the wet-bulb depression) becomes larger, the relative humidity of the air

Worked solution
$The wet-bulb thermometer cools due to evaporation; the drier the air, the more evaporation occurs and the lower the wet-bulb reading becomes.$ A larger depression means more evaporative cooling took place.
$More evaporation occurs when the air is farther from saturation, meaning the air holds less moisture relative to its capacity.$
$Therefore, a larger wet-bulb depression corresponds to a lower relative humidity.$
Answer: decreases

A greater wet-bulb depression indicates drier air, because more evaporative cooling of the wet bulb occurs when the surrounding air is farther from saturation, which corresponds to lower relative humidity.

Question 29 · 2 pts

A student uses a sling psychrometer and records a dry-bulb temperature of $22°C$ and a wet-bulb temperature of $18°C$. If the wet-bulb depression (the difference between the dry-bulb and wet-bulb readings) increases while the dry-bulb temperature stays the same, the relative humidity of the air will:

Worked solution
$Wet-bulb depression = 22°C - 18°C = 4°C$
$A larger depression means more evaporative cooling occurred at the wet bulb.$
$More evaporation happens when the surrounding air is drier, which corresponds to lower relative humidity.$
Answer: decrease

A greater wet-bulb depression indicates drier air, because more water evaporates (cooling the wet bulb more) only when the air is not already saturated, so relative humidity decreases as the depression increases.

Question 55 · 4 pts · constructed response

The air temperature at a weather station is 24°C. At this temperature, one kilogram of air can hold a maximum of 21 grams of water vapor. The air at the station actually contains 10.5 grams of water vapor per kilogram of air. Calculate the relative humidity of the air at this station. Show your work.

Model answer: 50%
Worked solution
$\text{Relative Humidity} = \dfrac{\text{actual water vapor content}}{\text{maximum water vapor capacity}} \times 100$
$\text{RH} = \dfrac{10.5 \text{ g}}{21 \text{ g}} \times 100$
$\text{RH} = 50\%$
Answer: 50%

Relative humidity compares how much water vapor the air actually contains to the maximum amount it could hold at that temperature; dividing the actual amount by the capacity and multiplying by 100 gives the percentage.

Question 62 · 4 pts · constructed response

At a certain air temperature, one kilogram of air can hold a maximum of $20$ grams of water vapor before becoming saturated. The air currently contains $12$ grams of water vapor per kilogram. Calculate the relative humidity of the air and state whether the air is close to saturation.

Model answer: The relative humidity is $60\%$. The air is moderately moist but not close to saturation, since relative humidity would need to approach $100\%$ for that.
Worked solution
$RH = \dfrac{\text{actual water vapor}}{\text{maximum water vapor capacity}} \times 100\%$ relative humidity formula
$RH = \dfrac{12 \text{ g}}{20 \text{ g}} \times 100\%$
$RH = 60\%$
Answer: The relative humidity is $60\%$. The air is moderately moist but not close to saturation, since relative humidity would need to approach $100\%$ for that.

Relative humidity compares the actual moisture content of air to the maximum amount the air could hold at that temperature; a value of 60% shows the air is unsaturated and can still absorb more water vapor before reaching the dew point.

Question 84 · 4 pts · constructed response

A parcel of air at a dry-bulb temperature of 22°C contains 15 grams of water vapor per kilogram of air. At 22°C, this air could hold a maximum of 20 grams of water vapor per kilogram of air before becoming saturated. Calculate the relative humidity of this air parcel. Show your work and express your answer as a percent.

Model answer: The relative humidity of the air parcel is 75%.
Worked solution
$RH = \dfrac{\text{actual water vapor content}}{\text{maximum water vapor capacity}} \times 100$ Relative humidity formula
$RH = \dfrac{15 \text{ g/kg}}{20 \text{ g/kg}} \times 100 = 75\%$
Answer: The relative humidity of the air parcel is 75%.

Relative humidity measures how close air is to being saturated by comparing the actual amount of water vapor present to the maximum amount the air could hold at that temperature. Dividing the actual content by the maximum capacity and multiplying by 100 gives 75%, meaning the air is holding three-quarters of the moisture it is capable of holding at 22°C.

Question 7 · 2 pts

The dry-bulb temperature and wet-bulb temperature at a weather station are both $18°C$. What is the relative humidity?

Worked solution
$\text{Wet-bulb depression} = \text{dry-bulb temp} - \text{wet-bulb temp} = 18°C - 18°C = 0°C$
$\text{A wet-bulb depression of } 0° \text{ means no evaporative cooling occurred.}$
$\text{This happens only when the air is already saturated, so relative humidity} = 100\%.$
Answer: 100%

When dry-bulb and wet-bulb temperatures are equal, the wet-bulb depression is zero, meaning no evaporation is cooling the wet thermometer because the surrounding air is already saturated with water vapor, which corresponds to a relative humidity of 100%.

Question 28 · 2 pts

A psychrometer measures relative humidity by comparing dry-bulb and wet-bulb temperature readings. When the dry-bulb temperature and wet-bulb temperature are equal, meaning the wet-bulb depression is 0°C, what is the relative humidity of the air?

Worked solution
$A wet-bulb depression of 0°C means no evaporative cooling occurred at the wet bulb.$
$No evaporation occurs when the surrounding air is already fully saturated with water vapor.$
$A saturated air mass corresponds to a relative humidity of 100%.$
Answer: 100%

Zero wet-bulb depression means no evaporative cooling took place, which only happens when the air is completely saturated, giving a relative humidity of 100%.

Question 34 · 2 pts

An air mass forms over land in northern Canada during winter. Based on its source region, this continental polar (cP) air mass would most likely have which characteristics?

Worked solution
$The prefix 'continental' indicates the air mass formed over land, which contributes low moisture content.$
$The term 'polar' indicates the source region is at high latitude, which contributes cold temperatures.$
$Combining these source-region characteristics gives an air mass that is cold and dry.$
Answer: Cold and dry

Continental polar air masses form over high-latitude land areas, so they inherit both the low moisture of a land source and the cold temperature of a polar source region.

Question 43 · 2 pts

As air temperature decreases toward the dew point while the amount of moisture in the air stays the same, the relative humidity of the air will

Worked solution
$Relative humidity compares the actual moisture in the air to the maximum amount the air could hold at its current temperature.$
$Cooler air can hold less moisture, so as temperature drops toward the dew point (with actual moisture unchanged), the air gets closer to its maximum capacity.$
$When air temperature equals the dew point, the air is fully saturated and relative humidity equals 100%.$
Answer: increase, approaching 100%

Relative humidity rises as air cools toward its dew point because cooler air has a lower moisture-holding capacity, bringing the actual moisture content closer to saturation; at the dew point itself, relative humidity reaches 100%.

Question 50 · 2 pts

A student uses a sling psychrometer and records a dry-bulb temperature of 22°C and a wet-bulb temperature of 22°C. What is the relative humidity of the air at the time of the reading?

Worked solution
$Wet-bulb depression = dry-bulb temperature - wet-bulb temperature = 22^{\circ}C - 22^{\circ}C = 0^{\circ}C$ calculate depression
$A wet-bulb depression of 0^{\circ}C means no evaporative cooling occurred from the wet-bulb thermometer's wick.$ no evaporation
$Using the Relative Humidity reference table with a dry-bulb temperature of 22^{\circ}C and a depression of 0^{\circ}C gives a relative humidity of 100\%.$ reference table lookup
Answer: 100%

When the wet-bulb and dry-bulb temperatures are equal, the wet-bulb depression is zero, meaning the air is holding all the water vapor it can at that temperature and no evaporative cooling can occur. A wet-bulb depression of 0 degrees always corresponds to a relative humidity of 100%, regardless of the actual temperature value.

Question 61 · 4 pts · constructed response

Two weather stations, A and B, are located 400 kilometers apart. Station A records a sea-level air pressure of 1016.0 millibars, and Station B records a sea-level air pressure of 996.0 millibars. Calculate the pressure gradient between Station A and Station B in millibars per 100 kilometers. Show your work.

Model answer: The pressure gradient is 5.0 millibars per 100 kilometers.
Worked solution
$\text{Pressure difference} = 1016.0 - 996.0 = 20.0\ \text{mb}$
$\text{Gradient} = \dfrac{20.0\ \text{mb}}{400\ \text{km}} \times 100\ \text{km} = 5.0\ \text{mb per 100 km}$
Answer: The pressure gradient is 5.0 millibars per 100 kilometers.

Pressure gradient is found by dividing the difference in pressure between two points by the distance between them, then scaling to a standard distance (100 km) so gradients can be compared consistently.

Question 80 · 4 pts · constructed response

Two weather stations lie 300 kilometers apart along a straight line. Station A records a temperature of 25 degrees Celsius, and Station B records a temperature of 13 degrees Celsius. Calculate the temperature gradient between the two stations in degrees Celsius per kilometer. Show your work.

Model answer: The temperature gradient is 0.04 degrees Celsius per kilometer.
Worked solution
$\text{gradient} = \dfrac{\text{difference in temperature}}{\text{distance}}$
$\text{gradient} = \dfrac{25^\circ C - 13^\circ C}{300\text{ km}} = \dfrac{12^\circ C}{300\text{ km}}$
$\text{gradient} = 0.04\ ^\circ C/\text{km}$
Answer: The temperature gradient is 0.04 degrees Celsius per kilometer.

A gradient is calculated as the change in a quantity divided by the distance over which that change occurs; here the 12 degree Celsius temperature difference divided by the 300 km separation yields a gradient of 0.04 degrees Celsius per kilometer.

Question 85 · 4 pts · constructed response

At an air temperature of 20°C, the maximum quantity of water vapor the air can hold (its water vapor capacity) is 15 grams of water vapor per kilogram of air. A sample of air at this same temperature actually contains 9 grams of water vapor per kilogram of air. Calculate the relative humidity of this air sample. Show the equation you used and your work.

Model answer: The relative humidity is 60%.
Worked solution
$\text{relative humidity} = \dfrac{\text{water vapor content}}{\text{water vapor capacity}} \times 100\%$ Relative humidity formula
$RH = \dfrac{9\ \text{g/kg}}{15\ \text{g/kg}} \times 100\%$ Substitute the given values
$RH = 0.60 \times 100\% = 60\%$ Multiply to find the percentage
Answer: The relative humidity is 60%.

Relative humidity compares how much water vapor is actually present in the air to the maximum amount the air could hold at that temperature; dividing the actual content by the capacity and multiplying by 100% gives the percent saturation, here 60%.

Question 4 · 2 pts

During a weather observation, the air temperature and the dew point temperature are found to be the same value. What is the relative humidity of the air at this time?

Worked solution
$Relative humidity depends on how close air temperature is to dew point.$ concept
$When air temperature = dew point, the air is completely saturated with water vapor.$ definition of dew point
$Saturated air has relative humidity of $100\%$.$
Answer: $100\%$

Dew point is the temperature to which air must be cooled to become saturated. When the actual air temperature equals the dew point, the air holds all the water vapor it can at that temperature, so relative humidity is 100%.

Question 32 · 2 pts

A sling psychrometer is used to measure air temperature. The dry-bulb thermometer reads 20°C and the wet-bulb thermometer also reads 20°C, giving a wet-bulb depression of 0°C. What is the relative humidity of the air?

Worked solution
$Wet-bulb depression = dry-bulb temperature - wet-bulb temperature = 20°C - 20°C = 0°C.$
$A wet-bulb depression of 0°C means no evaporative cooling occurred from the wet-bulb wick.$
$No evaporation occurs only when the air is already fully saturated with water vapor, so relative humidity = 100\%.$
Answer: 100%

Evaporation from the wet-bulb wick cools it below the dry-bulb reading unless the surrounding air is already saturated; a wet-bulb depression of exactly 0°C therefore corresponds to a relative humidity of 100%.

Question 58 · 4 pts · constructed response

At a weather station, the dry-bulb thermometer reads 18°C and the wet-bulb thermometer reads 14°C. Calculate the wet-bulb depression (dry-bulb temperature minus wet-bulb temperature).

Model answer: 4°C
Worked solution
$\text{Wet-bulb depression} = \text{dry-bulb temp} - \text{wet-bulb temp}$
$= 18^\circ\text{C} - 14^\circ\text{C} = 4^\circ\text{C}$
Answer: 4°C

Wet-bulb depression is found by subtracting the wet-bulb temperature from the dry-bulb temperature; a larger depression indicates drier air, since more evaporative cooling occurred at the wet bulb.

Question 62 · 4 pts · constructed response

At a weather station, the air temperature is 24°C and the dew point is 12°C. (a) State whether the air is currently saturated, and explain your reasoning. (b) If the air temperature drops to 12°C while the amount of moisture in the air remains unchanged, describe what will most likely happen to the water vapor in the air, and name the weather condition that could result near the ground.

Model answer: (a) The air is not saturated because the air temperature (24°C) is higher than the dew point (12°C); air is only saturated when temperature equals dew point. (b) If the air cools to 12°C without any change in moisture content, the relative humidity will reach 100 percent and the air becomes saturated, causing excess water vapor to condense on nuclei in the air. Near the ground, this condensation is likely to produce fog.
Worked solution
$Compare air temperature to dew point: 24^\circ C > 12^\circ C, so relative humidity is below 100%, meaning the air is unsaturated.$
$Cooling air to its dew point increases relative humidity to 100% without adding moisture.$
$At saturation, excess water vapor condenses into liquid droplets on condensation nuclei.$
$Condensation occurring near the ground surface is identified as fog.$
Answer: (a) The air is not saturated because the air temperature (24°C) is higher than the dew point (12°C); air is only saturated when temperature equals dew point. (b) If the air cools to 12°C without any change in moisture content, the relative humidity will reach 100 percent and the air becomes saturated, causing excess water vapor to condense on nuclei in the air. Near the ground, this condensation is likely to produce fog.

Dew point is the temperature at which air becomes saturated for its existing moisture content; cooling air down to its dew point causes condensation, which near the surface takes the form of fog.

Question 78 · 4 pts · constructed response

At a weather station, the air temperature is measured as 18°C and the dew-point temperature is also measured as 18°C. State the relative humidity of the air at this time. Explain, in terms of water vapor content, why the relative humidity has this value. Then state one type of weather condition that is likely to occur as a result.

Model answer: The relative humidity is 100%. When the dew-point temperature equals the air temperature, the air is holding all the water vapor it can hold at that temperature, meaning it is completely saturated. This condition often leads to fog, cloud formation, or precipitation.
Worked solution
$\text{Dew point} = \text{Air temperature} = 18^\circ C$ Given condition
$\text{Dew point} = \text{Air temperature} \Rightarrow \text{air is saturated}$ Apply saturation rule
$\text{Saturated air} \Rightarrow RH = 100\%$ State relative humidity
Answer: The relative humidity is 100%. When the dew-point temperature equals the air temperature, the air is holding all the water vapor it can hold at that temperature, meaning it is completely saturated. This condition often leads to fog, cloud formation, or precipitation.

Relative humidity compares the actual water vapor in the air to the maximum it could hold at that temperature; when temperature and dew point are equal, the air is saturated (100% RH), a condition commonly associated with fog, clouds, or precipitation.

Question 84 · 4 pts · constructed response

At 6:00 p.m., the air temperature at a weather station was 24°C and the dew point was also 24°C. State the relative humidity of the air at this time. Explain, in terms of the relationship between air temperature and dew point, why fog is likely to form at the surface at this time.

Model answer: The relative humidity is 100%. Because the air temperature equals the dew point, the air is completely saturated with water vapor; any further cooling causes water vapor to condense into tiny liquid droplets at the surface, forming fog.
Worked solution
$\text{Relative humidity} = 100\%\ \text{whenever air temperature} = \text{dew point}$ Definition of saturation
$\text{Saturated air holds the maximum water vapor possible at that temperature}$
$\text{Any additional cooling below this temperature forces condensation, producing fog near the ground}$
Answer: The relative humidity is 100%. Because the air temperature equals the dew point, the air is completely saturated with water vapor; any further cooling causes water vapor to condense into tiny liquid droplets at the surface, forming fog.

Relative humidity reaches 100% precisely when the air temperature equals the dew point, meaning the air is holding all the water vapor it can at that temperature. When saturated air near the ground cools even slightly further, water vapor condenses onto particles in the air, forming fog, which is essentially a cloud at ground level.

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