THERMO EXAMPLE PROBLEMS AND SOLUTIONS
 
METEOROLOGIST JEFF HABY
**denotes correct answer
1. What pressure (in millibars) is exerted by a layer of air having a thickness of 1,300 meters and an average
density of 1.15 kg/m^3? Use the Earth's gravitational constant.
Pressure = density * g * height
a. 97.78 mb
b. 146.51 mb**
c. 159.25 mb
d. 154.35 mb
e. 167.89 mb
P = (1.15 kgm^3)*(9.8 ms^2)*(1,300 m) = 14,651 Pa = 146.51 mb
2. Using the hypsometric equation given below, what is the thickness between the 1000 and 500 mb levels when the
average temperature of the layer is 22 degrees F?
Rd= 287 J/kg/K; Z= (RdTavg/g)LN(P1/P2)
a. 5,317.2 gpm
b. 5,385.8 gpm
c. 5,428.9 gpm**
d. 5,634.3 gpm
e. 5,178.3 gpm
22 F = 5.6 C = 267.4 K
Z = ((287 JK^1kg^1)*(267.4 K) / 9.8 ms^2)*LN(2) = 5,428.9 gpm
3. What is the density of water vapor that exerts a vapor pressure of 16.54 millibars at 40 degrees C? Use
vapor pressure formula below:
Rv= 461 J/K/kg; e*(specific volume)=Rv*T; e is in Pascals
a. 0.0115 kg/m^3**
b. 0.0121 kg/m^3
c. 0.0134 kg/m^3
d. 0.0143 kg/m^3
e. 0.0155 kg/m^3
16.54 mb = 1,654 Pa
specific volume = Rv*T/P = (461 JK^1kg^1)(313 K) / 1,654 Pa = 87.24 m^3/kg
density (inverse of specific volume) = 0.0115 kg/m^3
4. What is the virtual temperature if the vapor pressure is equal to 25.5 millibars with an outside pressure of
985 millibars and an ambient temperature of 104 degrees F?
The formula is given as: Tv= T / (1(e/p)(0.3774))
a. 313.5 K
b. 314.7 K
c. 315.8 K
d. 316.1 K**
e. 317.2 K
Tv = 313 K/(1  (25.5 mb/985 mb)(0.3774)) = 316.1 K
5. A sample of H2O molecules exerts a pressure of 17 millibars (1,700 Pa) at a temperature of +17 degrees C.
Calculate the specific volume of the water vapor.
R*=8314.3 J/kg/K, H2O=18 au
a. 78.79 m^3/kg**
b. 43.48 m^3/kg
c. 30.30 m^3/kg
d. 23.26 m^3/kg
e. 18.87 m^3/kg
R*/au H20 = Rv = 8,314.3/18 = 461.9 J/kg/K = gas constant for water vapor
specific volume = Rv*T/P = (461.9 JK^1kg^1)(290 K)/1,700 Pa = 78.79 m^3/kg
6. What is the potential temperature of an air parcel at the 925 millibar level which is initially at a
temperature of 70 degrees F?
PT = T(1000/P)^0.286
a. 276 K
b. 283 K
c. 301 K**
d. 208 K
e. 312 K
70 F = 21.2 C= 294.1 K
PT = 294.1 K*(1000 mb/925 mb)^0.286 = 301 K
7. A sample of aluminum having a mass of 200 grams, a specific heat capacity of 0.20 cal/g/C, and a temperature
of 200 degrees C is put into 500 grams of water at 10 degree C contained in a copper vessel which is in
thermal equlibrium with the water with a mass of 100 grams and a specific heat capacity of 0.095 cal/g/C.
What is the temperature of the system at thermal equilibrium?
c1M1(TF  T1) + c2M2(TF  T2) + c3M3(TF  T3) = 0
a. 18.3 degrees C
b. 19.4 degrees C
c. 21.2 degrees C
d. 23.8 degrees C**
e. 26.2 degrees C
[0.20(200g)x(TF200C)]+[1(500g)x(TF10C)]+[0.095(100g)x(TF10c)]
40Tf  8000 + 500Tf  5,000 + 9.5Tf  95
549.5Tf = 13,095
Tf = 23.8 C
8. How much energy does it take to evaporate 10 pounds of water? One pound equals 0.45 kilograms. Use value
for latent heat of vaporization.
a. 11250000 Joules**
b. 12250000 Joules
c. 13250000 Joules
d. 14250000 Joules
e. 15250000 Joules
2.5*10^6 J/kg * 4.5 kg = 1,1250,000 Joules
9. Using the ClausiusClapeyron equation, what is the saturation vapor pressure of air at 4 degrees C?
LN(es/6.11) = L/Rv(1/273  1/T); L = 2.453 * 10^6 J/kg; Rv= 461 J/kg
a. 7.1 mb
b. 8.1 mb**
c. 9.1 mb
d. 10.1 mb
e. 11.1 mb
4 C = 277 K
LN(es/6.11) = 5,321.0412(1/273  1/277)
LN(es/6.11) = 0.281458389
es/6.11 = e^0.281458389
es = e^0.281458389 * 6.11
es = 8.1 mb
10. What is the dewpoint in Fahrenheit when the actual vapor pressure is 13 mb?
LN(e/6.11) = L/Rv(1/273  1/Td) ; L = 2.453 * 10^6 J/kg; Rv= 461 J/kg
a. 37 F
b. 42 F
c. 47 F
d. 52 F**
e. 57 F
LN(13/6.11) = 5,321.0412(1/273  1/Td)
(LN(13/6.11) / 5,321.0412)  1/273 = 1/Td
Td = 284 K = 11 C = 52 F
11. Suppose the vapor pressure of the air is 18.0 millibars, what is the percentage of water vapor in the
atmosphere compared to total air using a surface pressure of 995 millibars?
a. 1.2 %
b. 1.5 %
c. 1.8 %**
d. 2.2 %
e. 2.6 %
Answer: 18/995 * 100%= 1.8%
12. What is the mixing ratio of a parcel of air that has a vapor pressure of 33.8 millibars at the 1020 mb level?
The formula is: w = (0.622e) / (P  e)
a. 21.3 g/kg**
b. 23.8 g/kg
c. 24.9 g/kg
d. 25.9 g/kg
e. 26.3 g/kg
Answer: w = (0.622*33.8 mb) / (1020 mb  33.8 mb) = 0.0213 kg/kg = 21.3 g/kg
13. What is the virtual temperature of air at the 1000millibar level at 300.0 K that has a vapor pressure of
13.7 millibars?
Tv = T(1 + 0.61w)
a. 301.6 K**
b. 302.6 K
c. 303.2 K
d. 304.0 K
e. 304.9 K
Answer: Tv = T(1 +0.61*(w)
w = (0.622*13.7 mb) / (1000 mb  13.7 mb) = 0.008639765 kg/kg
Tv = 300 K(1 + 0.61(0.008639765)) = 301.6 K
14. What is the relative humidity of air that has a temperature of 70 degrees F and a dewpoint of 60 degrees F?
Use ClausiusClapeyron equation.
RH = e/es
a. 57 %
b. 64 %
c. 71 %**
d. 79 %
e. 84 %
Answer: 60 F = 15.6 C = 288.555 K
70 F = 21.1 C = 294.1111 K
(es of 288.555 K) = 17.47 mb
(es of 294.111 K) = 24.75 mb
RH = (17.47/24.75)*100% = 71%
15. Using the ThetaE formula, what is the ThetaE of air at 650 millibars that has a temperature of 20 degrees F
and a mixing ratio of 2 g/kg?
ThetaE = T(1000/P)^0.286+3w
a. 338 K
b. 322 K
c. 307 K**
d. 297 K
e. 291 K
Answer: 20 F = 6.666 C = 266.333 K
THETAE = 266.333 K (1000 mb/650 mb)^0.286 + 3(2 g/kg) = 307 K
16. The 1000 to 500 mb thickness is 5,400 gpm. If this column of air exerts an air pressure of 500 mb, what is
the average density of this air?
a. 1.23 kg/m^3
b. 1.17 kg/m^3
c. 0.94 kg/m^3**
d. 0.81 kg/m^3
e. 0.64 kg/m^3
Answer: P = density * gravity * height
G = 9.8, H = 5,400 m, P = 50,000 Pa
50,000 Pa = density * 9.8 ms^2 * 5,400 m
density = 0.94 kg/m^3
17. What is the 1000 to 500 mb thickness when the average virtual temperature of that layer is 5 (negative 5) C?
a. 5,540 gpm
b. 5,440 gpm**
c. 5,340 gpm
d. 5,240 gpm
e. 5,140 gpm
Answer: Z = (Rd*Tv/g)*LN(p1/p2)
Z = (287 JK^1kg^1*268 K / 9.8 ms^2)*LN(2) = 5,440 gpm
18. What does the average 1000 to 500 mb average virtual temperature need to be in order to have a
1000 to 500 mb thickness of 5,300 gpm?
a. 18 C
b. 16 C
c. 14 C
d. 12 C**
e. 10 C
Answer: Z = (Rd*Tv/g)*LN(p1/p2)
5,300 m = (287 JK^1kg^1*Tv / 9.8 ms^2)*LN(2)
Tv = 261 K = 12 C
19. Water has a density of 1 g/cm^3. What is the specific volume of water?
a. 1 cm^3/g**
b. 4 cm^3/g
c. 9 cm^3/g
d. 16 cm^3/g
e. 25 cm^3/g
Answer: specific volume is inverse of density. Since density = 1/1, specific volume = 1/1
20. How much pressure in Pascals is exerted by liquid water on the bottom of a swimming pool that is 3.5 meters deep?
The density of water is 1000 kg/m^3.
a. 27,600 Pa
b. 30,800 Pa
c. 34,300 Pa**
d. 37,500 Pa
e. 38,900 Pa
P = density * gravity * height = 1000 kgm^3 * 9.8 ms^2 * 3.5 m = 34,300 Pa
21. The RH is 25% and the temperature is 68 degrees F. What is the actual vapor pressure?
a. 5.78 mb**
b. 5.98 mb
c. 5.18 mb
d. 5.38 mb
e. 5.58 mb
Answer: 68 F = 20 C= 293 K
es of 293 K = 23.11 mb
0.25 = e/23.11
e = 0.25*23.11 = 5.78 mb
22. The RH is 25% and the temperature is 68 degrees F. What is the dewpoint?
a. 38 F
b. 35 F
c. 31 F**
d. 27 F
e. 25 F
Answer: 68 F = 20 C= 293 K
es of 293 K = 23.11 mb
0.25 = e/23.11
e = 0.25*23.11 = 5.78 mb
LN(5.78/6.11) = 5,321.0412(1/273  1/Td)
LN(5.78/6.11) / 5,321.0412  1/273 = 1/Td
Td = 272.2 K = 0.8 C = 31 F
23. A sample of air has a RH of 65%, temperature of 32 F and a pressure of 1000 mb. Calculate its virtual
temperature.
a. 273.2 K
b. 273.4 K**
c. 273.6 K
d. 273.8 K
e. 274.0 K
Answer: Tv = T / (1  (e/p)*(0.3774))
32 F = 0 C = 273 K
es of 273 K = 6.11 mb
0.65 = e/6.11, e = 3.97 mb
Tv = 273 K / (1  (3.97 mb/1000 mb)*(0.3774)) = 273.4 K
24. A sample of air has a RH of 50% and a temperature of 32 F. What is the saturation vapor pressure and
actual vapor pressure of this air?
a. 12.8 mb and 6.4 mb
b. 9.5 and 4.8 mb
c. 10.0 mb and 5.0 mb
d. 6.1 mb and 3.1 mb**
e. 15.7 and 7.9 mb
Answer: es of 273 = 6.11 mb
e = es*0.5 = 3.11 mb
25. The temperature is 17 C with a dewpoint of 5 C. After a rain shower the temperature cools to 13 C and the
dewpoint rises to 11 C. By how many millibars did the actual vapor pressure rise in the time period to before
to after the shower?
a. 4.3 mb**
b. 5.3 mb
c. 6.3 mb
d. 7.3 mb
e. 8.3 mb
Use dewpoint to find actual vapor pressure
The dewpoint rises from 5 C to 11 C
5 C = 278 K, e of 278 = 8.68 mb
11 C = 284 K, e of 284 = 13.00 mb
13.00  8.68 = a change of 4.3 mb
26. Which temperature is an expression of absolute zero?
a. The temperature alcohol freezes
b. 173.16 ° C
c. 0 K**
d. The temperature water freezes
e. Both 2 and 3 are correct
Answer: 0 K is absolute zero, this is equal to 273.16 C
27. In the real atmosphere, why does an air parcel expand as it rises?
a. Because temperatures aloft are colder; the warmer air parcel expands into the colder air aloft
b. Because the force of gravity decreases; a smaller value of gravity causes mass to spread apart
c. Because the winds tend to be stronger; high wind causes divergence
d. Because pressure and air density decrease with height; The air parcel expands to equalize pressure with
surrounding air**
e. All of the above
Answer: When an air parcel is exposed to lower surrounding pressure, it expands. Since pressure
decreases with height, rising parcels expand and cool.
28. 2 kilogram of water at 35 degrees C is mixed with 4 kilograms of water at 3 degrees C, what is the
temperature of the water after the two water samples mix with each other and come to a temperature equilibrium?
a. 10.7 degrees C
b. 11.7 degrees C
c. 12.7 degrees C
d. 13.7 degrees C**
e. 14.7 degrees C
Answer: 2 kg(Tf  35) + 4 kg (Tf  3) = 0
2Tf  70 + 4Tf  12 = 0
6Tf = 82
Tf = 13.7 C
29. Which statement is false?
a. The value for the latent heat of freezing is 2.83 * 10^6 J/kg**
b. The value for the latent heat of condensation is 2.5 * 10^6 J/kg
c. The value for the latent heat of evaporation is 2.5 * 10^6 J/kg
d. The value for the latent heat of sublimation is 2.83 * 10^6 J/kg
e. The value for the latent heat of melting is 3.34 * 10^5 J/kg
Answer: sublimation and deposition have the largest latent heat value therefore answer a must be false.
30. The saturation vapor pressure over supercooled water is __________ than that over ice at the same
temperature. This causes ice crystals to grow at the expense of supercooled water droplets.
a. Higher**
b. Lower
Evaporation occurs more readily above a liquid surface since the water molecules have more motion. Thus,
there is a higher evaporation over a liquid surface and thus a higher vapor pressure over supercooled water
as compared to ice. Since motion is from high to low pressure, the vapor over the liquid water moves toward
the lower vapor pressure above the ice. Thus, the ice crystals grow at the expense of the supercooled
water droplets.
31. The temperature air WOULD have AFTER evaporation produces a relative humidity of 100%.
a. Absolute humidity
b. Mixing ratio
c. Specific humidity
d. Wetbulb temperature**
e. Equivalent potential temperature
Answer: This defines the wetbulb
32. In which case will the dewpoint equal the wetbulb temperature?
a. When the air is perfectly dry
b. When the air is saturated, relative humidity is 100%**
c. In both perfectly dry air OR when the relative humidity is 100%
d. The dewpoint can NEVER equal the wetbulb temperature
When air is saturated, T = Td = Tw
33. When air at 50 degrees F overruns a surface soil that is at 40 degrees F, heat travels in which direction?
a. From the soil to the overrunning air
b. From the overrunning air to the soil**
Answer: Heat travels from warm toward cold objects
34. How is LCL found on a SkewT LogP diagram?
a. From a pressure surface, it is where the mixing ratio through the dewpoint intersects the environmental sounding
b. From a pressure surface, it is half way between the dewpoint and temperature
c. From a pressure surface, it is the point on the sounding where the parcel temperature is warmer than the
environmental temperature
d. From a pressure surface, it is where the mixing ratio through the dewpoint intersects the dry adiabat
through the temperature**
e. From a pressure surface, it is where the MALR intersects the dewpoint
35. Condensation will result in the conversion of ______________________.
a. Latent to sensible heat**
b. Sensible to latent heat
Answer: The process of condensation warms the air, thus condensation releases latent heat to sensible
heat (heat energy that can be sensed).
36. Which value of the moist adiabatic lapse rate below would represent the moist
adiabatic lapse rate in a very warm and humid tropical environment?
a. 4 degrees C/km**
b. 6 degrees C/km
c. 7 degrees C/km
d. 9.8 degrees C/km
The release of latent heat warms the air. This changes the lapse rate from dry adiabatic to pseudoadiabatic.
Very warm and moist air will be able to release large amounts of latent heat, thus the lapse rate will be
small (less than 1/2 the DALR)
37. Which of these indices is found by raising a parcel from the 850millibar level and is used mainly for cool
season convection?
a. CAPE
b. Showalter Index**
c. Lifted Index
d. Slice method
e. Total Totals
38. From the hydrostatic equation dp/dz= g(density), The change is pressure with
height will decrease if:
a. Density of the air increases
b. The average temperature decreases
c. Density of the air decreases**
d. A deep layer of CAA occurs
dp/dz decreases if the right hand side of the equation decreases. This can be accomplished by decreasing
the density of the air. The other answers increase the density of air
39. A parcel of air with an initial pressure of 1000 mb is expanded from 5 m^3 to 10 m^3. How much heat energy,
in units of Joules, is taken by the air in this process? (du=0)
a. 250,000 Joules
b. 333,000 Joules
c. 500,000 Joules**
d. 755,000 Joules
e. 1,000,000 Joules
100,000 Pa * 5 m^3 = 500,000 Joules
40. The potential temperature of air at 850 mb is 283 K. What is the temperature of the air?
a. 18 (negative 18) C
b. 11 (negative 11) C
c. 7 (negative 7) C
d. 3 (negative 3) C**
e. 0 (zero) C
Answer: 283 = T(1000 mb/850 mb)^0.286
283 / (1000 mb/850 mb)^0.286 = T
T = 270 K = 3 C
41. All of the following are terms in the equation for the moist adiabatic lapse rate, except:
a. The latent heat of condensation
b. The gas constant for water vapor**
c. The change is saturation mixing ratio with temperature
d. The dry adiabatic lapse rate
e. The specific heat capacity
42. The mixing ratio is defined as the mass of water vapor divided by the mass of dry air. As a parcel of
air rises, the mixing ratio through the dewpoint of that parcel of air:
a. Increases
b. Decreases
c. Remains constant**
mixing ratio lines are drawn on the SkewT. When a parcel of unsaturated air rises, the mixing
ratio is conserved.
43. A parcel of air with a constant mixing ratio through the dewpoint of 10 g/kg rises from 1000 to 850 mb. How
much (absolute value) does the vapor pressure of this parcel of air change?
w = 0.622e / (P  e)
a. 0.7 mb
b. 2.4 mb**
c. 3.7 mb
d. 4.2 mb
e. The vapor pressure does not change
Answer: First the equation needs to be rearranged where e is on one side of the equation
w = 0.622e / (P  e)
w(P  e) = 0.622e
wP  We = 0.622e
wP = 0.622e + we
wP = e(0.622 + w)
e = wP / (0.622 + w)
second, find e when w= 0.010 kg/kg and P= 1000
e = (0.010*1000) / (0.622 + 0.010) = 15.82 mb
third, find e when w=0.010 kg/kg and P=850
e = (0.010*850) / (0.622 + 0.010)= 13.45 mb
change in e = 15.82  13.45 = 2.4 mb
44. As a parcel of unsaturated air rises in the atmosphere, the dewpoint of that parcel of air:
a. Increases
b. Decreases**
c. Remains constant
Answer: although w is conserved as a parcel of air rises, the dewpoint decreases by a lapse
rate of about 2 C/km
45. What are the units of heat?
a. Newtons
b. Pascals
c. Joules**
d. Kelvin
e. Watts
heat has the same units as energy which are Newtons times a meter = N*m = Joule


