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UNBC ENSC 201 SAMPLE FINAL LAB EXAM ANSWERS FOR STUDENT PRACTICE
UNBC ENSC 201
LAB FINAL Sample Exam ANSWERS
Name:
Lab Section Time:
Student #:
Lab Instructor: Name:
Alias:
Chris Jackson
Total
_____
XXXX
SAMPLE ENSC 201 Lab Final Answers: The format shown here (but not
the questions) is what you will see for the lab final exam.
•
This exam consists of X questions on X pages; charts, formulas and constants are provided on the
last pages of this exam. Check you have a complete exam.
•
You have 80 minutes to do this exam. It is marked out of X marks and worth X% of your grade.
•
It is a closed book exam. Any detached pages must be stapled together. This exam is not returned to you,
but you can review it with the Senior Lab Instructor. Indicate an alias if you want to see your grade posted.
•
Calculators are allowed. Graphing / programmable calculators must be shown to the instructor before the
exam starts to demonstrate that the memory has been cleared.
•
Answer in the spaces provided. Use the reverse side if more space is needed.
•
SHOW ALL YOUR WORK TO GET FULL MARKS – AN ANSWER WITH LITTLE OR NO
INDICATION OF HOW IT WAS DETERMINED IS WORTH ZERO.
EXAM Q & A: These are a random selection of questions used in labs or past exams.
It doesn’t represent
a complete summary of what types or weighting of questions you might see on a final lab exam. Expect an exam that
covers the entire terms lab work. For calculations, expect to show your work and units for full marks. The equation
sheet is the same as the one that will be given in the exam. As with the practice midterm additional constants and
information may be include with particular exam questions.
1) Give the concept/principles behind using a psychrometer to measure atmospheric humidity.
If the wet bulb temperature is 10 0C, and the air temperature is 15 0C, calculate the vapour
pressure and relative humidity. (5 marks)
Note this question has 2 parts: Solution to first part: The concept behind measuring
humidity with a psychrometer: When the air is unsaturated, evaporation occurs as air flows
over the wet wick of a wet-bulb thermometer. Evaporation requires heat, which comes from the
internal energy of the air. The heat loss from the air, which provides the energy to evaporate
water from the wet-bulb, can be measured as a decrease in the temperature of the wet-bulb
thermometer when compared with the air temperature on the dry bulb. If the air is very dry (ie.
far from saturation) there will be more evaporation from the wet-bulb and the wet-bulb
temperature will decrease more. Psychrometry, through the psycrhometric equation, equates
the loss of heat in the air (as measured by the wet-bulb temperature decrease) with the latent
heat required to evaporate water from the wet-bulb.
Solution to second part: Calculating the vapour pressure (e) and RH: requires application of
the psychrometric equation (see below), and use of the e* vs T curve / graph from the humidity lab.
Determining e using the psychrometric equation:
e = e*(Tw) − λ(T − Tw) here λ is the psychrometric constant = 66 Pa 0C-1 .
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UNBC ENSC 201 SAMPLE FINAL LAB EXAM ANSWERS FOR STUDENT PRACTICE
Chris Jackson
From the e* vs T graph, determine e* at Tw or e* at 10 0C = e*(10) = 12.3 hPa = 1230 Pa.
0
(To find this value use the Saturation vapour pressure curve (over water), find 10 C on the x-axis, follow
0
10 C and follow the graph vertically until it intersects the curve, and then follow the intersection point
horizontally to read the e* value on the y-axis.)
Using the e* value from the graph solve for e:
e = 1230 Pa – [66 Pa 0C-1 (15 0C -10 0C)] = 900 Pa
Determining RH:
The RH is the ratio of the actual vapour pressure to the saturation vapour pressure. The
saturation vapour pressure is found on the e* vs T graph this time using the air temperature
instead of the wet bulb temperature. So e* at T or e* at 15 0C = e*(15) = 17 hPa = 1700 Pa.
RH =
900 Pa
e
× 100% =
× 100% = 53%
*
1700 Pa
e
2) A cool winter Canadian air mass with a temperature of -4 0C and a relative humidity of 91% meets
a warm air mass sweeping northward from Texas with a temperature of 17 0C and a relative
humidity of 26%. In the mixing process will the southern border of the Canadian air mass gain or
lose water vapour content? Show your method of determination and explain. (5 marks)
Solution: Since RH is highly dependent upon temperature (not just water vapour content) we cannot tell from the RH alone which air mass (i.e. the Canadian or Texan) will actually have more moisture. To determine this you need to find the vapour pressure (e -­‐-­‐ the actual amount of moisture for each air mass) as vapour pressure doesn’t vary with temperature. Since you know the RH, and the air temperatures, you can use these to determine the vapour pressures for each air mass. The air mass with the highest moisture content (i.e. e value) will lose moisture when it mixes with a drier air mass. For the Canadian air: From the e* vs T graph, e* at T= -4 0C or e*(-4) = 4.4 hPa = 440 Pa.
Rearranging RH =
e
× 100% to solve for e we get: e = e* (RH)
e*
Solving for e for the Canadian air mass:
e = e* (RH)= (440 Pa)(0.91) = 400.4 Pa = 4 hPa
For the Texan air: From the e* vs T graph, e* at T= 17 0C or e*(17) = 19.4 hPa = 1940 Pa.
Rearranging RH =
e
× 100% to solve for e we get: e = e* (RH)
e*
Solving for e for the Texan air mass:
e = e* (RH)= (1940 Pa)(0.26) = 504.4 Pa = 5 hPa
So the Texas air mass is actually more humid (i.e. has more moisture content) so when they mix the southern part of Canadian air mass will increase in moisture. 2
UNBC ENSC 201 SAMPLE FINAL LAB EXAM ANSWERS FOR STUDENT PRACTICE
3) To answer these refer to Figure 1.
a) Explain the rather anomalous dip
in the QE curve (and the related
rise in QH) during July and
August. (2 marks)
Chris Jackson
Figure 1, Lab Final Review: Vancouver’s monthly averaged
annual energy budget.
Solution: The QE dip in July and August
occurs because availability of water for
evapotranspiration is limited. There is no
longer enough moisture available for
evapotranspiration to meet its full
potential. Since Q* is still high, this
decrease in QE is utilized with an
increase in QH, which results because it
is drier and energy now goes into heating
the air.
b) How is it possible for the heat used in evapotranspiration (QE) to exceed the available net
radiation (Q*) in the September to February period? (2 marks)
∗
Solution: Energy used in evapotranspiration comes not only from radiation (Q ), but when there is a deficit, the ground’s heat storage (negative or upward QG) and energy supplied via sensible heat (negative QH or heat going from the air toward the surface) can provide energy for
evapotranspiration.
4) Repeat the questions given in the midterm review.
Solution:See midterm review sample questions and answers posted on the course website.
5) Using the plotted weather station information for a surface weather map (figure given at the end
of this review), determine the: isobars at 4 hPa intervals (4 marks), identify any high (H) and/or
low (L) pressure areas (2 marks), and locate any associated fronts (3 marks). (total 9 marks)
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UNBC ENSC 201 SAMPLE FINAL LAB EXAM ANSWERS FOR STUDENT PRACTICE
Chris Jackson
Solution: The answer is the same as what was done for the weather lab. The L in red represents a
low pressure centre. Correct answers should show the 980 to 1008 hPa isobars and a cold and a warm
front which are correctly located and identified by their symbols. Note that to contour this properly you
need to add a leading digit and know where the decimal is for each station.
L
6) Given a temperature of 14 0C and a wet bulb temperature of 7 0C, find the following and explain
how you found them. (Do not use a tephigram to answer this question.) Be sure to show the units.
a) Vapour pressure and dew point temperature (3 marks)
Solution: Want: vapour pressure = e; and dew point temperature = Td
Know: T = 14 0C
T w = 7 0C
e = e*(Tw) - λ(T – Tw) where λ = 66 Pa 0C-1 and
e = e*(Td)
Determine vapour pressure using: e = e*(Tw) - λ(T – Tw) where λ = 66 Pa 0C-1
and from the saturation vapour pressure graph get e*(Tw) at 7 0C = 10 hPa or 1000 Pa
e = 1000 Pa - [66 Pa 0C-1 (14 0C – 7 0C)] = 538 Pa = 5.4 hPa
Determine Td using the e value above and the saturation vapour pressure curve but this
time reading it to get Td from e. (i.e. e = e*(Td))
e at 5.4 hPa = a Td of -1.5 0C
b) Saturation vapour pressure and relative humidity. (2 marks)
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UNBC ENSC 201 SAMPLE FINAL LAB EXAM ANSWERS FOR STUDENT PRACTICE
Chris Jackson
Solution: Determine e* from the saturation vapour pressure graph get e*(T) at 14 0C = 16 hPa
or 1600 Pa
Determine RH using e from question a) above and e*
RH =
e
5.4 hPa
× 100% =
× 100% = 33.75% = 34%
*
e
16 hPa
c) Which of the values in a) and b) can be used when comparing the air’s actual vapour content
between locations where temperature and moisture content might vary? Why? (3 marks)
Solution: Only e and Td of the humidity measures above (e, e*, Td or RH), can be used to
compare moisture content between areas with different temperature because neither of these
ways of determining humidity depends on temperature. The others depend on temperature
(i.e. the potential for moisture in the air (e*) rises and falls with temperature changes; and RH
changes with both moisture and temperature.)
d) Which of the values in a) and b) above could be used when comparing vapour content
between locations where elevation as well as moisture content might vary? (3 marks)
Solution: None of the above measure of humidity (e, e*, Td or RH) can be used to
compare moisture content in areas where elevation changes as pressure decreases with
increasing elevation and e and Td depend on pressure. (Not asked for but FYI – only
mixing ratio (r) is not dependent on pressure.)
7) Explain how you might recognize the frontal passage of:
a) A warm front from the perspective of a person at a location where the front is passing
over? (3 marks)
Solution: Initially the observer would see cirrus clouds increasing in amount and density
followed by altostratus which would become status clouds that are thicker and closer to the
ground and gradually precipitation would start from nimbostratus clouds. Steady winds would
gradually increase from the east or southeast and switch to south or south west as the front
passes. Temperatures would warm as the front passes. (Pressure would drop as the front
passes but this wouldn’t be visible to a ground observer so not really a good answer here.)
b) A cold front from a weather chart? In your answer make sure to indicate the type of chart
you are using. (3 marks) Assume that the cold front symbols aren’t on the chart.
Solution: The type of chart is a surface weather map. A cold front on a surface weather map
is indicated by the cold front symbols on the chart (but you are to assume this isn’t given). On a
surface weather map a cold front is identified by the winds shifting (usually from south or south
west to north or north west), a temperature decrease, a pattern of cumulus type clouds
associated with a front, and a pressure decrease and then increase as the front passes.
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UNBC ENSC 201 SAMPLE FINAL LAB EXAM ANSWERS FOR STUDENT PRACTICE
Chris Jackson
c) Sketch a cross-section diagram (not an actual cross-section) through a low pressure
system that represents the types of weather features you expect to see along the cross
section. Make sure to appropriately label it. (10 marks)
Solution:
8)
If the height of a water column in a water barometer is 9.70 m and the water temperature is
210C, what is the pressure at that location? (4 marks)
The density of water (ρw) at various temperatures is:
Temperature (°C)
15
17
19
21
23
25
-3
Density - ρw (kg m )
999.099 998.774 998.405 997.992 997.538 997.044
Solution: Want to determine the pressure from a water barometer
Know: Pa = ρw g ΔZw + e*(T)
where:
ρw = density of water for 21 0C = 997.992 kg m-3 (from the table above)
g = gravity = 9.80665 m s-2 (from the equation sheet)
e*(T) at 21 0C = 25 hPa = 2500 Pa (from the saturation vapour pressure curve)
ΔZw = 9.70 m (given in the question above)
To get these units to make
sense and complete the addition
you need to realize that
-1 -2
1 kg m s = 1 Pascal (Pa)
Pa = ρw g ΔZw + e*(T)
= (997.992 kg m-3)(9.80665 m s-2)(9.70 m) +2500 Pa
= 94933.495 (kg m-3)(m s-2)(m) + 2500Pa = 94933.495 (kg m-1 s-2) + 2500Pa
= 94933.495 Pa + 2500 Pa
= 97433.5 Pa
= 974.335 hPa = 974 hPa
9)
Answer all parts: A spherical alien spaceship has entered orbit around the earth, 3.0 x 107
m above the earth’s surface. The spaceship has a radius of 1 km and a surface temperature
of 3000 K. It is a radiative “black-body”.
Solution: Though it may not seem like it, this question is similar to one from the radiation lab
where you determined the solar constant. For this problem it may help to draw a picture of the
situation in this problem as this will enable you to understand the what you are being told.
a) What is the flux density (W m-2) leaving the spaceship’s surface? (2 marks)
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UNBC ENSC 201 SAMPLE FINAL LAB EXAM ANSWERS FOR STUDENT PRACTICE
Chris Jackson
S = spaceship in orbit around Earth. Spaceship’s radius is 1 km. Spaceship’s temperature = 3000 K
S
Earth
7
Radius from spaceship orbit to Earth = 3.0 x 10 m
Solution: Want to determine the flux density (Wm-2) leaving the spaceship’s surface.
Know E = σ T4 and know the space ship’s temperature
So: Espaceship = (5.67 x 10-8 Wm-2 K-4) (3000 K)4 = (5.67 x 10-8 Wm-2 K-4) (8.1 x 1013 K4)
= 4.59 x 106 Wm-2
b) What is the total flux (W) emitted by the spacecraft? (2 marks)
Solution:
Total fluxspaceship = (Espaceship)(Area of the spaceship)
Area of the spaceship = 4πr2 where the spaceship’s radius = 1 km
Area of the spaceship = 4π(1 km)2 = 4π(1000 m)2 = 1.26 x 107 m2
So: Total fluxspaceship = (4.59 x 106 Wm-2)( 1.26 x 107 m2) = 5.78 x 1013 W
c) What is the flux density (W m-2) reaching the earth’s surface from the spacecraft,
ignoring the effects the atmosphere might have on attenuating this radiation? (2 marks)
Solution:
Flux density reaching
= Total fluxspaceship
the earth’s surface
Surface area of shell at the radius of the spaceship to Earth orbit
from the spaceship
13
13
-3
-2
Flux density reaching = 5.78 x 1013 W
= 5.78 x 10 W = 5.78 x 10 W = 5.11 x 10 Wm
the earth’s surface
2
7 2
2
16
2
4πr for the radius of the spaceship to Earth orbit
4π(3 x 10 ) m
1.13 x 10 m
from the spaceship
d) How does this value compare to the solar constant? Would you be worried about the
intensity of this radiation? Explain. (2 marks)
Solution: The flux density reaching the earth from the spaceship is very small (5.11 x 10-3
Wm-2) when compared to the solar constant which is 1367 Wm-2.
Unless this spaceship emits a very harmful unknown type of radiation that hurts at very low
intensities this should not be a problem as the spaceship’s temperature of 3000 K indicates
this radiation should be in between long and short wave radiation.
e) BONUS QUESTION: What is the wavelength of maximum emission from the alien
spacecraft? (2 marks) Note this question was not directly covered in labs this year so
would not be expected other than as a bonus question.
Solution:
2.88 × 10 −3 m K 2.88 × 10 −3 m K
λmax =
=
= 9.6 × 10 − 7 m
T
3000 K
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UNBC ENSC 201 SAMPLE FINAL LAB EXAM ANSWERS FOR STUDENT PRACTICE
Chris Jackson
Lab Exam Constants and Formulas:
Distance from the earth to the moon: 3.84 x 108 m
Distance from Earth to Sun: 1.50 x 1011 m
Distance from Mercury to the sun: 5.79 x 1010 m
Radius of the earth: 6.37 x 106 m
Surface Area of a circle = π r 2
Surface Area of a sphere = 4 π r 2
Solar flux: 3.865 x 1026 W
Kirchoff’s Law: aλ = Єλ
Wein’s Law: λmax =
Note these equations
weren’t part of the labs
this year.
2.88 ×10 −3 m K
T
E = σ T4 where Є =1 as
~ air density: ρair = 1.2 kg m-3
mercury density at 00C: ρHg = 13595.1 kg m-3
g = 9.80665 m s-2
Radius of the sun: 7x108 m
Solar constant: 1367 W m-2
~ water density: ρwater = 1000 kg m-3
E = Є σ T4
Q* = K* + L*
Q* = K ↓- K ↑ + L ↓- L ↑
L* = L↓ - L↑
1 nautical mile = 1852 m
σ = 5.67 x 10 -8 W m-2 K-4
LV = 2.5 x 106 J kg-1
E of 1 mm h-1 is equivalent to QE of 680 W m-2
00C = 273.15 K
K = °C + 273.15
0
C=
0
5(0F − 32)
9
⎡ 9
⎤
F = ⎢ (0C )⎥ + 32
⎣ 5
⎦
I = Io cos Z
cos Z = sinø sinδ + cosø cosδ cosh
K* = K↓- K↑ = [K↓ (1 –α)]
L* = L↓ - σT04
α=
K↑
K↓
Q* = QH + QE + QG
P = Et + r + ∆ S
Q E = Lv Et
Q
β= H
QE
360(TJ + 10)
365
)
e = ρv Rv T where Rv = 462 J kg-1 K-1
(for water vapour)
e = ρd Rd T where Rd = 287 J kg-1 K-1
(for dry air)
vpd = (e*(T) – e)
e = e*(Td)
e = e*(Tw) - λ(T-Tw) where λ = 66 Pa 0C-1 (over water)
e = e*(Tw) - λ(T-Tw) where λ = 58.2 Pa 0C-1 (over ice)
Pa = ρw g ΔZw + e*(T)
ΔPL = ρL g ΔZL
g ΔZ p
VG =
f ΔN
f = 2Ωsin
δ = − 23.4 cos (
r=
0.622 × e
g
× 1000
P−e
kg
RH =
e
*
(T )
e
×100% =
r
×100%
rs
ø where Ω = 7.27 x 10-5 s-1
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UNBC ENSC 201 SAMPLE FINAL LAB EXAM ANSWERS FOR STUDENT PRACTICE
Chris Jackson
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