CIV300/ENV346TerrestrialEnergySystems,Fall2016 1
Test1QuestionSheet
-Test1
ClosedBook.ThetimeallowedforPartAplusPartBintotalis75minutes.
Answersaretobewrittenontheanswersheetsgiven(onesheetforPartAand
anotherforPartB).
PARTA
3
1. Lake Ontario has a volume of 1,640 km . What volume of gasoline would need to be
o
burned (assuming 35% combustion efficiency) to heat it up by 1 C? Gasoline contains 35
MJ of energy per litre.
(a) 5.62x10
11
litres (b) 6.88x10
10
litres (c) 1.28x10
11
litres
(d) 1.97x10
11
litres
ANS A
o
2. Earth’s angle of obliquity is 23.5 off the plane of the ecliptic. If an asteroid hits Earth and
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we are knocked to a new angle of 30 , which of the following will occur?
(a) Winter time will be longer in the northern hemisphere
(b) The pole star will no longer point north in the northern hemisphere
(c) The annual amount of energy reaching the Earth from the sun will change
(d) The equator will move further north
ANS B
3. From the deck of a cruise ship, which is 50m above sea level, a 2m tall person watches
the palm tree on a desert island slowly disappear over the horizon. How far away will the
ship be when the treetop, which is 15m above sea level at the top, disappears from view?
(a) 25.7 km
(b) 29.2 km
(c) 775.5 km
(d) 14.7 km
ANS B
4. The southern hemisphere has a greater percentage of its surface covered in ocean when
compared to the northern hemisphere. What is the impact on the southern hemisphere’s
average summer temperature?
(a) It is cooler than in the northern hemisphere summer
(b) It is warmer than in the northern hemisphere summer
(c) None as the Earth is closest to the sun during the southern hemisphere summer
(d) None as the Earth is furthest from the sun during the southern hemisphere summer
ANS A
th
5. Imagine that today was November 30 . Where would the sun rise on the horizon from a
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location with a latitude of 20 N?
(a) Due east
ANS C
(b) North of due east
(c) South of due east
(d) Due south
CIV300/ENV346TerrestrialEnergySystems,Fall2016 2
Test1QuestionSheet
6. A distant star of radius 100 million km has a surface temperature of 3,000K. What is the
2
energy output per second per km of surface area?
-8
−2
The Stefan Boltzmann constant, σ = 5.67x10 Wm K
6
(a) 4.59x10 J (b) 1.84x10
29
J
(c) 1.84x10
23
J
−4
(d) 4.59x10
12
J
ANS D
7. What is the maximum angle of elevation of the sun above the horizon (its zenith) at midday
st
in Toronto, (latitude 43.5ºN) on December 21 , the southern hemisphere mid summer’s
day?
(a) 46.5º
(b) 43.5º
(c) 23º
(d) 66.5º
ANS C
8. A Mars Bar (a type of chocolate bar) contains 964 kJ of energy. A typical lecture theatre
has dimensions of 10 m x 30 m x 5 m. A student eats the Mars Bar then stays in the
theatre all day. Assuming 100% of the chocolate bar’s energy is put into heating the room,
calculate the temperature rise in the room.
o
o
o
o
(a) 0.5 C (b) 0.001 C (c) 0.005 C (d) 0.01 C
ANS A
9. Jupiter has the largest gravitational impact on the solar system after the sun. If you were
able to stand on its surface, what acceleration due to gravity, g, would you experience if
you had a mass of 70 kg?
The mass of Jupiter is 1.9 x 10
-11
2
−2
G is 6.67×10 Nm kg .
(a) 1,815 N
27
kg and its radius is 69,900 km. The gravitational constant,
-2
(b) 25.9 ms (c) 6.48 ms
-2
(d) 1.6 ms
-2
ANS B
10. The perihelion of a planet occurs:
(a) Exactly at the point in time in between its two equinoxes
(b) At its furthest point from the star it orbits
(c) On the shortest day of the year
(d) At its closest point to the star it orbits
ANS D
11. At a certain depth in the ocean, which of the following is possible?
(a) That water of two different temperatures is present in adjacent water columns
(b) That water of two different salinities is present in adjacent water columns
(c) That water with different salinity AND temperature is present in adjacent water columns
(d) All of these answers are possible
ANS D
CIV300/ENV346TerrestrialEnergySystems,Fall2016 3
Test1QuestionSheet
st
12. The sun sets for good on December 1 in Cambridge Bay, Nunavut due to its high latitude.
The ground is frozen to a depth of 30 cm on that day. The average moisture content the
first meter of soil is 30% by mass AND volume, with the rest made of various solid
-1 -1
materials. The mixture has a combined thermal conductivity of 1 Wm K . If the average
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st
temperature for December is -20 C, how deep will the frost line be on January 1 ? The
-3
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density of ice is 916 kgm . Ignore any ground temperature drop below 0 C in your
calculations.
(a) 2.25 m
(b) 58 cm
(c) 1.95 m
(d) 88 cm
ANS A
13. Night temperatures in desert regions can fall to surprisingly low levels and frosts may
occur. The most likely reason for this is:
(a) Surfaces that heat quickly will also cool more quickly
(b) The high albedo of desert surfaces
(c) A lack of ground moisture
(d) A lack of cloud cover at night allows the ground and thus the air to cool quickly due to
radiant losses
ANS D
14. Where is the moon during a new moon?
(a) Between the Earth and the sun
(b) Behind the Earth from the sun
(c) Directly in the Earth’s shadow from the sun
(d) At 90° to the Earth from the sun
ANS A
CIV300/ENV346TerrestrialEnergySystems,Fall2016 4
Test1QuestionSheet
PARTB
1. Kiehl and Trenberth (1997) generated this amazingly comprehensive graphic showing what
happens to the sun’s energy on average when it reaches Earth. The Stefan Boltzmann
-8
−2 −4
constant, σ = 5.67x10 Wm K (8 marks total)
!
i)
Calculate the average albedo of the Earth (1 mark)
For the Earth as a whole, albedo (reflectance) = 107/ (342) = 0.31
ii)
Calculate the Earth’s average surface temperature in degrees Celsius (2 marks)
2
Use the Stefan Boltzmann equation and the 390 W/m surface radiation number,
E = σT
4
-8
390 = 5.67x10 T4
o
T = 288K = 15 C
iii)
2
Calculate the total annual energy per m of surface area leading to evaporation of
water from the surface. (1 marks)
2
78W/m contributes to evapo-transpiration
Annual that means Q = 78W x 3,600 seconds/hr x 24 hr/day x 365 days/year
9
Total energy intensity = 2.46 x 10 J/m
iv)
2
Then calculate the total annual rainfall on Earth in litres (2 marks)
9
2
Total energy contribution = 2.46 x 10 J/m x SA of Earth
9
2
2
= 2.46 x 10 J/m x 4 πr
9
2
= 2.46 x 10 x 4 π x (6,375,000)
CIV300/ENV346TerrestrialEnergySystems,Fall2016 5
Test1QuestionSheet
= 1.256 x 10
24
J
Now Q=mL where m = volume of rain, L = typical energy required to evaporate water
1.256 x 10
24
m = 5.0 x 10
= m x 2.5x10
17
6
litres
Note: can split problem into two parts using Q = mcT with average temperature from
6
(ii) above, then latent heat of evaporation of 2.26x10 J/kg
v)
2
How many meters of rain is this on average per m of surface area annually? (2
marks)
Average = total rainfall / SA
17
2
= 5.0 x 10 litres / 4 πr
17
2
= 5.0 x 10 litres / 4 π6,375,000
2
= 979 litres per m
= 0.98 meters of rain
Can also reach answer by going straight from answer to (iii) above
-2
2. On a sunny winter’s day, an average of 600 Wm falls onto the horizontal plane over a 5-hour
period. It strikes a pond whose frozen surface is at the point of melting and has a temperature
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of 0 C. (5 marks total)
i)
2
If the ice has an albedo 0.8 and a depth of 20 cm, calculate the mass of ice per m that
-3
will melt during the 5-hour period. The density of ice is 916 kgm (2 marks)
2
2
600 W/m x (100%-80%) = 120 W/m across the ice surface
Over 5 hours, total energy = 120 x 5 hrs x 60 mins x 60 seconds = 2.16 MJ
2
Ice that will melt, m = Q/L = 2.16 MJ / 334,000 = 6.47 kg per m of surface area
ii)
If the ice were 5 cm deep, would the ice surface totally disappear by the end of that 5hour time period? (1 mark)
Density of ice = 916 kg/m3. 6.4 kg represents 6.4/916 = 0.7% of the total equivalent of
3
1m of ice, or a depth of 0.7mm.
With a total of 5cm of ice present, the ice will still remain at the end of the day
iii)
One month later, all the ice has melted and the average solar intensity has increased
-2
to 800 Wm . The water now absorbs 100% of the energy. If 10% of that energy
results in surface evaporation, what mass of water is evaporated per hour if the lake is
10km x 20 km in size? (2 mark)
80 W/m2 x 10,000 m x 20,000 m x60 mins x 60 secs = 5.76x10
13
Q=mL so m = 5.76x10 J / 2.5 MJ/litre
7
= 2.3 x 10 litres (or kg)
13
J