ANSWER BOOKLET
School:
HWA CHONG INSTITUTION
Names of Team Members
Team No:
1
Class
NRIC_NO:
BOO KOH HUA JIE DAVID
2I4
S9701468Z
JUSTIN LO TIAN WEN
2I4
S9714051J
LEE BING QIAN RYAN
2I4
S9735116C
TAN JING LING
2I4
S9708074G
Name of Teacher-in-charge:
DR TAN KOK KIM
Date of submission: 8 APRIL 2011
IMPORTANT:
1. The answer or proposal to each problem should not exceed 6 pages of text.
Supporting figures and/or data could be in the Annex not exceeding 4 pages. Begin
the answer to each problem on a new page.
2. Use Times New Roman font size 12, single line spacing and left/right/top/bottom
margin of 1 inch.
3. The panel of judges will only evaluate the main text or annex that is within the
maximum number of pages specified.
4. Plagiarism is strictly prohibited. Such entries will be disqualified. All team
members must fill in their particulars and sign on the Declaration Form.
FOR OFFICIAL USE
Date Received:
S/N:
FOR OFFICIAL USE: S/N:
Problem 1 (Chemistry): [ANSWER]
Q1.
Assumptions made in the results:
1. CO2 emission after 2300 is at 5 gigatons per year
2. Annual increase in amount of fossil fuel (through natural process) is not taken into
account.
Workings:
Step 1: Creating a graph through linear interpolation to get an accurate result of carbon
released into the atmosphere through consumption of fossil fuels
CO2 emissions /Gton year -1
20
15
10
5
0
2500
2300
2100
2080
2060
2040
2020
2000
CO2 emissions
/Gton yr^-1
Step 2: Calculate total amount of fossil fuels burnt, assuming that the carbon released
into the atmosphere is in a 1:1 ratio to the amount of fossil fuels burnt.
Using the linear interpolation graph, we have found that the average CO2 emissions in a
certain each 10-year / 100-year period is the average of the emissions at the start point
and end point of each section.
Recoverable fossil fuels  4600 Gton
Fossil fuels consumed from 2000 to 2099 
= 10 × (8 + 10 + 12 + 14 + 16 + 16.5 + 15.5 + 14.5 + 13.5 + 12.5) 𝐺𝑡𝑜𝑛
= 10 × 132.5 𝐺𝑡𝑜𝑛
= 1325 𝐺𝑡𝑜𝑛
Fossil fuels consumed from 2100 to 2599 
= 100 × (9.5 + 6 + 5 + 5 + 5) 𝐺𝑡𝑜𝑛
= 100 × 30.5 𝐺𝑡𝑜𝑛
= 3050 𝐺𝑡𝑜𝑛
Fossil fuels remaining after 2599 
= 4600 𝐺𝑡𝑜𝑛 − 1325 𝐺𝑡𝑜𝑛 − 3050 𝐺𝑡𝑜𝑛
= 225 𝐺𝑡𝑜𝑛
Years taken to consume remaining fossil fuels 
= 225 𝐺𝑡𝑜𝑛 ÷ 5 𝐺𝑡𝑜𝑛/𝑦𝑟
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= 45 𝑦𝑟
Number of years taken to exhaustion of fossil fuels 
= (2099 − 2000 + 1)𝑦𝑟 + (2599 − 2100 + 1)𝑦𝑟 + 45 𝑦𝑟
= 100 𝑦𝑟 + 500 𝑦𝑟 + 45 𝑦𝑟
= 645 𝑦𝑟
Therefore, the fossils fuel reserves will be exhausted in the 2645 years.
Q2
There is 4600 Gtons of carbon dioxide in fossil fuels and 760 Gtons of it in the
atmosphere, with a concentration of 387 ppmv.
Mass of atmosphere 
= 760 𝐺𝑡𝑜𝑛 ÷ 387 𝑝𝑝𝑚
≈ 1963824 𝐺𝑡𝑜𝑛
Total mass of carbon dioxide in atmosphere and fossil fuels 
= 4600 𝐺𝑡𝑜𝑛 + 760 𝐺𝑡𝑜𝑛
= 5360 𝐺𝑡𝑜𝑛
New concentration of carbon dioxide in atmosphere 
= 5360 𝐺𝑡𝑜𝑛 ÷ (1963824 + 4600) 𝐺𝑡𝑜𝑛
= 5360 𝐺𝑡𝑜𝑛 ÷ 1968424 𝐺𝑡𝑜𝑛
≈ 2722 𝑝𝑝𝑚
Rise in atmospheric carbon dioxide concentrations 
= 2722 𝑝𝑝𝑚 − 387 𝑝𝑝𝑚
= 2335 𝑝𝑝𝑚
Q3a.
Total mass of carbon dioxide in atmosphere in 2011 
1
𝐺𝑡𝑜𝑛
5
= 760 𝐺𝑡𝑜𝑛 + 80 𝐺𝑡𝑜𝑛 + 2 𝐺𝑡𝑜𝑛
= 842 𝐺𝑡𝑜𝑛
= 760 𝐺𝑡𝑜𝑛 + 8 × 10 𝐺𝑡𝑜𝑛 + 10 ×
New concentration of carbon dioxide in atmosphere 
1
= 842 𝐺𝑡𝑜𝑛 ÷ (1963824 + 8 × 10 + 10 × ) 𝐺𝑡𝑜𝑛
5
= 842 𝐺𝑡𝑜𝑛 ÷ 1963906 𝐺𝑡𝑜𝑛
≈ 428 𝑝𝑝𝑚
Q3b.
Total mass of carbon dioxide in atmosphere in 2100 
= 760 𝐺𝑡𝑜𝑛 + 10
× (8 + 10 + 12 + 14 + 16 + 16.5 + 15.5 + 14.5 + 13.5 + 12.5) 𝐺𝑡𝑜𝑛
= 2085 𝐺𝑡𝑜𝑛
2
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New concentration of carbon dioxide in atmosphere 
= 2085 𝐺𝑡𝑜𝑛 ÷ (1963824 + 2085 − 760) 𝐺𝑡𝑜𝑛
= 2085 𝐺𝑡𝑜𝑛 ÷ 1965149 𝐺𝑡𝑜𝑛
≈ 1060 𝑝𝑝𝑚
Q3c.
As shown earlier, the carbon dioxide concentration the year in which the entire fossil fuel
reserves have been exhausted is 2722 ppm
Q4.
Using the results from question 2, the concentration in carbon dioxide in the atmosphere
when the entire fossil fuel reserves have been exhausted increased to 2722 ppm.
Number of times carbon dioxide concentration multiplied by 2 
2722 𝑝𝑝𝑚
= 𝑙𝑜𝑔2 (
)
387 𝑝𝑝𝑚
≈ 𝑙𝑜𝑔2(7.03)
≈ 2.81
Thus, given the mean temperature increase of x, it has a range of:
2.81 × 1.5℃ ≤ 𝑥 ≤ 2.81 × 4.5℃
4.215℃ ≤ 𝑥 ≤ 12.645℃
Q5.
Using the results from question 3b, the concentration in carbon dioxide in the atmosphere
by the end of the century increased to 1060 ppm.
Number of times carbon dioxide concentration multiplied by 2 
1060 𝑝𝑝𝑚
= 𝑙𝑜𝑔2 (
)
387 𝑝𝑝𝑚
≈ 𝑙𝑜𝑔2(2.74)
≈ 1.45
Thus, given the mean temperature increase of x, it has a range of:
1.45 × 1.5℃ ≤ 𝑥 ≤ 1.45 × 4.5℃
2.175℃ ≤ 𝑥 ≤ 6.525℃
Q7.
Using the results from question 5, the mean temperature will rise by from 2.175℃ −
6.525℃ by the end of the century.
∆𝑉⁄ = 𝛽∆𝑇 yields
𝑉
2.5 × 10−4 K −1 × 2.175℃ ≤ ∆𝑉⁄500 m × 3.6 × 108 km2
≤ 2.5 × 10−4 K −1 × 6.525℃
2.5 × 10−4 K −1 × 275.325 𝐾 ≤ ∆𝑉⁄1.8 × 108 km3 ≤ 2.5 × 10−4 K −1 × 279.675 𝐾
688.3125 × 10−4 ≤ ∆𝑉⁄1.8 × 108 km3 ≤ 699.1875 × 10−4
1238.9625 × 104 ≤ ∆𝑉 km−3 ≤ 1258.5375 × 104
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12389625 km3 ≤ ∆𝑉 ≤ 12585375 km3
Q8.
Through estimation and a bit of guess and check, using these two graphs we have
matched a pressure, a temperature and an ocean depth to allow for carbon dioxide to
become a liquid.
1 𝑎𝑡𝑚 ≈ 1 𝑏𝑎𝑟
And the pressure of the ocean increases by 1 atm per 33 ft in depth. Thus, using this, 78
bars of pressure gives us:
78 𝑎𝑡𝑚 × 33 𝑓𝑡 𝑏𝑎𝑟 −1
≈ 78 × 33 𝑚 ÷ 3.28
= 784 𝑚
According to the thermocline graph, this comes up to 8℃ which corresponds to about
281 Kelvins on the states of carbon” dioxide graph.
.
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References:
1. http://upload.wikimedia.org/wikipedia/commons/1/13/Carbon_dioxide_pressuretemperature_phase_diagram.svg
2. http://www.windows2universe.org/earth/Water/temp.html
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Problem 2 (Physics): [ANSWER]
Part 1: Thermal properties of water
a)
Specific heat capacity(S.H.C) of sea water
Apparatus used: Calorimeter, thermometer and sample(water), voltmeter, ammeter,
stopwatch, batteries, wires.
Formulae to remember: Q=IVt, c = 1m(Q(delta)(thethia))
Note: Change in temperature is measured through thermometer;
Energy input is measured through Q=IVt from heater; Mass is measured through a
weighing machine; Politically correct S.H.C of water is 4200 J/Kg/K
Experimental setup:
Batteries in series; Ammeter in series with batteries; calorimeter with heater in it is load;
Voltmeter parallel to batteries and calorimeter; Thermometer placed in calorimeter to
measure temperature.
TEST 1: 3 batteries
Initial temperature: 25 deg Celsius
Final temperature: 35.5 deg Celsius
Change in temperature:10.5 deg Celsius
Voltage:4.13
Amperes:5.21
Time: 360 secs
Amount of energy absorbed by set-up:7933.788
Mass of calorimeter: 26.5 grams(0.0265 kg)
Specific heat capacity of calorimeter: 900
Energy required to raise calorimeter by 1 deg celsius:23.85
Energy absorbed by calorimeter: 250.425
Mass of seawater: 111 grams (0.111 kg)
Specific heat capacity of seawater: 4519.58
Workings:
total energy absorbed - energy absorbed by calorimeter = 7933.788 - 250.425 = 7683.3
= energy absorbed by seawater
7683.55 joules of energy was absorbed by 0.111 kg of seawater to raise it by 10.5 deg
Celsius.
0.111 x 10.5 = 1.655
Energy absorbed by liquid divided by mass of liquid multiplied by change in temperature
= S.H.C
7683.3/ 1.7 = 4519.58 joules required to raise seawater by 1 deg Celsius
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TEST 2: 4 batteries
Initial temperature: 25 deg Celsius
Final temperature: 45 deg Celsius
Change in temperature: 19 deg Celsius
Voltage:5.2
Amperes:6.6
Time: 360 secs
Amount of energy absorbed by set-up:12355.2
Mass of calorimeter: 26.5 grams(0.0265 kg)
Specific heat capacity of calorimeter: 900
Energy required to raise calorimeter by 1 deg celsius:23.85
Energy absorbed by calorimeter: 477
Mass of seawater: 111g (0.111 kg)
Specific heat capacity of seawater(testing for): ???
Workings:
total energy absorbed - energy absorbed by calorimeter = 12355.2 - 477 = 11878.2
= energy absorbed by seawater
11878.2 joules of energy was absorbed by 0.111 kg of seawater to raise it by 20 deg
Celsius.
0.111 x 20 = 2.22
Energy absorbed by liquid divided by mass of liquid multiplied by change in temperature
= S.H.C
11878.2/ 2.22 = 5164.4 joules required to raise seawater by 1 deg Celsius
TEST 3: 5 batteries
Initial temperature: 25 deg Celsius
Final temperature: 52 deg Celsius
Change in temperature: 27 deg Celsius
Voltage:5.6
Amperes:7.2
Time: 360 secs
Amount of energy absorbed by set-up:14515.2
Mass of calorimeter: 26.48 grams(0.02648 kg)
Specific heat capacity of calorimeter: 900
Energy required to raise calorimeter by 1 deg celsius:23.832
Energy absorbed by calorimeter: 643.464
Mass of seawater: 111 grams (0.111 kg)
Specific heat capacity of seawater: 4628.5
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Workings:
Total energy absorbed - energy absorbed by calorimeter = 14515.2 - 643.464 = 13871.73
= energy absorbed by seawater
13871.73 joules of energy was absorbed by 0.111 kg of seawater to raise it by 27 deg
Celsius.
0.111 x 27 = 2.997
Energy absorbed by liquid divided by mass of liquid multiplied by change in temperature
= S.H.C
13871.73/ 2.997 = 4628.5 joules required to raise seawater by 1 deg Celsius
b) Latent heats of fusion and vaporisation
Test of latent Heat of Vaporization:
Apparatus Used: Rated electric Heater, Beaker, Stopwatch, Electric weighing Scale,
Water
Steps: Fill the beaker with 400ml of water; Place the rated electric heater in the beaker of
water; Place the beaker (with the rated electric heater in it) on the electric weighing scale;
Zero-off the weight when the water starts to boil; Record observations.
Formulae to remember: QMi-Mf=lf
Test 1:
Initial mass of water: 400g
Final mass of water: 350g
Difference in mass: 50g
Power: 600 watts
Time taken: 197 seconds
Workings:
Energy absorbed by water = 600x 197 = 118200 Joules
Amount of water evaporated = 50 ml (50 grams)
118200 joules was required to evaporate 50 grams of water
therefore 2364 joules was required to vaporise 1 gram of water
and 2364 kilojoules was required to vaporise 1 kg of water.
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Latent Heat of Fusion:
Apparatus used: Ice, Thermometer, Stirrer, Copper calorimeter, Jacket, Electronic
Balance
Steps:
I. Record the mass of the dry copper calorimeter and stirrer.
II. Fill half the calorimeter with warm water. Add some tap water until the III.
Temperature of the mixture is about 10 deg Celsius above room temperature.
IV. Record the total mass of the calorimeter, stirrer and warm water.
V. Record the initial temperature of warm water.
VI. Dry a piece of ice and drop it into the calorimeter. Stir the water continuously until
the ice melts completely.
VII. Repeat step VI until the temperature of water is 10 deg Celsius below room
temperature. Record the final temperature of water in the calorimeter.
VIII: Record the total mass of the calorimeter and its contents.
Formulae to remember: LF=lf*m
Test 2:
Mass of the given empty dry copper calorimeter and stirrer = 55.93 grams
Total mass of calorimeter, stirrer and warm water = 134.08 grams
Total mass of the calorimeter and its contents = 155.79
Mass of water: 78.15
Initial temp of water = 35 deg Celsius
Final temp of water = 15 deg Celsius
Workings:
Energy required to reduce the temp of water by 20 deg Celsius
= (134.08-55.93)(35-15)(4.20)
=6564.96
Energy required to raise temp of ice to 15 deg Celsius
= (155.79-134.08)(15)(4.20)
= 1367.73
Latent heat of fusion of 21.71 grams of ice
=6564.96 - 1367.73
= 5197.23
Latent heat of fusion of 1 gram of ice
= 5197.23 / 21.71
= 239.4 joules
Therefore specific latent heat of fusion of ice
= 239.4 kilojoules
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c) Rate of evaporation of water
Hypothesised factors that affect evaporation of water are:
Exposed surface area, Temperature of water, Humidity of surrounding air, Pressure,
Wind speed
Test 1: Exposed surface area
Apparatus: 3 water containers of different surface areas, tray, water.
Steps:
Each water container was filled with 200 ml of water, and left in the sun for 2 hours on
the tray.
Observations:
5 ml of water evaporated from container A which had the biggest surface area
1 ml of water evaporated from container B which had a slightly smaller surface area
No observable results were obtained from beaker C, which had the smallest surface area
of all three containers
Conclusion: The larger the surface area, the faster the rate of evaporation. Likewise, the
smaller the surface area, the slower the rate of evaporation.
Test 2: Temperature of water
Apparatus: Electric heater, 2 identical beakers, water.
Steps: Fill two identical beakers with 200ml of water. Place the electric heater (heated at
60 deg Celsius) in one of the beakers and observe.
Observations: After 2 hours, there was 196ml of water left in the beaker without the
electric heater. However, there was only 119ml of water left in beaker with the electric
heater.
Conclusion: Temperature affects the rate of evaporation proportionately. The higher the
temperature, the faster the rate of evaporation. Likewise, the lower the temperature, the
slower the rate of evaporation.
Test 3: Humidity of surrounding air
Apparatus: 2 humidity sensors, 2 identical beakers, water, and two rooms of different
humidity
Steps: Fill both beakers with 200 ml of water. Place them in two different rooms, of
which one has 44% humidity and the other has 68% humidity.
Observations:
After 2 hours, 1 ml of water evaporated from the beaker in the room with 44% humidity
while there was no observable change in volume of water for the beaker in the room with
68% humidity.
Conclusion: Humidity affects the rate of evaporation inversely. The lower the humidity,
the faster the rate of evaporation. The higher the humidity, the slower the rate of
evaporation.
Test 4: Pressure
Apparatus: Bell jar, 2 identical beakers, water
Steps: Fill both beakers with 200ml of water each. Place one of the beakers in an air-tight
bell jar and one in open air.
Observations: There was 198ml of water left in the beaker in the bell jar. There was no
observable change in water volume of the bell jar in open air.
Conclusion: The greater the pressure, the faster the rate of evaporation. The smaller the
pressure, the slower the rate of evaporation.
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Test 5: Wind speed
Apparatus:
2 beakers of similar sizes, anemometer, fan, water
Steps:
One beaker, as the control was filled with 200ml of water and placed in a windless
location. The other, also filled with 200ml of water was placed in front of a fan. The wind
speed of air over the exposed surface area of the water was at 2.3 m/s. The duration of the
setup was half an hour.
Observations: There was no visible change in water level for the control set-up, while the
water level dropped by about 2 ml for the setup placed in front of the fan.
Conclusion: Wind speed affects rate of evaporation proportionately. The greater the wind
speed, the faster the rate of evaporation. The slower the wind speed, the slower the rate of
evaporation.
d) Rate of melting of ice:
Hypothesised factors that affects rate of melting of ice: Exposed surface area of ice,
temperature, humidity of surrounding air, air pressure, wind speed
Test 1: Exposed surface area
Apparatus: 3 ices cubes of 8cm^3 surface area, but of different surface areas.
Steps: Leave the 3 ice cubes of different surface areas under the same environment for
30mins.
Observations: The ice cube with the largest exposed surface area has the smallest ice
cube left. The ice cube with the least exposed surface area has the largest ice cube left.
Conclusion: The larger the surface area, the faster the rate of melting of ice. Likewise, the
smaller the surface area, the slower the rate of melting of ice.
Test 2: Temperature of ice
Apparatus: Electric heater, 2 identical beakers, similar ice cubes
Steps: Place the two ice cubes in two dry beakers. Place the electric heater (heated at 60
deg Celsius) in one of the beakers and observe.
Observations: After 10 minutes, there was a larger piece of ice was left in the beaker
without the electric heater. However, there was a smaller piece of ice left in the beaker
with the electric heater.
Conclusion: Temperature affects the rate of melting proportionately. The higher the
temperature, the faster the rate of melting. Likewise, the lower the temperature, the
slower the rate of melting.
Test 3: Humidity of surrounding air
Apparatus: 2 humidity sensors, 2 identical beakers, 2 identical ice cubes, and two rooms
of different humidity
Steps: Place an ice cube in each beaker. Place the beakers in two different rooms, of
which one has 44% humidity and the other has 68% humidity.
Observations: After 30 minutes, the ice in the room with 44% humidity melted
completely, while the ice in the room with 68% humidity had only about half of it melted.
Conclusion: Humidity affects the rate of melting inversely. The lower the humidity, the
faster the rate of melting. The higher the humidity, the slower the rate of melting.
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Test 4: Pressure
Apparatus: Bell jar, 2 identical beakers, and 2 identical ice cubes
Steps: Place an ice cube in each of the 2 beakers. Place one of the beakers in an air-tight
bell jar and one in open air.
Observations: After 15 minutes, both ice melted. There was an insignificant difference in
their rate of melting.
Conclusion: The amount of pressure usually does not affect the rate of melting
significantly.
Test 5: Wind speed
Apparatus:
2 beakers of similar sizes, anemometer, fan, 2 identical ice cubes
Steps: Place an ice cube in one of the beakers, as the control, and placed in a windless
location. Place an ice cube in another beaker, and place in front of a fan. The wind speed
of air over the exposed surface area of the water was at 2.3 m/s. The duration of the setup
was 10 minutes.
Observations: The ice cube with the fan blowing it melted completely, while the ice cube
placed in a windless location melted about half of its volume.
Conclusion: Wind speed affects rate of melting proportionately. The greater the wind
speed, the faster the rate of melting. The slower the wind speed, the slower the rate of
melting.
Part 2: The greenhouse effect
Apparatus: 2 identical large bottles, 2 thermometers, 2 identical light bulbs, cardboard
Set-up: Cut off the bottom of both water bottles and paste a piece of cardboard on it.
Explanation:
Part 3: Model of the earth’s surface
Where are the highest and lowest points in Singapore? What are their heights above sea
level?
Lowest point: Singapore Strait, 0 m above sea level.
Highest point: Bukit Timah 164 m above sea level.
Construct a simple but realistic model of the earth’s surface.
Materials: Plasticine, blue tack, markers, ping pong ball (picture found in annex)
Part 4: Impact of the melting of ice in Antarctica on Singapore
Estimate the impact of the melting of ice in Antarctica in Singapore over time.
We assumed that the height of increase of water level over the different parts of the world
would be constant.
We found that the earth’s mean radius is 6371km.
Therefore, the total surface area of the earth is 4𝜋𝑟 2, which is 510064471.9𝑘𝑚2 .
However, the area of land on earth is approximately 150,000,000 km2 only.
29,300,000 km3 of grounded ice in the Arctic would produce 26,100,000 km3 of water.
However, about 2,100,000 km3 of the grounded ice in Antarctica is below sea level.
Therefore, 24,000,000 km3 of water would spread across the earth’s surface.
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Assuming this amount of water spreads over the earth’s surface evenly, there would be an
average increase of global water level of 0.04705287532km, which is approximately
47metres.
Singapore’s average land height is approximately 31m. Therefore, if the water level
increases by 47m, more than half of Singapore’s total land will be covered with water.
Although Singapore will not be fully submerged in water because of hills like Bukit
Timah Hill, she will be unable to support life.
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Problem 2 (Physics): [ANNEX]
Part 1:
Rate of Evaporation/ Melting
Part 3:
1
FOR OFFICIAL USE: S/N:
References:
Part 1:
http://arxiv.org/ftp/physics/papers/0407/0407090.pdf
http://en.wikipedia.org/wiki/Latent_heat
http://www.indianastandardsresources.org/admin/library/latent_heat_ice.pdf
http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html
http://en.wikipedia.org/wiki/Heat_capacity
http://www.blurtit.com/q316525.html
GCE ‘O’Level Physics Matters Textbook
Part 2:
http://oregonstate.edu/instruct/ch390/lessons/media/lesson20.pdf
http://staff.science.uva.nl/~bredeweg/pdf/thesis/03Dubbeldam.pdf
Part 3:
http://www.worldatlas.com/aatlas/infopage/highlow.htm
Part 4:
http://www.buzzle.com/articles/the-effects-of-melting-ice-glaciers.html
http://news.bbc.co.uk/2/hi/uk_news/4228411.stm
http://www.johnstonsarchive.net/environment/waterworld.html
http://hypertextbook.com/facts/2001/DanielChen.shtml
http://en.wikipedia.org/wiki/Geography_of_Singapore
I
FOR OFFICIAL USE: S/N:
Acknowledgment:
1. Hwa Chong Institution Science Research Centre Physics Lab for allowing us to
use their facilities to conduct our experiments
II
FOR OFFICIAL USE: S/N:
Problem 3 (Mathematics): [ANSWER]
ai)
The potential energy of the water will be converted into kinetic energy during the fall.
Thus, since:
The Principle of conservation of energy states that energy cannot be created or destroyed
in any process, therefore it can be converted from one form to another or transferred from
one body to another, even though the amount of energy remain constant.
Therefore, GPE of water = KE of water.
m = mass, g = gravity, h = height, v = velocity, Q = flow rate, A = area, V = volume
𝐺𝑃𝐸 = 𝑚𝑔ℎ
𝑚𝑣 2
𝐾𝐸 =
2
𝑚𝑣 2
𝑚𝑔ℎ =
2
𝑣 = √2𝑔ℎ
𝑆𝑖𝑛𝑐𝑒 𝑄 = 𝐴𝑣,
𝑄 = 𝐴 √2𝑔ℎ
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑡ℎ𝑒 𝑡𝑎𝑛𝑘 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑒𝑚𝑝𝑡𝑦 𝑖𝑛
𝑥𝑦𝑧
𝑥𝑦𝑧
=
𝐴 √19.62ℎ
𝐴√2𝑔ℎ
aii)
𝑥𝑦𝑧
The tank will empty in 𝐴
√19.62ℎ
Hence the maximum rate at which one can pour water into the tank without it
overflowing is to pour all 2𝑥𝑦𝑧 𝑐𝑛³ into the tank in exactly this amount of time.
Thus, the maximum rate is
2𝑥𝑦𝑧
(
𝑥𝑦𝑧
)
𝐴 √19.62ℎ
= 2𝐴 √19.62ℎ.
𝑥𝑦𝑧
A higher flow rate would cause the tank to overflow in𝐴
, while in the scenario
√19.62ℎ
above, the tank just fills up and then gets drained away.
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c)
.
Year
Total amount of rainfall
(mm)
2004
2136
2005
1931
2006
2753
2007
2886
2008
2325
2009
1921
Source: National Environment Agency
Maximum
Temperature
[daily] (deg
celcius)
31.7
31.9
31.5
31.1
31.1
31.7
Minimum
temperature [daily]
(deg celcius)
25.1
25.1
25.0
24.8
24.7
25.0
The higher the amount of rainfall, the lower the temperature. This is because the rain,
being water, has a lower temperature, thus more rain would decrease the average
temperature. This would help save energy, as less air-con needs to be used when it rains.
However, rain would also cause people to stay indoors, thus affecting sales.
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Problem 3 (Mathematics): [ANNEX]
The buildings are represented roughly in their shapes out of metal. And the “road” is
made out of acrylic. The road is created in a bowl shape to depict the ponding effect of
the orchard-Scott-Paterson road junction, where it is the lowest point in the roads leading
to the cross junction. This would mean that whatever water overflows would pool in the
junction. The bottoms of building would also collect water because the basements of
building are at an even lower elevation than the road. The pagoda like building represents
the Marriott hotel, the building with a cone in front of it represents Wheelock place with
its classical “Christmas tree”, the building with a curved quadrant is the Shaw house and
the remaining building would be orchard ion, which are the buildings closest to the
junction.
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References:
1. http://wildsingaporenews.blogspot.com/2010/06/blocked-drain-blamed-for-orchardroad.html
2. http://thetwophilo.files.wordpress.com/2010/07/1orchard-rd-flood-profile.jpg
http://thetwophilo.files.wordpress.com/2010/07/2scotts-orchard-paterson-profile.jpg
http://dreamsyntax.org/opencall/images/16june_how_the_orchard_flood_happened.jp
g
3. http://www.shawproperties.sg/Images/Property/13/ShawHse%20web%20pix.jpg
4. http://en.wikipedia.org/wiki/Volumetric_flow_rate
5. http://www.singstat.gov.sg/pubn/reference/yos10/statsT-climate.pdf
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Problem 4 (Biology): [ANSWER]
1. Flooding has a huge impact on the ecosystems present in Singapore. I think that the
most ecosystems which are most affected by flooding are those that are along rivers.
Some ecosystems which are present near drains are also very significantly affected. I
will explain in further detail a few of these examples and how flooding affected the
biodiversity of these areas.
Firstly, when flooding occurs, ecosystems near the river were affected. During
flooding, the massive and strong amounts of water caused animals to be swept away.
For example, marine life may be swept ashore where they are unable to survive, or
brought to other places where the conditions are unsuitable to live in. Also, insects
whose habitats are along the waters, for instance under dried leaves or tree logs will
also be picked up by the water where they cannot survive. Floods may just wipe out
something as small such as the producer (plants) in the food chain. However, this
would mean the rest of the food chain would be affected as the animals do not have
enough food and die, thereafter affecting other food chains. This also applied if a prey
or predator within the food chain. The ecosystem was be disrupted, causing
unbalanced ratios of animals. Therefore, there was a decrease in the biodiversity in
these areas.
Secondly, ecosystems near drains and canals were affected. The water from the floods
could not drain fast enough, therefore causing very high water levels. Naturally,
ecosystems near the drains and canals were flooded and animals who were living
there were unable to survive. As I have explained, having one type of animal removed
from our ecosystem caused a negative effect and decreased the biodiversity near
canals and drains.
2. More severe and frequent flooding events will affect the natural ecosystem in
Singapore greatly. The weather is an abiotic factor which plays a major part in the
ecosystem, as it affects some other abiotic factors. Abiotic factors ultimately
determine the organisms found in an area, thus, a change in abiotic factors would
mean a change in species of organisms available or present in the area.
A more erratic weather with more frequent and severe flooding affects the ecosystem
in the aspect of adaptability. Organisms that are not suited to the change of abiotic
factors would perish and those who are able to adapt would then have more space to
flourish and increase in population size. When there is more frequent and severe
flooding, the terrain would alternate between being very wet, and dry frequently and
drastically. Organisms that live in the ground would then be affected. For example, a
colony of ants requires an ant mound which is relatively moist, but not so wet as to
the structure of tunnels are unable to maintain their shape and collapse. What a
flooding does to ant mounds built at low elevation would swamp the mound, destroy
the nest, and perhaps kill the queen ant. If the ants are unable to adapt, for example to
build mounds at a higher elevation, they would perish. Other organisms that could not
move would have to adapt to change in dryness of terrain.
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When there is a change in population, in a habitat, the community is also changed in a
few aspects. Firstly, in the food chain and secondly, in the biodiversity. The food
change would be modified as there would be a change in the balance of different
producers and consumers, for example, a flourishing of weeds because there is an
increase in moisture, which would then lead to an increase in consumers of weeds
like smaller rodents or worms, and insects, as there is an increase in food source. This
means that the secondary consumers like birds or cats would also increase in
population size due to the abundance of food. The opposite could also happen where
constant change in habitat from wet to dry would make some organisms in the food
chain decrease in number which would lead to the species of organisms it consume to
increase in numbers and organisms that consume it to decrease in numbers.
Lastly, this would lead to a decrease in biodiversity because when there is a change of
environment, diversity usually decrease because of organisms there being unable to
cope with change, and leaving the area or going extinct in the area which lead to a
smaller amount of the species in the area.
3. When there are floods, a lot of the fishes in the water bodies will be washed ashore.
Therefore, fishermen around the region are unable to fish due to the change in coastal
lines. On top of that, they would be unable to sail on sea as their boats may have been
destroyed and the water levels may be inconsistent. Therefore, our supply of seafood
will be decreased, causing the price of such food to increase drastically due to the
high demand but the lack of supply, which will mean that we probably will be unable
to rely on the natural ecosystems for our food.
However, floods may also have positive implications on the population in the region.
Flooding clears the pollution which is present in water-based ecosystems. The sudden
appearance of floods can wash away the negative products from the water and bring
out fertile soil to the grounds along the water. For example, man-made products
which may have been littered into the water or waste chemicals which may be
detrimental to the growth of the living things will be washed out of the waters. In
addition, debris which may have fallen alongside the rivers will be washed away by
the water.
Plants which are on land will also be able to benefit from the floods. The soils which
are above ground will be able to soak up the water and the supply of water
underground can be refilled. The nutrients which are carried along by the floodwater
can improve the growth of plants, as well aid in germination of new seeds. Over an
extended period of time, the flood water is able to create a new food supplies. The
soil in the area will be very fertile which is effective to grow crops. Therefore, plants
and animals in the water or around the water will be able to survive and thrive. This
will allow more marine life to be able to prosper and therefore allow us to have more
food from the natural ecosystems.
4. Singapore has already had several measures in place in an attempt to reduce the
impact of the flooding. Firstly, they have put sandbags along the Orchard Road to
prevent water from floods to flow into the buildings. On top of that, they have raised
the grounds in Orchard Road so that the water will not accumulate on the roads, but
will instead flow to the side of the roads and drain out through the many drains in
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place. However, there are still more measures which can be taken to minimize the
impact the flooding.
The most important factor is to increase the awareness of these floods and have a
more efficient alert system in place. Therefore, when there is a flood which is
developing, people, especially shop owners, should be able to know about it fast so
that they can take necessary actions. For example, shop owners can quickly shift their
expensive goods which are closer to the ground to higher places so that the floods
will not destroy or affect them when it comes. This will save a lot of money on the
part of the shop owners. When the people of Singapore are informed, they also can
try to take cover by going onto higher grounds so that they will not be affected and
they will also know not to drive their cars to certain places so that they do not get
stuck. Therefore, I feel that it is very important to inform the Singapore population
when a flood is about to hit. This method can be applied to any country in the region
as it is simply increasing the efficiency of the alert system, and it will probably
benefit the countries.
Another way to reduce the impact of the flooding is to broaden the canals or increase
the drains. This will enable water to be able to be drained out at a quicker pace and
more effectively. For example, after the flooding in the Bukit Timah area, many cars
were drenched and unable to move. Therefore, canals were broadened to try to solve
this problem and I think this can be expanded to other places which are prone to
floods. This solution can also be applied to other countries in the region where they
broaden canals or add drains to places which may be prone to floods so as to prevent
any possible damage.
Thirdly, we can increase the amount of reservoirs in the country. This is killing two
birds with one stone, simply due to the fact that we are minimising the impact of
floods in the country and allowing our country to have more water supply. Reservoirs
are a very effective method to store the water from the floods and prevent them from
going to other places in the country and causing chaos. This is also a solution which
can and should be applied to other countries as it conserves our water supply at the
same time which is very important for the world we are currently living in.
Lastly, we should prevent deforestation in our country and instead build more trees in
Singapore. Trees and forests are vital to absorb and store great amounts of water at a
fast pace in the situation where there are floods. When forests are cut down, the trees
are no longer able to regulate the water, therefore causing floods to occur. By having
more trees, the water will be stopped more effectively. On top of this, forestation
helps us in the fight against global warming. By having more trees, they will
photosynthesize and take in the carbon dioxide and give out oxygen, therefore
reducing the carbon emission in our country. This solution can be especially useful in
countries where a lot of deforestation takes place.
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References:
1.
2.
3.
4.
5.
6.
http://wiki.answers.com/Q/How_do_floods_effect_the_ecosystem
http://sciencepolicy.colorado.edu/socasp/weather1/myers.html
http://en.wikipedia.org/wiki/Flood_control
http://www.ehow.com/list_7494478_effects-flood-ecosystem.html
http://en.wikipedia.org/wiki/Biodiversity
http://www.buzzle.com/articles/causes-and-effects-of-deforestation.html
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Acknowledgment:
1. MS JENNIFER TING (Teaching Staff, Hwa Chong Institution) for giving us advice and a few
pointers on flooding in Singapore.
II