BLM 10-1, Tilt and Turn — Data Tables/
Overhead Master
Answers
Sample data for Investigation 10-A: Tilt and Turn are provided in the SCIENCEFOCUS 10 Teacher’s Resource.
BLM 10-2, Climate Zones and Seasons/
Reinforcement
Goal: Students review the causes of seasons and climate zones.
Answers
1.2.
Zone tropical temperate polar
Summer
- Angle of Sun’s rays to Earth’s surface
- Average daily temperature
- Hours of daylight
- perpendicular to Earth’s surface at some latitude in this zone
- fairly constant
- 12 h at some latitude in this zone
- greatest angle to Earth’s surface that they will be all year; however, never perpendicular to Earth’s surface
- warmest it will be all year
- greater than 12 h at all latitudes
- greatest angle to Earth’s surface they will be all year; however, never as high as in temperate zone
- warmest it will be all year
- long; reaches 24 h at some latitudes in mid-summer
Winter
- Angle of Sun’s rays to Earth’s surface
- Average daily temperature
- Hours of daylight
- perpendicular to Earth’s surface at some latitude in this zone
- fairly constant
- 12 h at some latitude in this zone
- smallest angle to Earth’s surface they will be all year; low angle even at noon
- coldest it will be all year
- less than 12 h at all latitudes
- smallest angle to Earth’s surface they will be all year; extremely low
- coldest it will be all year
- short; dips to 0 h at some latitudes in mid-winter
3. Although Yellowknife experiences 24 h of daylight in mid-summer, Yellowknife is generally not as warm as locations at lower latitudes. Yellowknife is warmer in summer than it is in winter, since this is when it receives the most sunlight. Even in the summer, however, the Sun’s rays hit Earth at a relatively low angle in Yellowknife. Compared with the way that sunlight hits Earth at lower latitudes, the Sun’s rays are spread out over a much larger surface area in Yellowknife. Because “the land of the midnight sun” receives less solar energy per square metre than other regions, it is, on average, cooler than elsewhere. 4. (a) On June 21, Earth is at the position in its orbit
that is farthest from the Sun. Earth is oriented so that the tilt of the axis is greatest relative to the Sun. The north end of the axis is tilted toward the Sun.
(b) On December 21, Earth is at the position in its orbit that is closest to the Sun. Earth is oriented so that the tilt of the axis is greatest relative to the Sun. The south end of the axis is tilted toward the Sun.
(c) The Southern and Northern Hemispheres do not point directly away from or toward the Sun. Both are at the same distance from the Sun. Earth is at the narrower part of its orbit.
5. Regions at latitude 60ºN are in the Northern Hemisphere at the border of the temperate and polar zones. Therefore, regions at this latitude receive the most solar energy in June, when the Northern Hemisphere is pointed toward the Sun.
BLM 10-3, Specific Heat Capacity
Problems/Skill Builder
Goal: Students practise solving specific heat capacity problems.
Answers
Complete GRASP solutions are given for only the first two problems.
1.
Given
Mass of water, m = 250 mL
= 250mL 1.00 g 1.00 mL ×
= 250 g
Specific heat capacity of water, c =4.19 J g °C • Final temperature, T2 = 85.0°C
Initial temperature, T1 = 20.0°C
Change of temperature, ∆T = 65.0°C
Required
Energy required to heat the water from 20.0ºC to 85.0ºC, Q
Analysis
Use Q = mc∆T. Solve for Q.
Solution
Q = mc∆T=(250 g) 4.19 J
g ° C• (65.0° C
)
= 68 087.5 J
≅ 6.8 × 104 J
Paraphrase
It would take 6.8 × 104 J of energy to heat 250 mL of water from 20.0ºC to 85.0ºC.
2.
Given
Mass of water, m = 250 mL
= 250 mL 1.00 g 1.00 mL
×
= 250 g
Specific heat capacity of water, c =4.19 J g °C •
Final temperature, T2 = 75.0°C
Initial temperature, T1 = 85.0°C
Change of temperature, ∆T = –10.0°C
Required
Energy lost by cup of water as it cools from 85.0ºC to 75.0ºC, Q
Analysis
Use Q = mc∆T. Solve for Q.
Solution
Q = mc∆T= (250 g) 4.19 J
g•° C (–10° C
)
= –10 475 J ≅ –1.05 × 104 J
Paraphrase
The 250 mL of tea loses 1.05 × 104 J as it cools from 85.0ºC to 75.0ºC.
3.
Given
Q = 100.0 kJ = 1.000 × 105 J m = 500.0 g
c = 4.19 J g °C •
Solution
Q = mc∆T ∆T = Q/mc=
5 1.000 10 J
(500.0 g ×
J ) 4.19
g•° C
= 47.7327°C ≅ 47.8°C
4.
Given
Q = 35 kJ = 3.5 × 104 J m = 1.00 kg = 1.00 × 103 g c = 2.00 J
Solution
Q = mc∆T ∆T = Q/mc=
4 3.5 10 J×
3
(1.00 10 g× ) 2.00 J g °C
•
= 17.5°C ∆T = T2 – T1
T2 = T1 +∆T
= –25.0°C + 17.5°C = –7.5°C
5. The moist air will experience the greatest temperature change because it has the lowest specific heat
capacity. (Less energy is needed to heat it the same amount, or the same degrees).
6. The aluminium pot will keep the water warm the longest. It will store more energy while heating to 100°C because it has the highest specific heat capacity.
7.
Given
m = 5.0 g Q = 71 J T2 = 162°CT1= 125°C
Solution
∆T = T2 – T1∆T = 162°C – 125°C = 37°C
Q = mc∆T c = Q/m∆T = 71 J
(5.0g) (37 C) ° = 0.38378 J
g °C • ≅ 0.38 J
g °C •
Paraphrase
The specific heat capacity of the metal is similar to the specific heat capacity of copper, so the metal is most likely copper.
8.
Given
Q = –300.0 MJ = –3.000 × 108 J c = 3.89 J
g °C • T2 = 33°C
T1 = 75°C
∆T = 33°C –75°C = –44°C
Solution
Q = mc∆T m = Q/c∆T=
8 –3.000 10 J×
J 3.89
g ° C• ( –44° C
)
= 1 752 745.97 g ≅ 1.8 × 106 g
9.
Given
Q = 2.75 × 106 J c =4.19 J
g °C • T2 = 100.00°C
T1 = 0.00°C
∆T = 100.00°C
Solution
Q = mc∆T m = Q/c∆T=
6
2.75 × 10 J J
4.19 ° C
g• (100.00° C
)
= 6563.25 g = 6.56 × 103 g = 6.56 kg
10.
Given
m = 100.0 g
Q = 45 kJ = –4.5 × 104 J
T1 = 13.0°C
T2 = –15.0°C
∆T = –28.0°C
Solution
Q = mc∆T c = Q/m∆T=
(
)
4 –4.5 10 J 100.0g (–28.0°C)
×
= 16.071 43 J g °C • ≅16 J
g °C •
BLM 10-4, Heating the Air/Overhead
Master
BLM 10-5, Heat of Vaporization and Heat
of Fusion/Skill Builder
Goal: Students practise solving heat of fusion and vaporization problems.
Answers
Complete GRASP solutions are given for only the first two problems.
1.
Given
Number of moles of H2O(g), n = 2.05 mol
Heat of vaporization of water, ∆Hvap = –40.65 kJ
mol
Required
Energy released, QAnalysis
Use Q = n∆Hvap. Solve for Q.
Solution
Q = n∆Hvap= (2.05 mol) –40.65 kJ mol
= –83.3325 kJ ≅ –83.3 kJ
Paraphrase
When 2.05 mol of water vapour at 100°C condenses with no change in temperature, 83.3 kJ of energy is released.
2.
Given
Mass of water, m = 250 g
Heat of fusion of water, ∆Hfus = 6.01 kJ
mol
Number of moles of water, n = m M = 250 g
g 18.02
mol
= 13.873 47 mol ≅ 14 mol
Required
Energy needed to melt ice, Q
Analysis
Use Q = n∆Hfus. Solve for Q.
Solution
Q = n∆Hfus= (13.873 47 mol) 6.01 kJ mol
= 83.379 58 kJ ≅ 83 kJ
Paraphrase
It takes 83 kJ of energy to melt 250 g of ice at 0.0°C to liquid water at 0.0°C.
3. Given
m = 25 g
cube × 24 cubes
2 = 6.0 10 g×
∆Hfus= –6.01 kJ
mol n = m/M =
2 6.0 10 g×
g 18.02
mol
= 33.2963 mol ≅ 33 mol
Solution
Q = n∆Hfus= (33.2963 mol) –6.01 kJ mol
= –200.11 kJ ≅ –2.0 × 102 kJ
4.
Given
Q = 13.4 MJ = 1.34 × 104 kJ ∆Hfus = 40.65 kJ
mol
Solution
Q = n∆Hfusn = Q/∆Hfus
=
4 1.34 × 10 kJ
(
)
kJ 6.01
mol
= 2229.617 mol ≅ 2.23 × 103 mol
5.
Given
Q = 695 kJ ∆Hvap = 40.65 kJSolution
Q =n∆Hfusn = Q/∆Hfus
= 695 kJ kJ 40.65
mol
= 17.097 mol ≅ 17.1 mol n = m
M m = n M
= (17.097 mol) 18.02 g mol
= 308.088 g ≅ 309 g
Each cup of water is equivalent to 250 mL. Based on the
density of water 1 g mL
, each cup is equivalent to 250 g
of water.
Number of cups = 309 1 cup 250 g ×
= 1.236 cups ≅ 1.2 cups
6. Steam would cause a worse burn than water because it has gone through a phase change and absorbed an
additional 40.65 kJ
mol of energy, even though it is at the same temperature as the water.
7. Energy removed from water:
Given
m = 250 mL 1.00 g 100 mL ×
= 250 g
c = 4.19 J g °C• ∆T = –5.0°C
Solution
Q = mc∆T= (250 g) 4.19 J
g •° C (-5.0° C
)
= –5237.5 J ≅ –5.2 kJ
Amount of ice to melt:
Given
Q = –5.2375 kJ ∆Hfus = –6.01 kJmol
Solution
Q = n∆Hfusn = Q/∆Hfus
=–5.2375 kJ kJ –6.01
mol = 0.87146 mol ≅ 0.87 mol n = m
M m = n M
= (0.87146 mol) 18.02 g mol
= 15.70379 g ≅ 16 g
Paraphrase
Since 5.2 kJ of energy will have to be removed from the water and absorbed by the ice, 16 g of ice at 0.0°C will have to melt to liquid water at 0.0°C.
BLM 10-6, Thermal Energy Transfer/
Assessment
Goal: Students assess their understanding of the mechanisms of thermal energy transfer.
2. (a) convection (b) conduction (c) conduction (d) radiation
(e) conduction, convection, and radiation 3. The cornfield would absorb the most solar energy
directly from the Sun. The cornfield has a low albedo compared with snow, which is highly reflective. The cornfield has a higher specific heat capacity than the atmosphere, which solar energy does not heat significantly.
BLM 10-7, The Ocean’s
Currents/Enrichment
Goal: Students learn more about El Niño, the Gulf Stream, and deep ocean currents.
Answers
1. See straight, thick arrows in diagram below.
2. See diagram above.
3. (a) See straight, thin arrows in diagram above. With a switch in direction of the wind currents that normally blow warm water toward Australia and Indonesia, the winds would instead push the warm waters toward South America. As a result, Peru would receive much more rain than normal, and Australia would receive much less.
(b) Normally, the deeper cold water rises to the ocean’s surface as the warmer surface waters move west. During an El Niño, less cold water reaches the ocean’s surface and water
temperatures in the eastern Pacific are warmer than usual. The warm temperatures kill off marine life and disrupt the marine ecosystem. Also, the warmer surface waters allow more clouds to form and more rainstorms to occur. Although the climate in Peru is typically dry, Peru may endure massive flooding during an El Niño.
4. La Niña intensifies normal conditions, resulting in abnormally wet weather in Australia and extreme drought in South America.
5. During an El Niño, the winters in Canada may be significantly warmer than usual, especially in
Manitoba and western Ontario. Southern Canada tends to have drier winters and southern British Columbia receives less snow than normal. Warmer waters off the coast of British Columbia cause salmon to change their migration routes.
During La Niña, Canadian winters are colder than normal, particularly in western Canada. Southern Canada has wetter winters and southern British Columbia receives more snow than normal. Waters off the coast of British Columbia are especially cool. Also, the incidence of hurricanes on the Atlantic coast may increase.
BLM 10-8, Global Warming and the Gulf
Stream/Enrichment
Goal: Students research the potential effects of global warming on the Gulf Stream and the climate in Europe.
Answers
1. Titles of articles will vary.
2. References or sources should follow a standard format.
3. (a) The North Atlantic Drift moves warm waters past Western Europe. The warm water heats the air, which keeps the climate of western Europe relatively warm compared with the climate of Newfoundland and Labrador.
(b) If sea ice melts as a result of global warming, the fresh, cold water could dilute the Gulf Stream and disrupt thermohaline circulation in the Atlantic. (c) Waters along western Europe could become
cooler as a result of global warming. Ocean life, which would not be adapted to a cold
environment, could die off or move elsewhere. The climate in Europe itself could become very cold.
BLM 10-9, Chinooks/Information Handout
Goal: Students learn how Chinooks shape southern Alberta winters.
BLM 10-10, Movement of Air and Water/
Assessment
Goal: Students assess their understanding of the causes of movement of air and water near Earth’s surface.
Answers
1. Different substances have different albedos
(reflectivity). The higher a substance’s albedo is, the more the substance reflects light. Also, different substances have different specific heat capacities — the higher the specific heat capacity the more energy absorbed.
2. hot, cold 3. (a) jet stream
(b) Coriolis effect (c) convection (d) El Niño (e) low pressure (f) high pressure (g) North Atlantic Drift (h) thermohaline circulation (i) sea breeze
(j) land breeze (k) albedo (l) Gulf Stream
(m) orographic precipitation (n) rain shadow
(o) trade winds
BLM 10-11, Chapter 10 Test/Assessment
Goal: Students demonstrate their understanding of the information presented in Chapter 10.
Answers
1. F: A closed system allows the exchange of energy but not matter with its surroundings.
2. T
3. F: A substance with a high albedo will reflect much of the light energy that strikes its surface. 4. T
5. T 6. (c) 7. (b) 8. (b) 9. (c)
10. (a)
11. radiation budget 12. greenhouse effect 13. climate, weather 14. Coriolis effect
15. cools, becomes denser, descends 16. (b)
17. (c) 18. (a) 19. (d) 20. (c) 21. (a) 22. (a) 23. (d) 24. (b) 25. (d) 26. 38.0 27. 4.10 28. 1 4 6 8
29. The two principal factors that determine Alberta’s climate are latitude and the orientation of Earth. Because Alberta lies across relatively high latitudes, the Sun’s rays are spread over a larger area than they are at lower latitudes. As a result, Earth’s surface absorbs less solar energy and is colder in Alberta than in regions at lower latitudes. The tilt of Earth’s axis determines the clear differences in the seasons in Alberta. When the Northern Hemisphere is tilted away from the Sun, the length of the day in Alberta is shorter, and less solar energy reaches the region. When the Northern Hemisphere points toward the Sun, Alberta experiences summer.
