Water and Energy
What can thermal energy do?
• Change a substance’s
temperature
• Change a substance’s
state of matter
Any amount of thermal energy absorbed by a
substance can be used to do either one.
Whenever the water
is absorbing the
thermal energy from
the sun some of it is
used to change the
temperature of snow,
ice, water and vapour
Some of it is used to
evaporate or melt
some of the water
Consider this
Water has a relatively large heat
capacity
•It takes a lot of energy to raise the
temperature of water.
•This means that water can absorb
lots of the sun’s energy and it will
not change its temperature
dramatically
•Water’s high specific heat capacity
also means that it loses energy
relatively slowly
•So as the sun goes down and night
falls, the water will still remain a
relatively constant temperature
What if water had a low heat
capacity?
If water had a low heat capacity
•Only a little bit of thermal energy
would be needed to raise the
temperature
•So all day long as the Sun’s energy
was absorbed by the water, the
temperature would rise dramatically
•This would also mean that during
the night, water would lose its heat
very fast and the temperature would
drop quickly!!!
What can we take from this
• Water’s high specific heat capacity allows
it to moderate its temperature despite the
huge fluctuations of thermal energy that it
receives during the day and night.
• Without it, water’s temperature would
change drastically throughout the course
of a day
Recap - Heat
Energy that flows from something warm to
something cooler
A hotter substance gives Kinetic Energy to
a cooler one
When heat is transferred (lost or gained),
there is a change in the energy within the
substance
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Learning Check H1
A. When you touch ice, heat is transferred
from
1) your hand to the ice
2) the ice to your hand
B. When you drink a hot cup of coffee, heat
is transferred from
1) your mouth to the coffee
2) the coffee to your mouth
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Solution H1
A. When you touch ice, heat is transferred
from
1) your hand to the ice
B. When you drink a hot cup of coffee, heat
is transferred from
2) the coffee to your mouth
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Learning Check H2
When you heat 200 g of water for 1 minute, the
water temperature rises from 10°C to 18°C.
400 g
200 g
If you heat 400 g of water at 10°C in the same
pan with the same amount of heat for 1
minute, what would you expect the final
temperature to be?
1) 10 °C
2) 14°C
3) 18°C
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Solution H2
2) 14°C
Heating twice the mass of water using the
same amount of heat will raise the
temperature only half as much.
400 g
200 g
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The Specific Heat Capacity
• Specific heat is how much energy it takes to
raise the temperature of 1.0 g of a
substance by 1° C.
• We can find out how much energy is
needed to change an object’s temperature
with this equation:
Q = mcΔT
Effects of Mass on Heat
• Large object
mcΔT
•Q is bigger
• =
• Smaller object
• = mcΔT
Q is smaller
Conclusion: bigger objects need more energy to raise their temperature
Effect of Specific Heat on Heat Energy
• Large Specific Heat
• =m
cΔT
•Q is bigger
4.18
J/g·°C
• Smaller Specific Heat
• = mcΔT
Q is smaller
0.32
J/g·°C
Conclusion: objects with bigger specific heat need more energy to raise their
temperature by the same amount
Effect of Temp Change on Heat Energy
• Large Temp Change
• = mc
ΔT
•Q is bigger
• Smaller Temp change
• = mcΔT
Q is smaller
Conclusion: bigger temperature change will cause a bigger change in heat energy
Some Equalities for Heat
Heat is measured in calories or joules
 1 kcal = 1000 cal
 1 calorie = 4.18J
 1 kJ = 1000 J
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Specific Heat
 Why do some foods stay hot longer than
others?
 Why is the beach sand hot, but the water is
cool on the same hot day?
The answer to these questions is specific heat.
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Specific Heat
Different substances have different
capacities for storing energy
It may take 20 minutes to heat water to
75°C. However, the same mass of
aluminum might require 5 minutes and
the same amount of copper may take
only 2 minutes to reach the same
temperature.
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Specific Heat Values
Specific heat is the amount of heat needed to
raise the temperature of 1 g of a substance by
1°C
cal/g°C
J/g°C
water
1.00
4.18
aluminum
copper
silver
gold
0.22
0.093
0.057
0.031
0.90
0.39
0.24
0.13
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Learning Check H3
A. A substance with a large specific heat
1) heats up quickly
2) heats up slowly
B. When ocean water cools, the surrounding air
1) cools
2) warms 3) stays the same
C. Sand in the desert is hot in the day, and cool
at night. Sand must have a
1) high specific heat
2) low specific heat
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Solution H3
A. A substance with a large specific heat
2) heats up slowly
B. When ocean water cools, the surrounding air
2) warms
C. Sand in the desert is hot in the day, and cool
at night. Sand must have a
2) low specific heat
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Measuring Heat Energy
Requires
 Mass of substance
 Temperature change T
 Specific heat of the substance
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Calculating Heat Energy
mass
m
x specific heat x temp. change
x
c
x
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Heat Calculations
A hot-water bottle contains 750 g of water at
65°C. If the water cools to body temperature
(37°C), how many calories of heat could be
transferred to sore muscles?
x Sp. Ht. (H2O) x T
x 1.00 cal
x 28°C
g°C
= 21 000 cal
heat = mass
750 g
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Learning Check H4
How many kcal are needed to raise the
temperature of 120 g of water from 15°C to
75°C?
1) 1.8 kcal
2) 7.2 kcal
3) 9.0 kcal
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Solution H4
How many kcal are needed to raise the
temperature of 120 g of water from 15°C to
75°C?
2) 7.2 kcal
120 g x (75°C - 15°C) x 1.00 cal x 1 kcal
g°C
1000 cal
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Energy and Nutrition
1 Calorie (nutritional) = 1 kcal
1 Cal
= 1000 cal
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Caloric Food Values
Carbohydrate = 4 kcal/g
Fat = 9 kcal/g
Protein = 4 kcal/g
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Foods and Calories
Food
Carbo
Fat
carrots,
1 cup
11
0
1
50
banana
26
0
1
110
0
6
6
80
0
3
20
110
beef (3 oz) 0
5
22
130
egg
chicken
(no skin)
Protein Energy(kcal)
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Learning Check H5
1.0 cup of whole milk contains 12 g of
carbohydrate, 9.0 g of fat, and 9.0 g of
protein. How many kcal (Cal) are
obtained?
1) 48 kcal
2) 81 kcal
3) 165 kcal
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Solution H5
3) 165 kcal
12 g carbo x 4 kcal/g
=
48 kcal
9.0 g fat
=
81 kcal
=
36 kcal
=
165 kcal
x 9 kcal/g
9.0 g protein x 4 kcal/g
Total kcal
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1. Iron metal has a specific heat of 0.449
J/goC. How much heat is transferred to a
5.00 g piece of iron, initially at 20.0 oC,
when it is placed in a beaker of boiling
water?
C = 0.449 J/goC
Ti = 20.0oC
Q = mc
m = 5.00g
Tf = 100oC
T = mc(Tf – Ti)
Q = 5.00g × 0.449 J/goC × (100-20oC)
Q = 179.6 J
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2. Iron metal has a specific heat of 0.449 J/goC. How
many calories of energy are given off to lower the
temperature of 100.0 g of iron from 150.0 oC to 35.0 oC?
Q = mc(Tf-Ti)
Q = 100.0g × 0.449 J/goC × (150.0-35.0oC)
Q = 5163.5 cal
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Calculating Specific Heat Worksheet
1. A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its
temperature changes from 25°C to 175°C. Calculate the specific heat
capacity of iron.
m = 15.75 g
Q = 1086.75 J
∆T = 175 – 25 °C
∆T = 150 °C
Q
m
c
2. How many joules of heat are needed to raise the temperature of 10.0
g of aluminum from 22°C to 55°C, if the specific heat of aluminum is 0.90
J/g°C?
m=
∆T =
Q = mc ∆T = (10.0 g) x (0.90 J/g°C) x (33 °C)
Q = 297 J
c=
Q
m
c