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4. Heat
4.1 Thermal Equilibrium4.2 Thermometer4.3 Specific Heat Capacity
4.4 Application of Specific Heat Capacity4.5 Specific Latent Heat4.6 The Gas Law
Thermal Equilibrium
Card 1: Thermal Energy
Thermal Energy
Thermal energy is a measure of the total kinetic and potential
energy in all the molecules and atoms in a certain substance.
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Card 2: Thermal Equilibrium
Thermal Equilibrium
Thermal equilibrium is achieved between two bodies when there
is no net heat flow between the two bodies.
2 objects in contact will achieve thermal equilibrium when
the temperatures are the same.
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Card 3: Temperature
Temperature
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Temperature is a physical quantity which measures the degree of
hotness of an object.
Temperature is a measure of the average kinetic energy which
each molecule of an object possesses.
One object is at a higher temperature than another if the average
kinetic energy of each of its molecules is greater.
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Card 4: What is heat?
What is heat?
Heat is the flow of thermal energy.
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Card 5: Comparing Thermal Energy and Temperature
Comparing Thermal Energy and Temperature
Temperature Thermal Energy
Degree of hotness of an object A form of Energy
Unit:
Kelvin (K) / Degree Celsius (oC)
Unit:
Joule (J)
It is the average kinetic energy of
the particles
It is the sum of the kinetic energy
and potential energy of the
particles
It is a base quantity It is a derived quantity
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Card 6: Application of thermal equilibrium - Thermometer
Application of Thermal Equilibrium- Thermometer
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(The original file of this image can be founded in the wikipedia)
Thermometer is placed in contact with the patients
body.
If both the body temperature of the patient and that
of the mercury (or alcohol) in the clinical thermometer
have reached thermal equilibrium, then
the temperature of the thermometeris the same as the
body temperature, hence the reading of the thermometer
shows the body temperature of the patient.
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Card 7: Application of thermal equilibrium - Refrigerator
Application of Thermal Equilibrium- Refrigerator
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(The original file of this image can be founded in the wikipedia)
When food is put in the refrigerator, the heat from
the food is transferred into the air of the refrigerator.
This process is continue until the temperature of
the food is equal to the temperature of the air in the
refrigerator, when thermal equilibrium is reached
between the food and the refrigerator.
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Card 8: Application of thermal equilibrium - Oven
Application of Thermal Equilibrium- Oven
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When food such as meat or cake is put in the oven,
the heat of the oven is transferred into the food.
This process will continue until the food is in thermal
equilibrium with the air in the oven.
This happen when the temperature of the food
isequal to the temperature of the air in the oven.
Temperature and Thermometer
Card 1: Calibration of thermometer-determining the Ice Point
Calibration of thermometer-determining the Ice Point
The bulb of the uncalibrated thermometer is placed
in the ice.
The mercury level is marked. This is the taken as
the ice point and it is arbitrarily given as 0oC.
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Card 2: Calibration of thermometer-determining the Steam Point
Calibration of thermometer-determining the Steam Point
The bulb of the uncalibrated thermometer is placed
in the steam above a boiling water.
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The mercury level is marked. This is taken as the
steam point and its arbitrary given as 100oC.
Notes: Instead of placing in the boiling water, the thermometer is
placed in the steam above the boiling water.
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Card 3: Calibration of thermometer-Measuring the temperature.
Calibration of thermometer-Measuring thetemperature.
After marking the ice point (0oC) and steam point (100oC), the
temperature of a substance can be determined by the formula:
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Example 1
The lengths of mercury thread in a mercury thermometer are 10 mm when the bulb
is in melting ice; 130 mm when the bulb is in the steam above boiling water and 40
mm when the bulb is in a liquid A. What is the temperature of liquid A?
Answer:
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l0 = 10mm
l100 = 130mm
l = 40mm
\begin{gathered}
T = \frac{{l_\theta - l_0 }}
{{l_{100} - l_0 }} \times 100^o C \hfill \\
T = \frac{{40 - 10}}
{{130 - 10}} \times 100^o C \hfill \\
T = \frac{{30}}
{{120}} \times 100^o C \hfill \\
T = 25^o C \hfill \\
\end{gathered}
Card 4: Sensitivity of Thermometer
Sensitivity of Thermometer
The sensitivity of a thermometer can be increased by
Using a thermometer with a smaller bulb A
smaller bulb contains less mercury and hence absorbs
heat in shorter time.
A glass bulb with thinner wall Heat can be
transferred to the bulb easily.
Capillary with narrow bore produces a greaterchange in the length of the mercury column.
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Card 5: Increases the Accuracy of Thermometer
Accuracy of Thermometer
To increase the accuracy of a thermometer, the diameter of the
capillary tube of the thermometer must be constant throughout.
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Card 6: Advantages of Using Mercury as the Liquid in a Thermometer
Advantages of Using Mercury in a Thermometer.
It doesnt wet the side of the tube.
It makes a thread which can be seen easily.
It expand uniformly when heated
It can conduct heat well. Therefore it reponses
quickly to temperature changes
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Card 7: Disadvantages of Using Mercury as the Liquid in a Thermometer
Disadvantages of Using Mercury
It freezes at -39oC. Therefore it is not suitablefor
measuring temperature lower than -39oC.
It is poisonous
It is expensive
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Card 8: Advantages of Using Alcohol as the Liquid in a Thermometer
Advantages of Using Alcohol
It freezes at -115oC. Therefore it is suitable for low
temperature, including the Artic and Antarctic region.
It expands greaterthan mercury.
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Card 9: Disadvantages of Using Alcohol as the Liquid in a Thermometer
Disadvantages of Using Alcohol
It is transparent and difficult to be seen. It has to
be coloured in order to be seen easily.
It always clings to the sides of the tube.
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The thread has tendency to break.
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Card 10: What is absolute temperature?
Absolute Temperature
Absolute temperature is the temperature measured in Kelvin
scale, which it is a temperature reading made relative to absolute
zero.
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Card 11: What is Absolute Zero?
Absolute Zero
Absolute zero is the temperature where thermal energy is at
minimum. It is 0 on the Kelvin scale and -273 on the Celsius scale.
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Specific Heat Capacity
Card 1: What is Heat Capacity?
Heat Capacity
Heat capacity is the amount of heat required to increase the
temperature of an object by 1 oC (or 1 K).
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Card 2: Define specific heat capacity
Specific Heat Capacity
The amount of heat required to change the temperature of 1 kg of a
substance by 1oC.
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Card 3: Formula of Specific Heat Capacity
Formula of Specific Heat Capacity
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Example 1
How much thermal energy is required to raise the temperature of a 2 kg aluminium
block from 25 ?C to 30 ?C? [The specific heat capacity of aluminium is 900 Jkg-1 oC-1 ]
Answer:
Mass, m = 2kg
Specific heat capacity, c = 900 Jkg-1 oC-1
Temperature change, = 30 - 25 = 5 oC
Thermal energy required,
Q = mc = (2)(900)(5) = 9000J.
Card 4: Conversion Of Electrical Energy Into Thermal Energy
Conversion Of Electrical Energy Into Thermal Energy
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Example 2An electric heater supplies 5 kW of power to a tank of water. Assume all the energy
supplied is converted into heat energy and the energy losses to the surrounding is
negligible. How long will it take to heat 500 kg of water in the tank from 25 to 100
?C? [Specific heat capacity of water = 4200 J kg -1 oC-1 ]
Answer:
P = 5000W
m = 500kg
c = 4200 J kg-1 oC-1
= 100 - 25 = 75oC
t = ?
We assume,
all the electrical energy supplied = heat energy absorbed by the water
Pt = mc
(5000) t = (500)(4200)(75)
t = 31500s = 525 minutes = 8 hours 45 minutes
(Practically the time can be much longer than this because a lot of heat may be losses
to the surrounding.)
Card 5: Conversion Of Gravitational Energy Into Thermal Energy
Conversion of Gravitational Energy into Thermal
Energy
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Example 3
A lead shot of mass 5g is placed at the bottom of a vertical cylinder that is 1m long
and closed at both ends. The cylinder is inverted so that the shot falls 1 m. By how
much will the temperature of the shot increase if this process is repeated 100 times?
[The specific heat capacity of lead is 130Jkg -1 K-1 ]
Answer:
m = 5g
h = 1m 100 = 100m
g = 10 ms-2
c = 130Jkg-1 K-1
= ?
In this case, the energy conversion is from potential energy to heat energy. We
assume that all potential energy is converted into heat energy. Therefore
mgh = mc
gh = c
(10)(100) = (130)
= 7.69 oC
Card 6: Conversion Of Kinetic Energy Into Thermal Energy
Conversion Of Kinetic Energy Into Thermal Energy
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Example 4
A 2g lead bullet is moving at 150 m/s when it strikes a wooden block and is broughtto rest. Assuming all kinetic energy is converted into thermal energy and transferred
to the bullet, what is the rise in temperature of the bullet as it is brought to rest? [The
specific heat capacity of lead is 130 Jkg -1 K-1 ]
Answer:
m = 2g = 0.002kg
v = 150 m/s
c = 130 Jkg-1
= ?
We assume all the kinetic energy is converted into heat energy
\begin{gathered}
\frac{1}
{2}mv^2 = mc\theta \hfill \\
\frac{{v^2 }}
{{2c}} = \theta \hfill \\
\frac{{(150)^2 }}
{{2(130)}} = \theta \hfill \\
\theta = 86.54^o C \hfill \\
\end{gathered}
Card 7: Mixing 2 LiquidMixing 2 Liquid
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Example
What will be the final temperature if 500 cm3 of water at 0 ?C is added to 200cm3 of
water at 90 ?C? [Density of water = 1gcm-3 ]
Answer:
The density of water is 1g/cm3, which means the mass of 1 cm3 of water is equal to
1g.
Let the final temperature =
m1 = 500g = 0.5kg
c1 = c
1 = - 0 =
m2 = 200g = 0.2kg
c2 = c
2 = 90 -
m1c11 = m2c22
(0.5) c ( ) = (0.2) c ( 90 - )
0.5 = 18 - 0.2
0.5 + 0.2 = 18
0.7 = 18
= 25.71 oC
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Application of Specific Heat Capacity
Card 1: Car Radiator
Car radiator
Water is pumped through the channels in the
engine block to absorb heat.
Water is used as the cooling agent due to its high
specific heat capacity.
The hot water flows to the radiator and is cooled by
the air flows through the fins of the radiator.
The cool water flows back to the engine again to
capture more heat and this cycle is repeated
continuously.
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Card 2: Cooking utensils
Cooking utensils
Cooking utensils are made of metal which has low
specific heat capacity so that it need less heat to raise up
the temperature.
Handles of cooking utensils are made of
substances with high specific heat capacities so that its
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temperature wont become too high even if it absorbs
large amount of heat.
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Card 3: Thermal Radiator
Thermal Radiator
Thermal radiators are always used in cold country
to warm the house.
Hot water is made to flow through a radiator. The
heat given out from the radiator is then warm the air ofthe house.
The cold water is then flows back to the water tank.
This process is repeated continuously.
Water is used in the radiator because it has high
specific heat capacity.
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Card 4: Phenomena of specific Heat capacity Sea Breeze
Sea Breeze
Land has lower heat capacity than sea water.
Therefore, in day time, the temperature of the land
increases faster than the sea.
Hot air (lower density) above the land rises. Cooler
air from the sea flows towards land and hence produces
sea breeze.
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Card 5: Phenomena of specific Heat capacity Land Breeze
Land Breeze
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Land has lower heat capacity than sea water.
During night time, the temperature of the land drops
faster than the sea.
Hot air (lower density) above the sea rises. Cooler
air from the land blows towards sea and hence produces
land breeze.
Specific Latent Heat
Card 1:The Heating Curve
Heating Curve
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State of matter:
A-B: Solid
B-C: Solid and Liquid
C-D: Liquid
D-E: Liquid and Gas
E-F: Gas
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Card 2: The Heating Curve - Latent Heat
Heating Curve - Latent Heat
T1 is the melting point whereas T2 is the boiling
point.
From Q to R and S to T, the temperature remains
constant because the heat supplied to the object is
used to overcome the forces of attraction that hold the
particles together.
Heat obsorbs during Q-R is called the latent heat
of fusion.
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Heat obsorbs during S-T is called the latent heat
of vaporisation.
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Card 3: The Cooling Curve
Cooling Curve
States of matter:
P-Q: Gas
Q-R: Gas and Liquid
R-S: Liquid
S-T: Liquid and Solid
T-U: Solid
U-V: Solid
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Card 4: The Cooling Curve - Latent Heat
Cooling Curve - Latent Heat
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T1 is the condensation point, T2 is the freezing
point whereas T3 is room temperature.
During Q-R and S-T, the temperature remains
unchanged. This is because the energy
produced during the formation of bonds is equal to the
heat energy released to the surroundings during
cooling.
The heat energy released during Q-R is called
the latent heat of vaporization.
The heat energy released during S-T is called
thelatent heat of fusion.
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Card 5: What is Latent Heat?
Latent Heat
Latent heat is the heat absorbed or releases during a change of
state of matter.
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Card 6: Define the specific latent heat of vaporisation
Latent Heat of Vaporisation
The specific latent heat of vaporization is the heat needed
to change 1 kg of a liquid at its boiling point into vapour or vice
versa, without a change in temperature.
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Card 7: Define specific latent heat of fusion.
Latent Heat of Fusion
The specific latent heat of fusion is the heat needed to change 1
kg of a solid at its melting point into a liquid, or vice versa, without
a change in temperature.
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Card 8: Formula of Specific Latent Heat
Formula o Specific Latent Heat
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Example 1
How much heat energy is required to change 0.5 kg of ice at 0?C into water at 25?C?
[Specific latent heat of fusion of water = 334 000 J/kg; specific heat capacity of water
= 4200 J/(kg K).]
Answer:
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There are 2 processes involve when an ice is converted into water at 25C.
Ice at 0C ------> Water at 0C -------> Water at 25C
Energy absorbed to convert 0.5kg from Ice at 0C to Water at 0C
Q1 = mL = (0.5)(334000) = 167000J
Energy absorbed to convert 0.5kg from watyer at 0C to Water at 25C
Q2 = mc = (0.5)(4200)(25-0) = 52500J
Total energy required = Q1 + Q2 = 167000 + 52500 = 219500J
Example 2
How much energy is required to change exactly 1 g of ice at -20?C to steam at 120 ?C?
[Specific heat capacity of water = 4200J kg-1 oC-1 ; Specific latent heat of fusion of ice
= 334,000 Jkg-1 , specific heat capacity of steam is 2020 J/(kg ?C), Specific latent heat
of vaporization of water = 2,260,000 J/kg, specific heat capacity of ice = 2100 J/(kg
K)]
Answer:
The processes involved:
Ice (-20C) -----> Ice (0C) -----> Water (0C) -----> Water (100C)
-----> Steam (100C) -----> Water (120C)
Energy required:
Ice (-20C) to Ice (0C), Q1 = mc = (0.001)(2100)(20) = 42J
Ice (0C) to Water (0C), Q2 = mL = (0.001)(334000) = 334JWater (0C) to Water (100C), Q3 = mc = (0.001)(4200)(100) = 420J
Water (100C) to Steam (100C), Q4 = mL = (0.001)(2260000) = 2260J
Steam (100C) to Steam (120C), Q5 = mc = (0.001)(2020)(20) = 40.4J
Total energy required
= Q1 + Q2 + Q3 + Q4 + Q5
= 42 + 334 + 420 + 2260 + 40.4 = 3096.4J
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Card 9: What will happen to the melting point of ice if the ice contains impurities?
How impurities affect the melting point
If impurities are present in a substance, its melting point will be
lower than normal.
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Card 10: Factors affecting melting point
Factors affecting melting point
Pressure:
Applying pressure to ice, for example, lowers its melting point.
Presence of impurities:
Adding salt to melting ice, for example, can reduce its melting point
to as low as -18C.
The Gas Law
Card 1: State Boyle's law.
Boyle's Law
Boyle's law states that the pressure of a gas with constant mass is
inversely proportional to its volume provided the temperature of the
gas is kept constant.
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Card 2: Formula of Boyle's Law
Formula of Boyle's Law
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Example 1
Figure (a) above shows a capillary tube with a thread of mercury 2cm long. The
length of the air trapped in the tube is 10cm. Find the length of the trapped air if the
tube is inverted as shown in figure (b). [Atmospheric pressure = 76cmHg]
Answer:
In case (a), the pressure of the gas trapped in the capillary is equal to the
atmospheric pressure + the pressure caused by the mercury thread.
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P1 = 76 + 2 = 78cmHg
V1 = 10cm
Incase (b), the pressure of the gas trapped in the capillary is equal to the atmospheric
pressure - the pressure caused by the mercury thread.
P2 = 76 - 2 = 74cmHg
V2 = ?
By using Boyle's Lwa's formula
P1V1 = P2V2
(78)(10) = (74)V2
V2 = 10.54cm
The length of the trapped air in figure (b) is 10.54cmCard 3: Graph of Boyle's Law
Graph of Boyle's Law
According to Boyle's Law, gas pressure (P) is inversely proportional
to the volume of the gas (V) or P is directly proportional to 1/V.
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Card 4: State the Pressure Law of Gas
Pressure Law
Pressure states that for a fixed mass of gas, the pressure of the gas
is directly proportional to its absolute temperature if the volume
remains constant.
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Card 5: Formula of Pressure Law
Formula of Pressure Law
Notes: The temperature must be the absolute temperature (Kelvin
Scale)
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Example 2
A gas in a container with a constant volume has a pressure of 200,000Pa at a
temperature of 30oC. What is the pressure of the gas if the temperature is increased
to 60oC?
Answer:
At 30o
C,P1 = 200,000Pa
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T1 = 30oC = 273 + 30= 303K
(Note: the temperature must be in Kelvin scale)
At 60oC,
P2 = ?
T2 = 60oC = 273 + 60 = 333K
By using the formula of Pressure Law,
\begin{gathered}
\frac{{P_1 }}
{{T_1 }} = \frac{{P_2 }}
{{T_2 }} \hfill \\
\frac{{(200,000)}}
{{(303)}} = \frac{{P_2 }}
{{(333)}} \hfill \\
P_2 = \frac{{(200,000) \times (333)}}
{{(303)}} = 219802Pa \hfill \\
\end{gathered}
Card 6: Graph of Pressure LawGraph of Pressure Law
Pressure is directly proportional to the temperature.
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Card 7: State the Charle's Law of Gas
Charle's Law
Charless Law states that for a fixed mass of gas, the volume of the
gas is directly proportional to the absolute temperature provided the
pressure remains constant.
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Card 8: Formula of Charle's Law
Formula of Charle's Law
http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#6http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#1http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#8http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#7http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#1http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#9http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#6http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#1http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#8http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#7http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#1http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#98/2/2019 bab 4 heat
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Notes: The temperature must be the absolute temperature (KelvinScale)
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Example 3
A balloon is filled with 2000cm3 of gas at 27oC. The balloon is immersed in a
container filled with water and the water is then heated. If the pressure in the balloon
remain constant, find the volume of the gas when its temperature is 57 oC.
Answer:
At 27oC,
V1 = 2000cm3
T1 = 27oC = 27 + 273 = 300K
At 57oC,
V2 = ?
T2 = 57oC = 57 + 273 = 330K
By using the formula of Charle's Law,
\begin{gathered}
\frac{{V_1 }}
{{T_1 }} = \frac{{V_2 }}
http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#8http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#1http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#10http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#8http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#1http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#108/2/2019 bab 4 heat
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{{T_2 }} \hfill \\
\frac{{(2,000)}}
{{(30)}} = \frac{{V_2 }}
{{(33)}} \hfill \\
V_2 = \frac{{(2,000) \times (330)}}
{{(300)}} = 2200cm^3 \hfill \\
\end{gathered}
Card 9: Graph of Charle's Law
Graph of Charle's Law
Volume of gas is directly proportional to the temperature of the gas.
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http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#9http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#1http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#11http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#9http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#1http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#118/2/2019 bab 4 heat
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Card 10: Universal Gas Law - Formula
Formula of Universal Gas Law
http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#11http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html#11Recommended