heat form 4 bab 4
TRANSCRIPT
![Page 1: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/1.jpg)
HEAT
Chapter 4
![Page 2: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/2.jpg)
4.1 Thermal Equilibrium
![Page 3: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/3.jpg)
Thermal Energy
Thermal energy is a measure of the total kinetic and potential energy in all the molecules and atoms in a certain substance.
![Page 4: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/4.jpg)
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.
![Page 5: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/5.jpg)
![Page 6: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/6.jpg)
Temperature
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.
![Page 7: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/7.jpg)
Comparing Thermal Energy and Temperature
![Page 8: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/8.jpg)
Application of Thermal Equilibrium--Thermometer
![Page 9: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/9.jpg)
Thermometer is placed in contact with the patient’s 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 thermometer is the same as the body temperature, hence the reading of the thermometer shows the body temperature of the patient
![Page 10: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/10.jpg)
Application of Thermal Equilibrium- Refrigerator
![Page 11: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/11.jpg)
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.
![Page 12: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/12.jpg)
Application of Thermal Equilibrium - Oven
![Page 13: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/13.jpg)
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 is equal to the temperature of the air in the oven.
![Page 14: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/14.jpg)
Calibration of thermometer
![Page 15: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/15.jpg)
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.
![Page 16: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/16.jpg)
Calibration of thermometer-determining the Steam Point
The bulb of the uncalibrated thermometer is placed in the steam above a boiling water.
The mercury level is marked. This is taken as the steam point and it’s arbitrary given as 100oC.
Notes: Instead of placing in the boiling water, the thermometer is placed in the steam above the boiling water.
![Page 17: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/17.jpg)
Calibration of thermometer-Measuring the temperature.
After marking the ice point (0oC) and steam point (100oC), the temperature of a substance can be determined by the formula:
![Page 18: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/18.jpg)
example 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: l0 = 10mm
l100 = 130mmlθ = 40mm
![Page 19: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/19.jpg)
T = 40 - 10 130 – 10 30 20 25oC
X 100 oC
X 100 oC
Answer:l0 = 10mml100 = 130mmlθ = 40mm
![Page 20: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/20.jpg)
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 greater change in the length of the mercury column.
![Page 21: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/21.jpg)
Accuracy of Thermometer
To increase the accuracy of a thermometer, the diameter of the capillary tube of the thermometer must be constant throughout.
![Page 22: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/22.jpg)
Advantages of Using Mercury in a Thermometer.
It doesn’t 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
![Page 23: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/23.jpg)
Disadvantages of Using Mercury
It freezes at -39oC. Therefore it is not suitable for measuring temperature lower than -39oC.
It is poisonous It is expensive
![Page 24: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/24.jpg)
Advantages of Using Alcohol
It freezes at -115oC. Therefore it is suitable for low temperature, including the Artic and Antarctic region.
It expands greater than mercury
![Page 25: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/25.jpg)
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.
The thread has tendency to break.
![Page 26: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/26.jpg)
Absolute Temperature
Absolute temperature is the temperature measured in Kelvin scale, which it is a temperature reading made relative to absolute zero.
![Page 27: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/27.jpg)
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.
![Page 28: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/28.jpg)
4.2 specific heat capacity
![Page 29: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/29.jpg)
Heat Capacity
Heat capacity is the amount of heat required to increase the temperature of an object by 1 oC (or 1 K).
![Page 30: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/30.jpg)
Specific Heat Capacity
The amount of heat required to change the temperature of 1 kg of a substance by 1oC.
![Page 31: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/31.jpg)
Formula of Specific Heat Capacity
![Page 32: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/32.jpg)
example How much thermal energy is required to
raise the temperature of a 2 kg aluminium block from 25 oC to 30 oC? [The specific heat capacity of aluminium is 900 Jkg-1 oC-1] Answer:
Mass, m = 2kgSpecific heat capacity, c = 900 Jkg-1 oC-1
Temperature change, θ = 30 - 25 = 5 oC Thermal energy required,
Q = mcθ = (2)(900)(5) = 9000J.
![Page 33: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/33.jpg)
Conversion Of Electrical Energy Into Thermal Energy
![Page 34: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/34.jpg)
example An 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 oC? [Specific heat capacity of water = 4200 J kg-1 oC-1] Answer:
P = 5000Wm = 500kgc = 4200 J kg-1 oC-1
θ = 100 - 25 = 75oCt = ?
![Page 35: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/35.jpg)
We assume,all the electrical energy supplied = heat energy absorbed by the waterPt = mcθ(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.)
![Page 36: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/36.jpg)
Conversion of Gravitational Energy into Thermal Energy
![Page 37: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/37.jpg)
example 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-1K-1]
Answer: m = 5g
h = 1m × 100 = 100mg = 10 ms-2
c = 130Jkg-1K-1
θ = ?
![Page 38: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/38.jpg)
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 = mcθgh = cθ(10)(100) = (130) θθ = 7.69 oC
![Page 39: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/39.jpg)
Conversion Of Kinetic Energy Into Thermal Energy
![Page 40: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/40.jpg)
example A 2g lead bullet is moving at 150 m/s when it
strikes a wooden block and is brought to 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-1K-1] Answer:
m = 2g = 0.002kgv = 150 m/sc = 130 Jkg-1
θ = ? We assume all the kinetic energy is converted
into heat energy
![Page 41: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/41.jpg)
![Page 42: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/42.jpg)
Mixing 2 Liquid
![Page 43: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/43.jpg)
example What will be the final temperature if 500 cm3 of
water at 0 oC is added to 200cm3 of water at 90 oC? [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 = θ
m1 = 500g = 0.5kgc1 = cθ1 = θ - 0 = θm2 = 200g = 0.2kgc2 = cθ2 = 90 - θ
![Page 44: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/44.jpg)
m1c1θ1 = m2c2θ2
(0.5) c ( θ ) = (0.2) c ( 90 - θ )0.5 θ = 18 - 0.2 θ0.5 θ + 0.2 θ = 180.7 θ = 18θ = 25.71 oC
![Page 45: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/45.jpg)
Application of Specific Heat Capacity 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.
![Page 46: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/46.jpg)
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 temperature won’t become too high even if it absorbs large amount of heat.
![Page 47: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/47.jpg)
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 of the 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.
![Page 48: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/48.jpg)
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
![Page 49: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/49.jpg)
Land Breeze
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
![Page 50: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/50.jpg)
4.3 specific latent heat
![Page 51: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/51.jpg)
Heating Curve State of matter:A-B: SolidB-C: Solid and LiquidC-D: LiquidD-E: Liquid and GasE-F: Gas
![Page 52: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/52.jpg)
Heating Curve - Latent Heat
![Page 53: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/53.jpg)
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.
Heat obsorbs during S-T is called the latent heat of vaporisation.
![Page 54: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/54.jpg)
Cooling CurveStates of matter:P-Q: GasQ-R: Gas and LiquidR-S: LiquidS-T: Liquid and SolidT-U: SolidU-V: Solid
![Page 55: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/55.jpg)
Cooling Curve - Latent Heat
![Page 56: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/56.jpg)
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 the latent heat of fusion.
![Page 57: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/57.jpg)
Latent Heat
Latent heat is the heat absorbed or releases during a change of state of matter
![Page 58: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/58.jpg)
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
![Page 59: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/59.jpg)
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.
![Page 60: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/60.jpg)
Formula of Specific Latent Heat
![Page 61: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/61.jpg)
example
How much heat energy is required to change 0.5 kg of ice at 0oC into water at 25oC? [Specific latent heat of fusion of water = 334 000 J/kg; specific heat capacity of water = 4200 J/(kg K).]
![Page 62: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/62.jpg)
Answer: There are 2 processes involve when an ice is converted
into water at 25°C. Ice at 0°C ------> Water at 0°C -------> Water at 25°C Energy absorbed to convert 0.5kg from Ice at 0°C to
Water at 0°C Q1 = mL = (0.5)(334000) = 167000J Energy absorbed to convert 0.5kg from watyer at 0°C to
Water at 25°C Q2 = mcθ = (0.5)(4200)(25-0) = 52500J Total energy required = Q1 + Q2 = 167000 + 52500 =
219500J
![Page 63: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/63.jpg)
example How much energy is required to
change exactly 1 g of ice at -20oC to steam at 120 oC? [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 oC), Specific latent heat of vaporization of water = 2,260,000 J/kg, specific heat capacity of ice = 2100 J/(kg K)]
![Page 64: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/64.jpg)
Answer: The processes involved: Ice (-20°C) -----> Ice (0°C) -----> Water (0°C) -----> Water
(100°C)-----> Steam (100°C) -----> steam (120°C)
Energy required:Ice (-20°C) to Ice (0°C), Q1 = mcθ = (0.001)(2100)(20) = 42JIce (0°C) to Water (0°C), Q2 = mL = (0.001)(334000) = 334JWater (0°C) to Water (100°C), Q3 = mcθ = (0.001)(4200)(100) = 420JWater (100°C) to Steam (100°C), Q4 = mL = (0.001)(2260000) = 2260JSteam (100°C) to Steam (120°C), Q5 = mcθ = (0.001)(2020)(20) = 40.4J
Total energy required= Q1 + Q2 + Q3 + Q4 + Q5 = 42 + 334 + 420 + 2260 + 40.4 = 3096.4J
![Page 65: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/65.jpg)
How impurities affect the melting point
If impurities are present in a substance, its melting point will be lower than normal.
![Page 66: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/66.jpg)
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 -18°C.
![Page 67: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/67.jpg)
4.4 gas law
![Page 68: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/68.jpg)
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
![Page 69: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/69.jpg)
Formula of Boyle's Law
![Page 70: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/70.jpg)
example
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]
![Page 71: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/71.jpg)
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.
P1 = 76 + 2 = 78cmHgV1 = 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 = 74cmHgV2 = ?
By using Boyle's Law's formula P1V1 = P2V2
(78)(10) = (74)V2V2 = 10.54cm
The length of the trapped air in figure (b) is 10.54cm
![Page 72: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/72.jpg)
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.
![Page 73: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/73.jpg)
![Page 74: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/74.jpg)
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.
![Page 75: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/75.jpg)
Formula of Pressure Law
Notes: The temperature must be the absolute temperature (Kelvin Scale)
![Page 76: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/76.jpg)
example
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?
![Page 77: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/77.jpg)
Answer: At 30oC,
P1 = 200,000PaT1 = 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
![Page 78: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/78.jpg)
Graph of Pressure Law
Pressure is directly proportional to the temperature.
![Page 79: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/79.jpg)
Charle's Law
Charles’s 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
![Page 80: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/80.jpg)
Formula of Charle's Law
Notes: The temperature must be the absolute temperature (Kelvin Scale)
![Page 81: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/81.jpg)
example
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 57oC.
![Page 82: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/82.jpg)
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,
![Page 83: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/83.jpg)
Graph of Charle's Law
Volume of gas is directly proportional to the temperature of the gas
![Page 84: HEAT Form 4 Bab 4](https://reader035.vdocument.in/reader035/viewer/2022081420/5520bbaf497959892f8b4e7d/html5/thumbnails/84.jpg)
Formula of Universal Gas Law