HEAT

Chapter 4

4.1 Thermal Equilibrium

Thermal Energy

Thermal energy is a measure of the total kinetic and potential energy in all the molecules and atoms in a certain substance.

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.

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.

Comparing Thermal Energy and Temperature

Application of Thermal Equilibrium--Thermometer

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

Application of Thermal Equilibrium- Refrigerator

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.

Application of Thermal Equilibrium - Oven

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.

Calibration of thermometer

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.

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.

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:

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

T = 40 - 10 130 – 10 30 20 25oC

X 100 oC

X 100 oC

Answer:l0 = 10mml100 = 130mmlθ = 40mm

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.

Accuracy of Thermometer

To increase the accuracy of a thermometer, the diameter of the capillary tube of the thermometer must be constant throughout.

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

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

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

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.

Absolute Temperature

Absolute temperature is the temperature measured in Kelvin scale, which it is a temperature reading made relative to 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.

4.2 specific 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).

Specific Heat Capacity

The amount of heat required to change the temperature of 1 kg of a substance by 1oC.

Formula of Specific Heat Capacity

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.

Conversion Of Electrical Energy Into Thermal Energy

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 = ?

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.)

Conversion of Gravitational Energy into Thermal Energy

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 = ?

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

Conversion Of Kinetic Energy Into Thermal Energy

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

Mixing 2 Liquid

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

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

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.

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.

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.

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

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

4.3 specific latent heat

Heating Curve State of matter:A-B: SolidB-C: Solid and LiquidC-D: LiquidD-E: Liquid and GasE-F: Gas

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.

Heat obsorbs during S-T is called the latent heat of vaporisation.

Cooling CurveStates of matter:P-Q: GasQ-R: Gas and LiquidR-S: LiquidS-T: Liquid and SolidT-U: SolidU-V: Solid

Cooling Curve - Latent Heat

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.

Latent Heat

Latent heat is the heat absorbed or releases during a change of state of matter

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

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.

Formula of Specific Latent Heat

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).]

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

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)]

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

How impurities affect the melting point

If impurities are present in a substance, its melting point will be lower than normal.

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.

4.4 gas 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

Formula of Boyle's Law

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]

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

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.

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.

Formula of Pressure Law

Notes: The temperature must be the absolute temperature (Kelvin Scale)

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?

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

Graph of Pressure Law

Pressure is directly proportional to the temperature.

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

Formula of Charle's Law

Notes: The temperature must be the absolute temperature (Kelvin Scale)

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.

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,

Graph of Charle's Law

Volume of gas is directly proportional to the temperature of the gas

Formula of Universal Gas Law