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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.

Top | Next

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.

Previous | Top | Next

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.


Card 4: What is heat?

What is heat?

Heat is the flow of thermal energy.


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


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.


Card 7: Application of thermal equilibrium - Refrigerator

Application of Thermal Equilibrium- Refrigerator
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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.


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.


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:


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 100ô C \hfill \\

T = \frac{{40 - 10}}

{{130 - 10}} \times 100ô C \hfill \\

T = \frac{{30}}

{{120}} \times 100ô C \hfill \\

T = 25ô 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.


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


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


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.


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.


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.


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.

Previous | Top

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).

Top | Next
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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.


Card 3: Formula of Specific Heat Capacity

Formula of Specific Heat Capacity


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.

Top | Next

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.


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.


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.


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
http://www.myhometuition.com/lessons/105

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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

Top | Next

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.


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


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.


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.


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.


Card 8: Formula of Specific Latent Heat

Formula o Specific Latent Heat


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.


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.

Top | Next
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Card 2: Formula of Boyle's Law

Formula of Boyle's Law


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.


Card 5: Formula of Pressure Law

Formula of Pressure Law

Notes: The temperature must be the absolute temperature (Kelvin

Scale)


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.


Card 8: Formula of Charle's Law

Formula of Charle's Law
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Notes: The temperature must be the absolute temperature (KelvinScale)


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 }}
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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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Card 10: Universal Gas Law - Formula

Formula of Universal Gas Law
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