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Tyas Ajeng Puspitasari, S.Pd

Junior High School 1 Malang

Grade VII Semester 2

2010/2011

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Standard competency and

Basic Competency

Physics

Grade/ Semester : VII/2

Standard Competency : Comprehending states of matter

and its change

Basic Competency : Describe role of heat in changes

states of matter and

temperatures substance also its

application in daily life

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Content

What is HEAT ?

What is HEAT ?

Heat and the change of temperature

Heat and the change of temperature

Heat and the change of matter state

Heat and the change of matter state

The Blacks Principle

The Blacks Principle

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Are heat energy and temperature just the same thing?

No! Let's get this clear:

Heat is the energy stored inside something.

Temperature is a measurement of how hot or cold

something is.

An object's temperature doesn't tell us how much heat

energy it has.

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Before the 9th century, many scientist

believed that heat was a fluid.

Heat as fluid firstly stated by Antoine

Laurent Lavoisier, He said that when a

hotter object was in contact with the

colder object, then the fluid would flow

Some scientist disagreed with the concept

of Heat as fluid. Those scientist proved that

Heat was not a fluid, but a form of Energy Antoine Lavoiser

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What is heat?

Heat energy is the total energy of particles composing amatter.

Heat is one form of energy which is flowing from an object with

higher temperature to another object with lower temperature

The unit of Heat is Joule (J)Heat is also expressed in units of calories.

One calorie defined as the amount of heat needed to heat 1

gram of water until its temperature increasing 1 C .1 calorie = 4.2 joules

1 joule = 0.24 calories

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Heat and the change

of temperatureNot all substances have the same ability in absorbing

heat.

The ability to absorb heat determined by the nature ofa substance called specific het capacity.

Specific Heat Capacity (c)The specific heat capacity of a substance is

defined as the heat required to raising the

temperature of 1 kg of the substance by 1C.

The unit of specific heat capacity is joule per kilogramper Celsius degree (J/kg C)

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This proved that heat needed to raise the

temperature of 1C alcohol is smaller

than the heat needed to raise water

temperature at 1C. That is, alcohol heatfaster than water.

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The amount of heat (Q) required by an object is proportional to

the mass of the object (m), depending on the specific heat capacity

(c), and comparable to the increase in temperature(T).

The change of heat (received heat or released heat) of a

substance cannot be measured directly, but it can be calculated

using the following equation.

Q = m x c x T

Notes:

Q = heat received or released (Joules)

m = mass (kg)

c = specific heat capacity (J/kg C)T = changes of temperature (C)

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Example

Calculate heat needed to increasing temperature of 500 g

water from 20 C to 100 C,if the specific heat of water is

4200 J/ kg C?

Given:

m= 500 g =0.5kg c = 4200 J/kg CT = 100C - 20C = 80 C

Question: Q = ?

Answer:Q = m x c x T

= 0.5 kg x 4200 J/kgC x 80 C

= 168,000 JSo the heat received by the water is 168,000 J

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Problems of Heat and the

change of temperatureAnswer these following questions!

1. Two containers of the same size are filled with water and sand

respectively. If they are exposed to the sun heat, which one getshotter more quickly? Why?

2. Youre heating a 3.0 kg glass block, specific heat capacity of 840J/(kg.C), raising its temperature by 60C. What heat do you have

to apply?3. 2 kg of iron is heated from 15C to 30C. If the heat required is

13,500 J, what is specific heat capacity of iron?

4. 42 kilojoules of heat released from the 2 kg of ice at temperatures -15 C. What is the temperature finally, if the specific heat of ice

is 2100 J/kg. C

5. To raise the temperature of an object from 10C to 40C, it

required heat 60,000 J. Calculate the heat capacity of its object!

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Heat and the change

of matter state When a matter change its state, the

temperature does not increase although the

heat is continuously given. The heat is used to changing the state of matter.

The heat used for changing the state of matter is

called Latent heat.

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Melting and FreezingMelting and FreezingMelting and FreezingMelting and Freezing Melting point is the temperature

when a matter is starting to melt.

Example: melting point of ice is 0C

Freezing point is the temperature

when a matter is starting to freeze.

Example: freezing point of water is

0C

Melting point = Freezing point

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Meltingpoint

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15Melting heat and Freezing heatThe amount of heat absorbed by every 1 kg of matter formelting at its melting point is called Melting heat(L).

The amount of heat released by 1 kg of matter for freezing atits freezing point is called

Freezing heat(L).

SI Unit of Melting Heat and Freezing Heat: J/kg

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AMOUNT OF HEAT(Q) RECEIVED OR RELEASED BYAMATTER

WHEN IT MELTS OR FREEZES

Where:

Q = amount of Heat received or released (J)

m = mass of matter (kg)

L = Melting heat or Freezing Heat (J/kg)

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boiling and condensation Point

Boiling point is the temperature when a matter is

starting to boils.

Example: boiling point of water is 100CCondensation point is the temperature when a

matter is starting to condenses.

Boling point = Condensation point

Boilingpoint

Boiling point of a liquid is affected by air

pressure

The greater the air pressure the greater theboiling point of liquid

On the sea level, at air pressure of 1 atm, the

boiling point of water is 100 C.

At higher place, the air pressure is decreaseand boiling point of water is also decrease.

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Vaporing heat and condensation heat

The amount of heat absorbed by every 1 kg of matter forvaporize at its boiling point is called vapouring heat (U).

The amount of heat released by 1 kg of matter for condense

at its condensation point is calledcondensation heat

(U).

SI Unit of Vapouring Heat of Condensation Heat: J/kg

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AMOUNT OF HEAT(Q) RECEIVED OR RELEASED BYAMATTER

WHEN IT EVAPORATES OR CONDENSES

Where:

Q = amount of Heat received or released (J)

m = mass of matter (kg)

L = Vaporing Heat or Condensation Heat (J/kg)

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Substance Melting

point (C)

Melting

Heat (J/kg)

Boiling point

(C)

Vapouring Heat

(J/kg)

Helium -269,65 5,23 x 103 -268,93 2,09 x 104

Nitrogen -209,97 2,25 x 104 -195,81 2,01 x 105

Oxygen -218,79 1,38 x 104 -182,97 2,13 x 105

Ethyl alcohol -114 1,04 x 105 78 8,54 x 105

Water 0 3,33 x 105 100 2,26 x 106

Sulfur 119 3,81x 104 444,6 3,26 x 105

Lead 327,3 2,45 x 104

1750 8,70 x 105

Aluminum 660 3,97 x 105 2450 1,14 x 107

Silver 960,8 8,82 x 104 2193 2,33 x 106

Gold 1063 1,34 x 105 2660 1,58 x 106

Copper 1083 5,23 x 103 1187 5,06 x 106

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Heating

When we heated liquid, the molecules near the surface vibrate

faster, which enable them to release from liquids surface

Blowing air through the liquids surface

If we blow air through the surface of a hot water, the air near the

water surface would carry the waters molecules at the surface.

Extending the surface

To make hot tea in a cup cold faster, you should pour on a saucer,

because the surface area of the saucer is larger than on the cup,

so the molecules have more chances to leave the water surface Decreasing the pressure on the surface

If the pressure decreasing, the space between air molecules near

the surface become looser and it make easier for the molecules ofwater to fill the empty space among the molecules of air.

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1. If ice melting, the state change from . to 2. For melting, ice . heat energy

3. During ice melted, temperature is.. and it iscalled

4. Amount of heat energy used to melt you cancalculate with the formula

5. During the water boil, the temperature is andit is called

6. During the water boil, the state change from become and is called

7. During evaporation the water heat energy.8. Quantity heat energy used to evaporation can

calculate by the formula

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Example24

1. Calculate the quantity of heat required to melt 3 kg of iceon 0C ! (L = 3.36 x 105 J/kg)

Answer:

Q = 3 kg x 336,000 J/kg

Q = 1,008,000 J = 1,008 kJ

Known: m= 3 kgL = 3.36 x 105 J/kg = 336,000 J/kg

Question: Q ?

Q = m x L

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25Example

2. Calculate the quantity of heat required to change 500 gramsof water at a temperature of 100C into water vapor at atemperature of 100C! (U = 2270 kJ/kg)

Answer:

Q = m x U

Q = 0.5 kg x 2,270 kJ/kg

Q = 1,135 kJ

Known: m= 500 grams = 0.5 kgU = 2270 kJ/kg

Question: Q ?

Q = 1,135,000 J

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26Example

3. Calculate the heat required to change 1 kg of water attemperature of 80C into 1 kg of water vapor at atemperature of 100C (specific heat of water is 4200 J/kgC

and vapouring heat of water is 2270 kJ/kg)

Answer:

Known: m= 1 kgT = (100-80) C = 20 Cc = 4200 J/kg C

U = 2270 kJ/kg = 2,270,000 J/kgQuestion: Q total

100C

80C

Temperature

Heat (J)

Q2 = m x UQ

1= m x c x T

Q1

Q2

Q total = Q 1 + Q 2=(m x c x T) + (m x U)=(1 kg x 4200 J/kg C x 20 C)+(1 kg x 2,270,000J/kg)

= 84,000 J + 2,270,000 J = 2,354,000 J

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Example

=(100 g x 80 cal/g ) + (100 g x 1 cal/g C x 20 C )

4. Ice cube has mass 100 gram at 0C. If it heated until 20Cthe all ice become water. How much heat energy absorb?(melting heat of ice = 80 cal/g ; specific heat of water 1 cal/g c)

Answer:

Known: m= 100 gT = (20-0) C = 20 CL = 80 cal/g

c = 1 cal/gCQuestion: Q total

Q2 = m x c x TQ1 = m x L

= (m x L) + (m x c x T)

= 8,000 cal + 2000 cal= 10,000 cal

Q total = Q 1 + Q 2

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Exercise of Heat

1. Calculate heat needed to change 100 gram of ice at 30C into

water vapor at 120C.

cice = 2100 J/kgCcwater = 4200 J/kgC

cwater vapor = 2010 J/kgC

Lice = 336,000 J/kg

Uwater = 2,270,000 J/kg

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Joseph Black (16 April 1728 6

December 1799) was a Scottishphysician, known for his

discoveries of latent heat, specific

heat, and carbon dioxide.

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When two substances or more are mixed, the

amount of heat released by the substance

with higher temperature is equal to the

amount of heat absorbed by the substance

with lower temperature.

Q released = Q received

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Qreleased= Q receivedm1. cwater . T1 = m2. cwater . T2

m1. cwater . (T1 Tf) = m2. cwater . (Tf T2)

m1

. (T1

Tf

) = m2

. (Tf

T2

)

100 g(40 Tf) = 200 g. (Tf 10)

4000 100Tf= 200 Tf 2000

4000+2000 = 200 Tf+ 100 Tf

6000 = 300 TfT = 6000/300 = 20C

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100 grams of hot water at 40C is mixed with 200 grams of water at

10C . What is the final temperature of this mixture?

Given: m1 = 100 grams T1 = 40Cm2 = 200 grams T2 = 10C

Question: Tf(Final Temperature)

Solution

Example

Tf

T1

T2

T1 = T1 - Tf

T2 = Tf T2

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1. The volume of water in glass B is half of the volume of

water in glass A. The water is then mixed in glass C. The

final temperature of the water in glass C is

2. kg of hot water at 80C is mixed with 2 kg of cold

water. What is the temperature of cold water if the final

temperature of mixture is 24C?

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THANKS!