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CHAPTER 6: TEMPERATURE AND HEAT

6.1HEAT AND TEMPERATURE

6.1.1 HEAT

Definition- heat is another name for ‘internal energy’ or energy that is stored inside a substance. It is a form of energy that moves from one place to another when its temperatures are different.

Heat is not the same as temperature. Temperature is a measure of how fast molecules are

moving. Heat is the movement energy of all molecules added together.

Heat measured in Joules (J)

6.1.2 TEMPERATURE

Definition – temperature is a measure of how fast molecules are moving. Temperature is measured in degrees on a temperature scale. The most common temperature

scales are Celcius (Centigrade), Fahrenheit and Kelvin.

On the Celcius scale, the freezing point of water is 00C and its boiling point is 1000C. On the

Fahrenheit scale, these two points are at 320F and 2120F.

BB 101 - ENGINEERING SCIENCE 158


6.2HEAT TRANSFER

Heat spreads in three ways:

i) Conduction Conduction is the spread of heat from atom to atom. The atoms in a hot substance

move fast and hit each other. These collisions pass energy on to nearby atoms, and those atoms pass their energy on to the next atoms, and so on.

If one end of a metal rod is at a higher temperature, then energy will be transferred down the rod toward the colder end because the higher speed particles will collide with the slower ones with a net transfer of energy to the slower ones.

Example: Heat transfer between two plane surfaces,

ii) Convection Convection spreads heat in gases ( and liquids).When they are heated their

molecules speed up, collide more often and spread out further apart. The gas (or liquid) becomes lighter than the surrounding, cooler gas (or liquid) and floats upwards.

Convection above a hot surface occurs because hot air expands, becomes less dense, and rises Hot water is likewise less dense than cold water and rises, causing convection currents which transport energy.

Examples: Rain process and boiling water

iii) Radiation Radiation spreads heat by means of invisible infrared (heat) rays. When electromagnetic waves come in contact with an object, the waves transfer the

heat to that object. Electromagnetic waves travel through empty space. The sun warms the earth

through the radiation of electromagnetic waves. Example: sun radiation and home heating lamp

6.3HEAT CAPACITY AND SPECIFIC HEAT CAPACITY BB 101 - ENGINEERING SCIENCE 159

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/temper.html#c1


6.3.1 HEAT CAPACITY, C

Definition – the amount of heat energy needed to increase the temperature of an object by 1 o C

Measured in Joule per Kelvin (J/K) or Joule per degrees Celsius (J/°C)

Formula:

6.3.2 SPECIFIC HEAT CAPACITY, c

Definition – the quantity of heat energy required to increase its temperature of 1 kg of the substance by 1 o C.

Measured in (J/kgK) or (J/kg°C) Formula:

Example 1:


Heat Capacity= Quantity Of HeatChange∈temperature

C=Qθ

C= Qt2−t1

@ Heat Capacity=mass x Specific Heat Capacity C=m x c

Specific Heat Capacity= Quantity of HeatMass xChange∈Temperature

c= Qmx θ

c= Qm x ¿¿


An object of mass 3.5 kg has absorbed 4500 J of thermal energy when it is heated from 40°C to 48 °C. Calculate:

a) Specific heat capacity, b) heat capacity

Solution :a) Specific heat capacity

m = 3. 5 kg

Q = 4500 J

t 1 = 40 ° Ct2 = 48 ° Cθ = 48 − 40 = 8 ° C

b) Heat capacity

Q = 4500 J

t 1 = 40 ° Ct2 = 48 ° Cθ = 48 − 40 = 8 ° C

OR

m = 3. 5 kg C = m x c

c = 160 .714 J /kg°C C= 3 . 5 x 160 .714 C= 562 .499 J /°C

Example 2:

The specific heat capacity of metal X is 1200 J/kg°C. What is the heat capacity of a piece of metal X of mass 300 g?

m = 300 g kg C = m x c = 0 .3 kg C= 0 . 3 x 1200c = 1200 J /kg° C C= 360 J /° C


c = Qm x θ

c = 4500

3 . 5 x 8 c = 160 .714 J /kg°C

C = Qθ

C = 4500

8 C = 562 .5 J /° C


Exercise 1:

What is the heat capacity for a 5 kg water? Assume the specific heat capacity of water is 4200 J/kg°C.

Exercise 2:

If 20.5 J of energy is needed to raise the temperature of 150 g of copper from 25 °C to 60 °C, find the specific heat capacity of copper?

Exercise 3:

The heat energy required to raise the temperature of 400 g of copper by 5 °C is X Joule. If the specific heat capacity of copper is 380 J/kg°C, calculate the value of X?



6.4QUANTITY OF HEAT, Q

Since heat is a form of energy, the units in which a quantity of heat is measured are fundamentally units of energy.

Formula:

Example 1:

The specific heat capacity of an object of mass 1.5 kg is 0.90 kJ/kgoC. Determine the heat energy of the object to raise the temperature by 1 oC.

Solution :

m = 1. 5 kg Heat energy, Q = m x c x θc = 0 . 9 kJ /kg°C = 1.5 x 0.9 x 1θ = 1 °C =1 .35 kJQ = ? =1350 J

Example 2:

How much heat is lost when a 640 g piece of copper cools from 26 oC to 15 oC? (The specific heat of copper is 385 J/Kg oC)

Solution :

m = 640 g kg

= 0 .64 kg

c = 385 J /kg ° C

t 1 = 26 °C , t2= 15 ° Cθ = 26 − 15 = 11 °C

Q = ?


Q=m x c xθ

Where is,Q = quantity of heat (J)m = mass (kg)c = specific heat capacity (J/kg°C)θ = change in temperature (°C)

Heat lost, Q = m x c xθ = 0 .64 x 385 x 11 = 2710.4 J


Example 3:

The heat energy required to raise the temperature of 0.85 kg iron by 20 °C is 2500 J. What is the specific heat capacity of iron?

Solution :

m = 0 .85 kgc = ?θ = 20 ° CQ = 2500 J

Exercise 1:

Water at 40 °C needs 20 kJ of energy to heat the water until 65 °C. Calculate the mass of water. (Specific heat capacity of water = 4200 J/kg-1C-1)

Exercise 2:

What would be the final temperature of a 300 g sample of copper with an initial temperature of 80 °C, after it loses 6000 J? (Specific heat capacity of copper = 390 J/Kg°C )


Heat required , Q = m x c x θ 2500= 0 . 85 x c x 20 2500= 17 x c

c = 147 .059 J /kg °C

mc ( temperature ) mc ( temperature )


Exercise 3:

The initial temperature of a glass block with a mass of 2 kg is 45 oC. If it received 20.5 kJ of heat energy, find the final temperature of glass block? Assume specific heat capacity for glass is 670 J kg-1C-1.

6.5THERMAL EQUILIBRIUM

6.5.1 HEAT ENERGY LOST OR GAINED THROUGH CHANGE IN TEMPERATURE

The terms ‘loss’of heat or ‘gain’ of heat is frequently used in connection with heat energy

transfer from and to a body.

This is expressed mathematically by the statement that the quantity of heat given out (‘lost’)

or received (‘gained’) is equal to the product of the mass of the substance, the specific heat

capacity of the substance, and its change in temperature.

Formula:


Hot water

m = 200 g kg = 0.2 kg

c = 4200 J/kg°C

t1 = 80 °Ct2 = ?θ = 80 – t2

Cold water

m = 100 g = 0.2 kg

c = 4200 J/kg°C

t1 = 20 °Ct2 = ?θ = t2 –20

Mixture

t2 = ?+ =


Example 1: 200 g of hot water at 80 °C is mixed with 100 g of cold water at 20 °C. What is the final temperature of the water?

Heat lost byhot water=Heat gained by cold water(m x cθ ) hot water=(m x c x θ ) cold water 0.2 x 4200 x (80−t2 )=0.1 x 4200 x ( t2−20 )

840 x ( 80−t 2 )=420 x (t 2−20 ) 67200−840t 2=420t 2−8400

67200+8400=4200 t 2+840t 2

75600=1260 t 2

756001260

=t2

t 2=60 °C


mixture

t2 = ?

copper

m = 1.5 kg

c = 0.378 kJ/kg°C = 1378 J/kg°C t1 = 150 °Ct2 = ?θ = 150 – t2

waterm = 0.85 kg

c = 4200 J/kg°C t1 = 25 °Ct2 = ?θ = t2 - 25

+ =


Example 2:

A piece of copper having a mass of 1.5 kg is heated to a temperature of 150oC and is then dropped into a Thermos flask containing 0.85 kg of water at an initial temperature of 25oC. What will be the final temperature of water? Assume specific heat capacity of copper is 0.378 kJ/kgoC and no heat is absorbed by Thermos flask.

Solution


Heat lost by copper = Heat gained by water(m x c x θ ) copper = (m x c x θ ) water

1 .5 x 1378 x (150 − t2) = 0 .85 x 4200 x ( t2 − 25 )

2067 x (150 − t 2) = 3570 x ( t 2 − 25 )310050 − 2067 t 2 = 3570 t 2 − 89250

310050 + 89250 = 3570 t 2 + 2067 t2

399300 = 5637 t2

3993005637 = t 2

t 2 = 70 . 836 °C


Example 3:

An 18 kg piece of metal which has a specific heat capacity of 500 J/kg°C is immersed in oil. The initial temperature of metal is 150 °C and the initial temperature of oil is 18 °C. Calculate the mass of oil if the final temperature of the mixture is 40 °C. Assume specific heat capacity of oil is 1580 J/kgoC.

metal + oil = mixturem = 18 kg m = ? t2 = 40 ° Cc = 500 J /kg ° C c = 1580 J /kg ° C

t2 = 40 ° C t 2 = 40 ° Cθ = 150 − 40 θ = 40 − 18 = 110 °C = 22 ° C

Exercise 1:

A 130 g sample of zinc at 75 °C is placed in 105 g of water at 20 °C and the final temperature of the mixture is 35 °C. What is the specific heat capacity of zinc?(specific heat capacity of water = 4200 J/kg°C )


Heat lost by metal = Heat gained by oil(m x c x θ ) metal = (m x c x θ ) oil18 x 500 x 110 = m x 1580 x 22

990000 = m x 3476099000034760 = m

m = 28 . 481 kg

Ct 1501


Exercise 2:

A 320 g of water at 15 oC is poured into a Thermos flask containing 600 g of water at 75 oC. Assuming no heat loss to the surroundings, what is the final temperature of the mixture? [specific capacity of water = 4200 J/kg oC ]

Exercise 3:

A copper sphere of mass 5 kg is heated for a several minutes before being immersed into 2 kg of water at 30 oC. The final temperature of the water and the copper sphere is 50 oC. Calculate the temperature of the copper before it was dropped into the water. ( specific capacity of copper = 400 J/kg oC and specific capacity of water = 4200 J/kg oC)


=+

m = 16 g kg = 0.016 kg

c = ?

t1 = 100 oCt2 = 39 oCθ = 100 – 39 = 61 oC

Wooden stick

The mixture

t2 = 39 oC

m = 70 g kg = 0.07 kg

c = 400 J/kg oC

t1 = 28 oCt2 = 39 oCθ = 39 – 28 = 11 oC

calorimeter Water&

m = 30 g kg = 0.03 kg

c = 4200 J/kog oC

t1 = 28 oCt2 = 39 oCθ = 39 - 28 = 11 oC


6.6.2 MORE THAN 2 SUBSTANCES

Formula :

Example 1:

A metal spoon of mass 16 g is heated in boiling water and subsequently placed in a 70 g copper calorimeter at a temperature of 28 oC. The calorimeter contains 30 g of water. Upon stirring, the temperature of the mixture is 39 oC. Determine the specific heat capacity of metal spoon. Specific heat of water and copper are 4200 J/kg oC and 400 J/kg oC respectively.

Solution:


Qlost by object = Q gained by object + Q gained by object

( m c θ ) lost = ( m c θ )gained + ( m c θ )gained

Qlost by object = Q gained by object

( m c θ ) stick = [ ( m c θ )calorimeter ] + [ ( m c θ )water ]

0.016 x 61 x c = [ 0.07 x 400 x 11 ] + [ 0.03 x 4200 x 11]

0.976 c = 308 + 1386

c =

16940 .976

c = 1736 J/kg °C

=+

m = 0.4 kg

c = 4200 J/kg oC

t1 = 15 oCt2 = ? oCθ = t2 – 15

m = 0.5 kg

c = 390 J/kg oC

t1 = 15 oCt2 = ? oCθ = t2 – 15

Container Water 1 Water 2&

The mixture

m = 0.3 kg

c = 4200 J/kog oC

t1 = 60 oCt2 = ? oCθ = 60 – t2

t2 = ? oC


Example 2:

A 0.5 kg container is filled with 0.4 kg water and its temperature is 15°C. If 0.3 kg water at 60°C is added into the container, what is the final temperature of the mixture? Assume Specific Heat Capacity of water and container are 4200 J/kg°C and 390 J/kg°C respectively.

Solution:


Q lost by the object = Q gain by the object

(m x c x θ ) Water 2 = [ (m x c x θ ) container ] + [ (m x c θ ) Water 1 ]

[0 . 3 x 4200 x (60 − t2 )] = [0 . 5 x 390 x (t2 − 15) ] + [ 0. 4 x 4200 x ( t2 − 15 )]

1260 x (60 − t2 ) = [195 ( t2 − 15 ) ] + [1680 ( t2 − 15 ) ]

75600 − 1260 t2 = (195 t2 − 2925 ) + (1680 t 2 − 25200 )

75600 + 2925 + 25200 = 195 t 2 + 1680 t 2 + 1260 t2

103725 = 3135 t2

1037253135 = t 2

t 2 = 33 . 086 °C


Exercise 1:

A 0.1 kg block of iron which is 100 °C is transferred immediately into a cuprum calorimeter contain of 0.12 kg of water at 14 °C. Given that the mass and the specific heat capacity of cuprum calorimeter is 0.05 kg and 400 Jkg-1C-1 respectively. If the temperature of calorimeter and water have increased to 21 °C, determine the specific heat capacity of iron? ( Specific heat capacity of the water is 4200 J kg-1C-1)



Exercise 2:

A piece of metal which mass is 0.25 kg is heated to 100°C. Then, the metal is dipped into a calorimeter filled with water at 30°C. If the final temperature of the mixture are 75°C, calculate the specific heat capacity of the metal. Given:

mass of water = = 0.4 kg mass of calorimeter = 0.3 kg Specific Heat Capacity of water = 4200 J/kg°C Specific Heat Capacity of calorimeter = 400 J/kg°C



Exercise 3:

A piece of metal bar which mass is 200 g is heated until its temperature reached 100oC and subsequently placed in a 500 g calorimeter filled with 300 g of water at 28 oC. Assuming specific heat of water, calorimeter and metal as 4200 J/kg°C, 400 J/kg°C and 390 J/kg°C respectively, find the final temperature of the mixture?



SELF ASSESSMENTSELF ASSESSMENT

Question 1Define temperature and heat.

Question 2Give the definition and formula for heat capacity and specific heat capacity.

Question 3Explain three methods of Heat Transfer.

Question 4How much heat must be supplied to a 1.0 kg iron pan to raise its temperature from 20 oC to 100 oC. Given the specific heat capacity of iron is 450 J/kg oC.

Question 5The heat energy required to raise the temperature of 400 g of copper by 5 oC is 760 J. What is the specific heat capacity of copper?

Question 6A 0.25 kg sample of water with an initial temperature of 98.8 oC loses 7500 J of heat. What is the final temperature of the water? ( Specific heat capacity of water = 4200 J/kg oC)

Question 7A polystyrene cup is filled with 200 g of water at a temperature of 50 oC. A 10 g of lead block at 100 oC is placed in the water. What is the temperature of the water when thermal equilibrium is reached?

Specific heat capacity of water = 4200 Jkg-1 oC-1

Specific heat capacity of lead = 130 Jkg-1 oC-1

Question 8In an experiment, 50 g of water at 0 oC is mixed with 50 g water at 20 oC. The final temperature of the mixture is ……………?

Question 9A container which mass is 0.4 kg is filled with 0.2 kg cold water and their temperature are 10 oC. If 0.3 kg of hot water at 80 oC is added into the container, what is the final temperature of the mixture? Given the specific heat capacity of the container and water are 390 J/kg oC and 4200 J/kg oC respectively.

Question 10A block of metal which mass is 0.5 kg is put into a 0.4 kg calorimeter filled with 0.3 kg water. The initial temperature of the metal is 90 oC and the initial temperature of



the calorimeter and water are 10 oC. If the final temperature of the mixture are 60 oC, determine the specific heat capacity of the metal?

Specific heat capacity of calorimeter = 400 J/kg oCSpecific heat capacity of water = 4200 J/kg oC

Question 11The quantity of heat released from 1 kg of liquid is 1.68 x 105 J when its temperature changes by 40 °C. Calculate specific heat capacity of the liquid.

Question 12The water in the ice maker of arefrigerator has a mass of 0.4 kg and temperature of water of 22 C. What is the temperature of the water after 33600 J of heat has been removed from it?

Question 13500 g of liquid A at the temperature of 70 °C is mixed with the liquid B with mass, M at the temperature 15 °C. If the final temperature of the mixture is 45 °C, ( specific heat capacity of liquid A = 850 J/kg°C and specific heat capacity of liquid B = 350 J/kg°C)

i) calculate the heat released by liquid Aii) determine the mass, m of liquid B

Question 14Calculate the heat quantity needed to raise the temperature of 8 kg water by 40°C. Specific heat capacity of water is 4.2 kJ/kg°C.

Question 15Calculate the specific heat capacity of a metal if 880 kJ of heat quantity is needed to raise 2.5 kg of the metal by 250 °C.


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