Physics - Ch6 Temperature and Heat
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DAS 12603 TECHNICAL SCIENCE I
CHAPTER 6: TEMPERATURE AND HEAT NORAIHAN BINTI SALLEH HUDIN DEPT OF SCIENCE AND MA MATHEMA THEMATICS TICS CENTRE FOR DIPLOMA STUDIES, UTHM
LEARNING OBJECTIVES The objectives of this chapter are: i.
ii.
To impart students with the basic knowledge in science especially heat and thermal properties of matter. To apply the concept of heat and thermal properties in science and engineering courses.
LEARNING OUTCOMES After completing this chapter, chapter, students should be able to: explain the definition of temperature and heat. describe the thermal equilibrium and the zeroth law of thermodynamics.
convert the temperature unit scales. explain the specific heat (sensible heat) and latent heat. calculate the problem of heat by using the calorimetric principle. differentiate latent heat of fusion and latent heat of vaporization. solve problem of heat when a matter change from one phase to another phase. List down three types of mechanisms of heat transfer.
3.1 TEMPERA TEMPERATURE TURE
Definition: the measure of the degree of hotness and coldness of matter whether in solids, liquids or gases.
When two or more objects are in contact with each other and their temperature is equal, they are said to be in thermal equilibrium.
3.1.1 THERMAL EQUILIBRIUM
When matters in thermal contact to each other, other the to energytwo (heat) from are matter with higher temperature will ,flow the matter with lower temperature. As energy continues to flow, flow, temperature of the two matters matters will approach each other. other.
At equal temperature, there are no longer net flow of energy between them. The two matters are said to be in thermal equilibrium. thermal equilibrium
hotter
cooler
hotter
cooler
3.1.2 THE ZEROTH LAW OF THERMODYNAMICS
States that: “If two objects are in thermal equilibrium equilibrium with a third object, then the two objects are in thermal equilibrium with each other.” C
C
A
B
A in thermal equilibrium with C
C in thermal equilibrium with B
A
B
Therefore, A in thermal equilibrium with B
T h er er m o m e t r y p r o p e r t y of a material: property of a material that is of
temperature-dependent (change linearly with the changing of temperature-dependent temperature) Thermometer
Thermometry properties
Mercury thermometer
The change in expansion of mercury in glass column with temperature
Ideal gas thermometer
The change in glass pressure with temperature at constant volume OR the change c hange in gas volume with temperature at constant gas pressure
Thermocouple
The change in electric motion force (emf) through thin wires of different metals (welded together at the ends to form two junctions) with temperature
Electric resistance thermometer (thermistor)
Electrical resistant changes with temperature
Thermogram
The intensity of the radiation emitted by an object (infrared radiation) increases substantially as the temperature is raised
Types Ty pes of thermometer with their thermometry properties
3.1.3 TEMPERATURE SCALES Temperature scales were built to determine the temperature of an object quantitatively. To built a temperature scale, several standard fixed points were used as calibration points, which are:
Ice point: the point where an ice (solid) and water (liquid) are in equilibrium phase (atmospheric pressure, p = 1 atm) Steam point: the point where the water (liquid) and steam (gases) (gases) are in equilibrium phase ((atmospheric atmospheric pressure, p = 1 atm)
There are three temperature scales:
Celsius (°C) Fahrenheit (°F) Kelvin (K)
Temperature scale
Symbol
Ice Point
Steam Point
Difference
Celsius
°C
0
100
100 °C
Fahrenheit
°F
32
212
180 °F
Kelvin
K
273.15
373.15
100 K
Ice point and steam point of temperature scales
The Kelvin temperature scale has been chosen as the international standard temperature scale.
Thus, Kelvin (K) is the SI unit for temperature.
The temperature 0 K represents the absolute zero of temperature (there are no temperature below belo w 0 K)
Therefore, there are no negative temperature number in Kelvin scale.
The three temperature scales can be related by the following equations:
Celsius to Fahrenheit:
T F 1.8T C 32
Fahrenheit to Celsius:
T C T F 32 1.8
Celsius to Kelvin:
T K T C 273.15
Example:
A friend suffering suffering from flu has a slight fever fever.. His body temperature temperature is 37.20°C. What is his temperature in i)
Kelvin
ii)
degree Fahrenheit, °F.
Solution: i)
Celsius to Kelvin conversion
T K T C 273.15
37.2 273.15 310.35 K i)
Celsius to Fahrenheit conversion
T F 1.8T C 32
1.8 1. 8(3 (37. 7.20 20)) 32 98.96 F
3.1.4 DETERMINING TEMPERATURE BY USING THERMOMETRY PROPERTIES
Thermometry property of a material can be exploited to determine its temperature.
Firstly, the thermometer scale must be calibrated with the melting point of 0°C and boiling point of 100°C of water at atmospheric pressure of 1 atm.
The value of parameter X 0 and X 100 are assigned to the melting and boiling point respectively.
Plot a graph of a straight line between these two points.
Temperature, T 100°C
T°C 0°C
X 0
X T
X 100
Thermometry property
Any unknown unknown temperature T with with their corresponding corresponding parameter X T that lies along the straight line can be determined by the graph or by using the equation:
T X T X 0 100C X 100 X 0
Example: A resistance o off thermometer giv gives es a reading of 3 30.00 0.00Ω Ω at melting point, 41.58Ω 41.58Ω at boiling point and 34.59Ω 34.59Ω when immersed into a beaker filled with oil. Calculate the oil’s temperature. Solution: X 0 30.00 X 100 41.58 X T 34.59
X T X 0 T 100C X 100 X 0 34..59 30 30..00 34 41 41.5 .58 8 30 30.0 .00 0 100C 39.64C
3.2 HEA HE AT AND INTERNAL ENERGY ENERGY
Heat: an energy that flows from a higher-temperature object to a lower-temperature object because of the difference differenc e in temperature.
SI unit for heat: Joule (J) Internal energy (thermal energy): the sum of the molecular kinetic energy, molecular potential energy and other kinds of molecular energy in a substance.
Molecular kinetic energy: originates from the random motion (or vibration) of the molecules. Molecular potential energy: due to forces that acts between the atom of a molecule and between molecules.
HEAT 3.3 HEA T & TEMPERA TEMPERATURE TURE CHANGE CHANGE .
3.3.1 HEAT CAPACITY CAPACITY
Heat capacity, capacity, C : the amount of heat required to raise the temperature of a substance by 1 unit (JK-1)
The equation of heat capacity which relates heat, Q and the temperature change, Δ change, ΔT is is given by:
C
Q
T
Specific heat, c : heat capacity per unit mass (Jkg-1K-1)
c
Q mT
Substance
Specific Heat, c (Jkg-1 °C-1)
Aluminium
910
Brass
380
Copper
390
Lead
130
Mercury
140
Iron
470
Steel
450
Silver
270
Glass
700
Ice Water
2100 4200
Steam
2000
Glycerine
2500
Alcohol 2430 Specific heat, c of of some common substances
Example: How much heat is required to raise the temperature of mercury = 140 200g of mercury from 20°C to 100°C? Given c Jkg-1 °C-1.
Solution: Q mcT l mc T final Tinitial initia (0.2)((140 140)( )(100 100 20) (0.2)
2240 J
Example: How much heat is required to raise the temperature of mercury = 140 200g of mercury from 250K to 330K? Given c Jkg-1 °C-1.
Solution:
Example: A 1.5 kg materia materiall requires 20.25 kJ heat to raise its temperature from 30°C to 80°C. State the name of the material. Solution: Q mcT 20250 J (1.5kg )(C )(80 30C )
c 270 Jkg 1 C 1
Substance
Specific Heat, c (Jkg-1 °C-1)
Aluminium
910
Brass
380
Copper
390
Lead
130
Mercury
140
Iron
470
Steel
450
Silver
270
Glass
700
Ice Water
2100 4200
Steam
2000
Glycerine
2500
Alcohol
2430
Specific heat, c of of some common substances
Example: A 1.5 kg materia materiall requires 20.25 kJ heat to raise its temperature from 30°C to 80°C. State the name of the material. Solution: Q mcT 20250 J (1.5kg )(C )(80 30C )
C 270 Jkg 1 C 1
Therefore the material is silver.
Principle of Conservation Energy: the heat the hotter object must equal toofthe heat gained bylost theby cooler object.
Heat lost = Heat gained
.
This principle is based on two assumptions:
No thermal energy lost from the system
No external energy comes into the system
Example: An aluminium cylinder is heated to 98°C and then dropped into a glass filled with 120g of water with initial temperature 20°C. The equilibrium temperature of the mixture is 31°C. Calculate the mass of the aluminium cylinder? Neglect the glass heat capacity.
Solution: heat lost by aluminium = heat gained by water
m Al c Al TAl mwcw Tw m Al (9 (910 10)( )(31 31 98) (0.1 (0.12) 2)(4 (420 200) 0)(3 (31 1 20) 20) 60970m Al 5544 J m Al 0.091kg
What if the glass is not negligible?
To be continued.
3.4 HEA CHANGE HE AT AND PHASE CHANGE
There forms of material: solid, liquid or gases.
Solid: atoms or molecules are strongly binding each other in a rigid crystalline structure by their mutual attraction.
Liquid: molecules have more energy, thus more free to move.
Gases: molecules molecules have even more energy and are free of one another.
Heating a material: temperature of the material will gradually increase until it reaches a certain stage at which the phase of the material changes:
Melting point: phase change from solid to liquid
Boiling point: phase change from liquid to gases
Cooling a material: temperature of the material will gradually decrease until it reaches a certain stage at which the phase of the material changes:
Condensing point: phase change from gases to liquid liq uid
Freezing point: phase change from liquid to solid
At melting/freezing melting/freezing point and boiling/cond boiling/condensing ensing point, the temperature of the phase does not change even if the heat is supplied continuously.
This is because the energy absorbed by the material is used to separate the molecules further apart than its previous phase.
Latent heat: the amount of heat, Q required to change the phase of a unit mass of a substance
Latent heat of fusion, Lf : heat per unit mass for the solidliquid phase transition (in either direction) Latent heat of vaporization, Lv: heat per unit mass for the liquid-gases phase transition (in either direction)
L
SI unit for latent heat: Jkg-1
Q m
Substance
Melting
Latent heat of
Boiling
Latent Heat of
Point (°C)
fusion,-1Lf (kJkg )
Point (°C)
Vaporization, Lv (kJkg-1)
Aluminium
660
377
2467
11390
Copper
1083
134
2595
5075
Lead
327
25
1620
871
Mercury
-39
11.7
357
293
Silver
961
88
2193
2340
Water
0
333
100
2264
Oxygen
-219
13.8
-183
214
Nitrogen
-210
25.5
-196
200
Table of melting and boiling points of few common substance with their latent heat of fusion f usion and vaporization
Temperature, T
100 °C
Water + steam 0 °C
-20 °C
Ice + Ice only water
Steam only
Water only
Quantity heat, Q of
Example: Calculate the energy required to change 100g of ice at -15°C to steam at 110°C. Explanation: Calculate the energy according to this order: • To increase temperature from -15°C to 0°C. • To melt the ice at 0°C. • To increase temperature from 0°C to 100°C. • To vaporize the water into steam at 100°C. • To raise the temperature from 100°C to 110°C. Hint • To calculate energy required to increase temperature, use specific heat equation, Q = mc ΔT. •
To calculate energy require to change phase, use latent heat Q equation,
L
m
3.5 HEAT TRANSFER
Three principle methods by which heat transfer can occur:
Conduction Convection Radiation
3.5.1 CONDUCTION
Definition: Conduction Conduction is the process of heat flow that occurs in all forms of material whether in solids, liquids or gases.
Dominantly occurs in solids. When cylindrical bar is heated at one end, heat will flow from the hot end to the cold end by means of the atomic/molecular collision.
Atoms/molecules that are vibrating Atoms/molecules vibrating at the hi higher gher temperature end collide with atoms/molecules atoms/molecules at the lower temperature end resulting in a net transfer of heat.
In metal, conduction process is aided by the motion of free electrons within the substance. Thus, Thus, metals are good heat conductor. Liquid conduct less heat than solids because the forces between the atoms are weaker.
Gases are even less efficient as heat conductor because the atoms of gases are further apart.
Hence, in order from the most conductive conductive to the least conductive:
Solids liquids gases
Direction of heat flow T hot
A
T cold d
A cylindrical bar with length d and and uniform cross-section A is maintained at temperature T hot and T cold at its both end. Evidences from experiments show that the quantity of heat, Q, transferred per unit time, t , is:
Directly proportional to the temperature difference between the two ends; ΔT= T hot - T cold ) ( Δ
Directly proportional to the cross-section area, A. Inversely proportional to the length, d .
The above statements can be expressed in form of equation:
Thot T cold kA t d
Q
= thermal conductivity coefficients of material (Wm-1K-1) k =
Thot T cold = temperature gradient.
.
d
Substance
Thermal Conductivity Conductivity,, k (Wm-1K-1)
Silver
418
Copper Aluminium
385 238
Brass
122
Iron
80
Steel Lead
46 38
Mercury
8
Concrete
1.7
Glass (pyrex) Brick
1.1 1
Water
0.6
Asbestos
0.17
Dry soil
0.14
Wood
0.13
Air
0.023
Example: 2
A block of co concrete ncrete has a c cross ross section of 5m and a thickness of 10cm. One side is at 40°C and the other is at 20°C.What is the rate of heat transfer? Given k concrete = 1.7 Wm-1K-1. Solution: Q t
Thot T cold kA
d
40 20 (1.7)(5) 0.1 1700W
3.5.2 CONVECTION
Convection of thermal energy occurs in a fluid (liquids or gases) when warm fluids flow, carrying energy with them as to replace cooler fluids.
Cannot happen in solids since atoms/molecules in solids are tightly bound together in their position.
Convection occurs in two forms:
Natural convection, e.g. atmospheric convection and daily weather variations. Forced convection, e.g. by means of pumps, fans, etc. to propel fluids and create artificially induced in duced convection current.
Heated water at the bottom of the t he beaker will expand and become more buoyant due to the increase of volume per unit mass. This will cause the density of the hot water to decrease. Hot water will rise towards the upper side of the beaker beaker.. The cooler and denser water from the top will go down to replace the hot water. So the pattern of circulation is formed within the beaker of water.
RADIATION TION 3.5.3 RADIA
Radiation is the heat by thecarry emission of away from the electromagnetic (EM)transfer waves which energy emitting object. In contrary to the conduction and convection, radiation requires no medium to transfer heat. The radiation can travel through vacuum, e.g. from the sun to the earth. The rate, P , at which an object emits energy via EM radiation is given by: 4
P AT (unit: Watt,W) A = surface area (m2) T = absolute temperature (K) -2 -4 σ = Stefan-Boltzmann constant = 5.67 x 10-8 Wm K . ε = emissivity of the emitting object which has the value
between 0 and 1.
An surface with with the maximum emissivit emissivity y of 1 is said to be
a blackbody radiator, but such surface is ideal and does not occur in nature. Any object with with absolute temperature temperature above 0K emits emits thermal radiation, at the same time it absorbs thermal
energy from surroundings surroundings.. Thus, net energy radiated per second by such object is given by: 4 4
P A T T E
Example: A spherical body with of 6 6.0 .0 cm is maintained at 200°C. Assuming that diameter the spherical body is ma an intained ideal blackbody radiator, at what rate (in Watt) is energy radiated from the sphere? Solution: Temperature, T = 200°C = 473.15 K Surface area, A = 4r 2 = 4(0.03m)2 = 1.13 x 10-2 m2 Emissivity for blackbody,
ε
=1
P AT 4
(5.6 5.67 108 )(1)(1. 1.13 13102 )(473 (473)) 32W
EXERCISE 1. 2.
3.
4. 5.
6. 7. 8. 9.
How much heat must be added to 3kg of water to raise its temperature from 20°C to 80°C? A total of 0.8kg of water at 20°C is placed in a 1.2kW 1.2kW electric kettle. How long a time is needed to raise the temperature of the water to 100°C? In preparing tea, 600g of water at 90°C is poured into a 200g china pot (c pot = 0.84 kJkg-1°C-1) at 20°C.What is the final temperature of the water? Five kilograms of water at 40°C is poured on a large block of ice at 0°C. How many kilograms of ice melts? 500 kcal of heat is added to 2kg of water at 80°C. How much steam is produced? Find the minimum amount of ice at -10°C needed to bring the temperature of 500g of water at 20°C down to 0°C. A 30g ice cubes at 0°C is dropped into 200g 200g of water at at 30°C. What is the final temperature? How much steam at 150°C is needed to melt 1kg of ice at 0°C? A copper ball of 2cm radius heated in a furnace at 400°C. If its emissivity is 0.3, at what rateisdoes it radiate energy?
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