Temperature qualitatively describes the hotness & coldness of objects. An object is considered hot when its temperature is higher than a reference temperature; it is considered cold when its temperature is lower than said reference temperature.

Temperature is related to the average kinetic energy of the molecules comprising the material of the object.

Two common temperature scales are the Celsius Scale (i.e. 300C) & the Fahrenheit Scale (i.e. 50F).

The lowest temperature that has been extrapolated for the Celsius scale is -273.150C, which has been designated the zero temperature (absolute zero) of the Kelvin scale, the absolute temperature scale of the metric system.

Kelvin scale temperature is not expressed in degrees (i.e. 100K).

When 2 objects or system have the same temperature, they considered to be in Thermal Equilibrium.

1000C

00C320F

2120F

T F=95

TC+320

T K=T C+273.15

When talking about temperature difference, the temperature unit is slightly different for the Celsius & Fahrenheit scales.

whereL = change in lengthL0 = original lengthT = change in temperature = coefficient of linear

expansion

T=T fT i=50 C150C=10C0

It has been observed that most materials expand when their temperatures increase. And when their temperatures decrease, these materials contract. This phenomenon is called Thermal Expansion.

When looking at how the length of an object changes due to changes in temperature, we have Linear Expansion.

L= L0T

When looking at how the area (surface or cross-section) of an object changes, we have Area Expansion.

Heat is energy associated with changes in temperature.

Heat is energy in transition. Heat (Q) flows from one object to another.

A= A0T

where = coefficient of area expansion

And for Volume Expansion,

V=V 0T

where = coefficient of volume

expansion

The 3 expansion coefficients are related to each other as follows

=2

=3=32

COEFFICIENTS OF LINEAR EXPANSIONMaterial

AluminumBrassCopperGlassInvar (nickel-iron alloy)Quartz (fused)Steel

[K-1 or (C0)-1

2.4 x 10-5

2.0 x 10-5

1.7 x 10-5

0.4-0.9 x 10-5

0.09 x 10-5

0.04 x 10-5

1.2 x 10-5

A BQ

Even though no object possesses heat, we still say object A losses heat while object B gains heat when heat flows from A to B.

Unit of heat:1 joule = 1 J1 calorie = 1 cal = 4.186 J1 food calorie = 1 Cal = 1kcal1 British thermal unit = 1 Btu = 252 cal

Heat flows from hot to cold. That is, heat flows from the object with higher temperature to the object with lower temperature.

When an object gains heat, it's temperature increases and vice-versa.

whereQ = heat gained or lostm = mass of the objectc = specific heat of the material

comprising the objectT = change in temperature

Qgain = + (positive)Qloss = - (negative)

dQ=mcdTQ=mcT

APPROXIMATE SPECIFIC HEATSMaterial c (J/kg-K)

Aluminum 910Berylium 1970Copper 390Ethanol 2428Ethylene glycol 2386

2100Iron 470Lead 130Mercury 138Silver 234Water 4190

Ice (near 00C)

A liquid object willvaporize when it gainsheat and it is as itsboiling point temperature. While agaseous object willcondense when it losesheat and it is at itscondensation pointTemperature.

When an object at the right temperature gains the right amount of heat, the object changes in phase.

A solid object will melt when it gains heat and it is at its melting point temperature. While a liquid object will freeze when it loses heat and it is at its freezing point temperature.

T melting=T freezingQ=mL f

whereQ = heat gained (+) or lost (-)m = mass of the objectLf = heat of fusion of the material

T boiling=T condensationQ=mLv

whereLv = heat of vaporization of the

material

For water,Tfreezing = 0

0C Tboiling = 1000C

Lf = 334 x 103 J/kg

Lv = 2,256 x 103 J/kg

It must be noted that temperature change and phase change do not happen at the same time.

A solid substance can sometimes transition directly to gaseous phase. This phenomenon is referred to as sublimation.

Heat is a form of energy and as such is subject to the Principle of Conservation of Energy.

What is lost by one part of the system is gained by another part of the system.

Since heat energy may be converted into other forms of energy and vice-versa,

Qgain+Qloss=0mcT+ (mL)=0

mcT+ (mL)+E+W done=0

Heat is transferred via different mechanisms: Conduction, Convection, and Radiation.

When heat flows within a body or between 2 bodies in thermal contact, heat transfer via conduction is said to occur. This particularly true with solid objects.

Heat transfer is measured in terms of Heat Transfer Rate.

whereH = heat transer rate or heat currentk = thermal conductivity of the

material comprising the objectA = cross-section area of the objectL = length of the objectTH = hot temperatureTC = cold temperature

H= dQdt=kA

T HT CL

THERMAL CONDUCTIVITIESSubstance k (W/m-K)

Aluminum 205.000Brass 109.000Copper 385.000Lead 34.700Steel 50.200Concrete 0.800Cork 0.040Glass 0.800Ice 1.600Styrofoam 0.010Air 0.024Helium 0.140Oxygen 0.023

H 1=H 2=H

k1 A1T HT

L1=k2 A2

TT CL2

When 2 materials are connected one after another, the temperature at their connection point adjusts such that the 2 will have the same heat transfer rate.

And when 2 materials are arranged parallel to each other, their heat transfer rates add together.

H=H 1+H 2

H=k1 A1T H1T C1

L1+k2 A2

T H2T C2L2

While conduction also occurs with fluids, heat transfer via convection is more dominant. Heat convection is heat transfer via the physical motion of fluids.

Heat radiation is the transfer of heat via the motion of electromagnetic waves like visible light, infrared, and ultraviolet radiation.

Every object emits energy in the form of electromagnetic waves. Only at absolute zero temperature will an object stop emitting energy.

H=AeT 4

where A = area of the emitting surfacee = emissivity of the surface = Stefan-Boltzmann constant = 5.67 x 108 W/m2K4T = temperature (in Kelvin) of the

emitting object

0 e 1

Since surfaces are both emitting and absorving energy in the form of electromagnetic radiation,

H net=H emitH absorbH net=Aemit eT obj

4 Aabsorb eT surrounding4

Example 1The Humber Bridge in England has the world's longest single span, 1410 m in length. Calculate the change in length of the steel deck of the span when the temperature increases from 5.00C to 18.00C. steel = 1.2 x 10

5 K1

T=180 C(50 C )=23C0=23KL0=1,410msteel=1.210

5 K1

L=steel L0T=(1.2105 K1)(1,410m)(23K)=0.38916m

L=0.389m

Example 2A copper cylinder is initially at 20.00C. At what temperature will its volume be 0.150% larger that it is at 20.00C?copper = 1.7 x 10

5 K1

T=T f293.15 Kcopper=1.710

5 K1

copper=3copper=5.1105 K1

VV 0

=0.0015 V=0.0015V 0

V=V 0T T=0.0015

T f293.15 K=0.0015

T f=0.0015

5.1105 K1+293.15 K

=322.5617647KT f=322.562K

Example 3A machinist bores a hole of diameter 1.350 cm in a steel plate at a temperature of 250C. What is the cross-sectional area of the hole when the temperature of the plate is increased to 1750C? steel = 1.2 x 10

5 K1

T=1750 C250 C=150C0=150Kd0=1.350 cm=0.0135 msteel=1.210

5 K1

steel=2steel=2.4105 K1

A0=d0

2

4=0.000143138 m2

A= steel A0T=(2.4105 K1)(0.000143138 m2)

(150K)=0.000000515 m2

A=A0+ A=0.000143653m2

A=1.437104 m2

Example 4Steel train rails are laid in 12-m-long segments placed end-to-end. The rails are laid on a winter day when their temperature is 2.00C. How much space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 33.00C?steel = 1.2 x 10

5 K1

L0=12mT=330 C(20 C)=35C0=35Ksteel=1.210

5 K1

D= L01

2+ L02

2= L

=steel L0T=(1.2105 K1)(12 m)(35K)=0.00504 m

D=5.04103 m

Example 5While running, a 70-kg student generates thermal energy at a rate of 1200W. If this heat could not be removed by perspiration or other mechanisms to maintain a constant body temperature of 370C, for what amount of time could the student run before irreversible body damage occurs? Protein structures in the body are irreversibly damaged if body temperature rises to 440C or higher. The specific heat of a typical human body is around 3480J/kgK.

m=70kgP=1,200WT=440 C370 C=7C0=7Kc=3,480 J /kgKPt=Q Pt=mcT

t=mcTP

=(70 kg)(3,480 J /kgK )(7K)

1,200W=1,421 s

t=1,421.00 s

Example 6While painting the top of an antenna 225m in height, a worker accidentally lets a 1.00-L water bottle fall from his lunchbox. The bottle lands in some bushes at level ground and does not break. If a quantity of heat equal to the magnitude of the change in mechanical energy of the water goes into the water, what is its increase in temperature?

H2O=1,000 kg/m3

V=1.00 L=1,000 cm3=1103 m3

H2O=mV

m=H2O V=1kg

h=225 m c=4,190 J /kgKU=Q mgh=mcT

T=ghc=(9.8m / s2)(225 m)

4,190 J /kgK=0.526252983 K

T=0.526 K

Example 7An ice cube tray of negligible mass contains 0.350kg of water at 18.00C. How much heat, in Btu, must be removed to cool the water to 0.00C and freeze it?

Example 8A copper pot with mass 0.500kg contains 0.170kg of water at 20.00C. A 0.250-kg block of iron at 85.00C is dropped into the pot. Find the final temperature, assuming no heat loss to the surroundings.ccopper = 390 J/kgKciron = 470 J/kgK

m=0.350kgT=0.00 C18.00 C=18.0C0

=18.0Kc=4,190 J /kgKLf=33410

3 J /kgQremove=Qloss=(mcTmLf )=(0.35)(4,190)(18)

+(0.35)(334103)=143,297J=135.8265403 Btu

Qremove=135.827 Btu

m1=0.500 kg m2=0.170 kgm3=0.250kg c1=390 J /kgKc2=4,190 J /kgK c3=470 J /kgKT 1=T2=20.0

0 C=293.15KT 3=85.0

0 C=358.15 KQgain+Qloss=0m1 c1(TT 1)+m2 c2(TT 2)

+m3 c3(TT 3)=0

T=m1 c1 T 1+m2 c2 T2+m3 c3 T 3

m1 c1+m2 c2+m3 c3T=300.603 K

Example 9A thirsty mechanic cools a 2.00-L bottle of softdrink (mostly water) by pouring it into a large aluminum mug with mass 0.257kg and adding 0.120kg of ice initially at 0.00C. If the softdrink and mug are initially at 20.00C, what is the final temperature of the system?caluminum = 910 J/kgK

V=2.00 L =1,000 kg /m3

m1=2kgm2=0.257 kg m3=0.120 kgT 1=T 2=20.0

0 C=293.15 KT 3=0.0

0 C=273.15 Kc1=4,190 J /kgK c2=910 J /kgKc3=c1 L3=33410

3 J /kgQgain+Qloss=0m3 L3+m3 c3(TT 3)+m1 c1(TT 1)+m2 c2(TT 2)=0

T=m1 c1 T 1+m2 c2 T2+m3 c3 T 3m3 L3

m1 c1+m2 c2+m3 c3=287.6506236 K

T=287.651 K

Example 10One end of an insulated metal rod is maintained at 0.00C by an ice-water mixture. The rod is 60.0cm long and has a cross-sectional area of 1.25cm2. The heat conducted by the rod melts 8.50g of ice in 10min. Find the thermal conductivity of the metal.

T H=1000 C=373.15 K

TC=00 C=273.15 K

Lf=334103 J /kg

L=60.0cm=0.60mA=1.25cm2=0.000125 m2

m=8.50 g=0.0085kgt=10min=600 s

H=Qt=

mLft

H=kAT HTC

L=

mLft

k=mLf L

At (T HTC)

=(0.0085)(334103)(0.6)(0.000125)(600)(100)

=227.12W /mK

k=227.12W /mK

Example 11The emissivity of tungsten is 0.35. A tungsten sphere with radius 1.50cm is suspended within a large evacuated enclosure whose walls are at 290K. What power input is required to maintain the sphere at a temperature of 3,000K if heat conduction along the supports is neglected?

e=0.35r=1.50 cm=0.015 mA=4 r2=0.002827433m2

T=3,000 KT S=290 KH=AeT 4AeT S

4

=4,544.546804 W=heat loss rate

Pinput needed=4,544.547W

Example 12If solar radiation energy incident per second on the frozen surface of a lake is 600W/m2 and 70% of this energy is absorbed by the ice, how much time will it take for a 2.50-cm-thick layer of ice to melt? The ice and the water beneath it are at a temperature of 00C.HA=(0.7)(600W /m2)=420 W /m2

H=(420W /m2)A

L=2.5cm=0.025m=1,000 kg /m3

=mV=m

LAm= LALf=33410

3 J /kg

H=Qt=

mL ft

=LA Lf

tLA Lf

t=(420W /m2)A

t=L Lf

420W /m2=19,880.95238 s

=331.349064 min

t=331.349min