1. No category

How to Solve Equations in Physics
Most equations in FY2 involve quantities that are multiplied and divided with each other. It is easy to
solve unknowns when you remember the following rule:
When you move quantities from one side of equation to the other side of
the equation, multipliers become divisors and divisors become multipliers.
If a quantity is not under a division line, it is a multiplier.
Examples
pµV = nµRµT
Solve T
Change sides to get the unknown to the left
nµRµT = pµV
multipliers n and R go to the right side as divisors
pµV
T=
nµR
1
m v2 = m µ g µ h
2
Solve v
2µmµgµh
v2 =
= 2µgµh
m
v=
2µgµh
If the equation has more than one term on each side then leave only the
term with unknown quantity to the left:
1
m v2 - F µ s = m µ g µ h
2
Solve m
1
m v2 - m µ g µ h = F µ s
2
1 2
m
v - gµh = Fµs
2
Fµs
m=
1 2
v - gµh
2
2
FY2 materiaali.nb
Chapter 5
Work, Heat, Kinetic Energy and Potential Energy
All these quantities are measured with same units, Joule
You are already familiar with quantities like
Kinetic Energy (liike-energia)
KE =
1
2
m v2
(1)
m = mass in kg
v = velocity in
@KED =
m
s2
kg m2
=J
s2
Read : unit of @KED is joule
Potential Energy (potentiaalienergia)
PE = m µ g µ h
(2)
† m = mass in kg
† g = acceleration of gravity = 9.81
m
s2
† h = height in meters
Note, that you get Joules only by using SI-system units: kg and më s2 .
@PED =
kg m
s2
µm = J
Simple Definition of Work (työ)
W = Fµs
† F = force in Newtons
† s = distance in meters
@WD = N µ m = J
Note, that force must be in the direction of motion.
Energy Conversions
(3)
FY2 materiaali.nb
3
Energy Conversions
Work done can be converted to kinetic energy, potential energy or heat. In such situations the
formulas above can be used to write an equation to solve unknown quantities.
Examples
1. If a man lifts a 20.0 - kg bucket from a well and does 6.00 kJ of work, how deep is the well?
Assume that the speed of the bucket remains constant as it is lifted.
m = 20.0 kg; g = 9.81
m
s2
; h = ?;
W = 6000 J;
Equation : increase in potential energy = Work
m g h = W H = PEL
W
6000
h=
=
= 30.6 m
m g 20 µ 9.81
2. A 50 - kg pole-vaulter running at 10 m/s vaults over the bar. Her speed when she is above
the bar is 1.0 m/s. Neglect air resistance, as well as any energy absorbed by the pole, and
determine her altitude as she crosses the bar.
m = 50 kg; v1 = 10
m
s
; v2 = 1.0
m
s
; h = ?; g = 9.81
m
s2
Equation : change in KE = increase in PE
1
2
or
m v1 2 =
mgh=
h=
1
2
1
2
1
2
m v2 2 + m g h
m v1 2 -
m v1 2 mg
1
2
1
2
m v2 2
m v2 2
=
1
2
50 µ 102 -
1
2
50 µ 12
50 µ 9.81
= 5.04587 º 5.0 m
3. A 2 000-kg car moves down a level highway under the actions of two forces. One is a 1000N forward force exerted on the drive wheels by the road; the other is a 950-N resistive
force. Use the work-kinetic energy theorem to find the speed of the car after it has moved a
distance of 20 m, assuming it starts from rest.
4
FY2 materiaali.nb
m = 2000 kg
Net force SF = 1000 N - 950 N = 50 N
s = 20 m
Equation : Work done by net force = increase in KE
Fµs =
1
2
m v2 OR
1
m v2 = F µ s
2
m v2 = 2 µ F µ s
2µFµs
v2 =
m
2µFµs
v=
m
2 µ 50 µ 20
=
2000
= 1.0
m
s
OR
1
2
1
m v2 = F µ s
2000 v2 = 50 µ 20
2
1000 v2 = 1000
v=1
4. A book is thrown at 11 m/s along a corridor floor. After sliding for 2.5 meters its speed has
decreased to 8.3 m/s. Find the average frictional force on the book. Mass of the book is
150g.
1
2
m vf 2 -
Fµs =
F=
1
2
1
2
1
2
m vi 2 = F µ s
m vf 2 -
m vf 2 -
1
2
s
Ans. F = 1.6 N
1
2
m vi 2
m vi 2
=
1
2
0.15 µ 112 -
1
2
0.15 µ 8.32
2.5
Power (teho)
Power gives the rate of doing work. Same work in less time needs more power.
P=
W
t
(4)
FY2 materiaali.nb
P=
W
t
=
Fµs
t
= Fµ
s
t
5
= Fµv
P = Fµv
(5)
P = power in Watts
W = work in joules
t = time in seconds
@PD =
J
s
= W HWattL
Examples
1. The electric motor of a model train accelerates the train from rest to 0.620 m/s in 21.0 ms.
The total mass of the train is 875 g. Find the average power delivered to the train during its
acceleration.
m = 0.875 kg
m
v = 0.620
s
-3
t = 21 µ 10 s
P=
W
1
2
m v2
=
t
t
P = 8.00 W
=
1
2
0.875 µ 0.6202
21 µ 10-3
6
FY2 materiaali.nb
Chapter 9
Density (tiheys)
r=
m
(6)
V
kg
† density r, @rD =
m3
, (NOTE r is Greek letter rho, it is not p, which means pressure.)
† volume V, @VD = m3
† mass m, @mD = kg
Some terms explained:
Liquid = The state of matter in which a substance exhibits a characteristic readiness to flow, little or
no tendency to disperse, and relatively high in-compressibility
Fluid = A continuous, amorphous substance whose molecules move freely past one another and
that has the tendency to assume the shape of its container; a liquid or gas.
Examples
1. Density of air at normal pressure and 20°C is 1.293
kg
m3
. Calculate the mass of air in a
classroom that is 9.0 m long, 6.0 m wide, and 3.0 m high.
m = r V = 1.293
kg
m3
µ H9 µ 6 µ 3L m3 = 209.466 kg º 210 kg
2. Find the volume of 55.0 g of aluminum. Density of aluminum is 2700
density first to
2700
V=
kg
m3
m
r
=
=
g
cm3
1 000 000 cm3
2.7
g
cm3
. Convert given
.
2 700 000 g
55 g
kg
m3
= 2.7
g
cm3
= 20.37 cm3 º 20 cm3
Pressure (paine)
p=
F
(7)
A
† pressure p, @pD =
N
m2
= Pa,
FY2 materiaali.nb
7
† Some old units:
1 bar = 105 Pa,
1 atm = normal atmospheric pressure = 101.325 kPa = 1.01325 bar = 760 mmHg
Examples
1. Laske paine auton renkaiden alla, kun paino jakautuu neljälle renkaalle. Auton massa on
1600 kg ja yhden renkaan kontaktipinta maahan on 20 cm × 20 cm.
2
A = H20 cmL2 = I20 µ 10-2 mM = 0.04` m2
m = 1600 kg
m
w = m g = 1600 kg µ 9.81
= 15 696 N
s2
F
15 696 N
98 100 N
P=
=
=
º 98 100 Pa = 98 kPa
4 A 4 µ 0.04 m2
m2
2. Lääkeruiskun männän halkaisija on 12 mm. Kuinka suuren paineen mäntä aiheuttaa
ruiskussa olevaan veteen, kun mäntää työnnetään 7,5 N:n voimalla.
A = p r2 = p
12
2
mm
= p µ H0.006 mL2 = 0.000113097 m2
2
F = 7.5 N
F
7.5 N
P= =
=
A 0.000113097 m2
66 314.756 N
º 66 000 Pa = 66 kPa
m2
Variation of Pressure with Depth
In liquid the total pressure (called also absolute pressure) at depth h is made up of atmospheric
pressure and hydrostatic pressure:
p = pa + r g h,
(8)
† pa = atmospheric pressure = 101.325 kPa,
† r = density of the fluid,
† g = 9.81
m
s2
,
† h = depth in the fluid
ph = r g h is called hydrostatic pressure.
Hhydrostaattinen paineL
Pressure does not depend on the shape of the container.
Examples
(9)
8
FY2 materiaali.nb
Examples
1. Find the total pressure in sea at the depth of 120 meters. Density of seawater = 1027
kg
m3
Solution
p = pa + r g h
= 101.325 kPa + I1027 kg ë m3 µ 9, 81 m ë s2 µ 120 mM =
101.325 kPa + 1208.98 kPa = 1310.31 kPa = 1.31 MPa
Buoyant Force and Archimedes’s Principle
Archimedes' principle
The magnitude of the buoyant force always equals the weight of the fluid displaced by the
object
Noste nesteessä on yhtä suuri kuin syrjäytetyn nestemäärän paino.
Buoyant force (Noste)
B = rVg
(10)
† where r = density of the fluid (gas or liquid) and V = Volume of the displaced fluid.
Examples
1. Find the buoyant force on a body totally submerged in water when the volume of the body
is a) 3 dm3 b) 2 m3
B = r * V * g = 1.000
B = r * V * g = 1000
kg
dm3
kg
m3
* 3 dm3 * 9.81
* 2 m3 * 9.81
m
s2
m
s2
= 29.4 N
= 19 620 N
2. A floating boat weighs 37500 N. a) What is the buoyant force? b) What is the weight of
displaced water? c) What is the volume of displaced water?
† Buoyant force must equal the weight of the boat so, B = 37500 N.
Weight of displaced water is equal to the buoyant force, w = 37500 N.
Volume of water:
FY2 materiaali.nb
9
B = r*V*g
B
37 500
V=
=
= 3.82 m3
r µ g 1000 µ 9.81
weight of water = m g = 37 500 N;
37 500 N 37 500 N
mass of water m =
=
= 3822.6 kg
g
9.81 m2
s
r=
V=
m
V
m
r
=
3822.6 kg
1000
kg
m3
= 3.8226` m3 º 3.82 m3
Floating (kelluminen)
The weight of a floating object equals the weight of displaced fluid = the buoyant force
(Jos kappale kelluu, niin noste = kappaleen paino = syrjäytetyn nestemäärän paino).
robj
rfluid
=
Vdisplaced fluid
(11)
Vobj
Examples
1. A block of wood is kept under water. Calculate the buoyant force, mass of the object
(rwood = 600
kg
m3
), and the weight of the object.
If the block is let to float freely in water, to what depth will it sink?
Repeat the same for a block of aluminum (rAl = 2700
kg
m3
)
How much lighter (in newtons) aluminum is in water than in air?
(rwood = 600
m3
), and the weight of the object.
If the block is let to float freely in water, to what depth will it sink?
10
FY2 materiaali.nb
Repeat the
same for a block of aluminum (rAl = 2700
kg
m3
)
How much lighter (in newtons) aluminum is in water than in air?
For wood:
rwood = 600
kg
; rwater = 1000
kg
;
m3
m3
V = 40 µ 60 µ 50 cm3 = 120 000 cm3 = 0.120 m3
V = 0.4 µ 0.6 µ 0.5 m3 = 0.12 m3
kg
m
B = rwater µ V µ g = 1000
µ 0.12 m3 µ 9.81
m3
s2
B = 1177 º 1200 N
m = rwood µ V = 600
kg
m3
µ 0.12 m3 = 72 kg
weight = m g = 72.` kg µ 9.81
m
s2
= 706.32 N º 710 N
Depth of the wooden block:
rwood
rwater
600
1000
=
=
Vdisplaced water
Vwood
Vdisplaced water
0.120
1.
600 µ 0.120
Vdisplaced water =
= 0.072 m3
1000
V = Ah
h=
V
A
=
0.072 m3
0.4 µ 0.6 m2
For aluminum:
= 0.3 m under water
FY2 materiaali.nb
rAl = 2700
kg
; rwater = 1000
11
kg
;
m3
m3
B = rwater µ V µ g º 1200 N Hsame as aboveL
kg
m = rAl µ V = 2700
µ 0.12 m3 = 324 kg
3
m
m
weight = m g = 324 kg µ 9.81
= 3178 N º 3200 N
s2
The weight of aluminium is reduced in water by 3200 N - 1200 N = 2000 N
Hydraulic Press (hydraulinen puristin)
Pressure is the same throughout the liquid, so
p=
F1
F
= 2.
A1
A2
F1
F2
=
Usually this is written as:
A1
A2
† http : // auto.howstuffworks.com/auto - parts/brakes/brake - types/brake3.htm
(12)
12
FY2 materiaali.nb
Chapter 10
Thermometers
Many physical properties can be used
•volume of a liquid
•length of a solid
•pressure of a gas held at constant volume
•volume of a gas held at constant pressure
•electric resistance of a conductor
•color of a very hot object
Kelvin Scale
When the pressure of a gas goes to zero, its temperature is –273.15°C
This temperature is called absolute zero
This is the zero point of the Kelvin scale –273.15°C = 0 K
http://zonalandeducation.com/mstm/physics/mechanics/energy/heatAndTemperature/heatAndTemp
erature.html
Thermal Expansion of Solids and Liquids
If heat flows into a solid object, its molecules will vibrate faster and take more space. This causes
the phenomenon called thermal expansion.
Change in length
DL = a L DT
(13)
Where a is called coefficient of linear expansion (pituuden lämpölaajenemiskerroin), L is original
length and DT is change in temperature in Kelvins or degrees Celcius (because changes are the
same in both units).
In the same way areas and volumes expand. Formulas are
FY2 materiaali.nb
13
Where a is called coefficient of linear expansion (pituuden lämpölaajenemiskerroin), L is original
length and DT is change in temperature in Kelvins or degrees Celcius (because changes are the
same in both units).
In the same way areas and volumes expand. Formulas are
DA = g A0 DT
(14)
for area expansion and
DV = b V0 DT
(15)
for volume expansion.
g = 2 a and b = 3 a if the values are not given. These are only approximate values.
Note that Finnish books use b for area and g for volume!
Examples
1. (11. chapter 10) The New River Gorge bridge in West Virginia is a 518-m-long steel arch.
How much will its length change between temperature extremes of – 20 °C and 35 °C?
The Concept of Ideal Gas
It is difficult to make an exact mathematical model for all different gas molecules. So called ideal gas
model works quite well for most gases.
An ideal gas is defined as one in which all collisions between atoms or molecules are perfectly
elastic and in which there are no intermolecular attractive forces. One can visualize it as a collection
of perfectly hard spheres which collide but which otherwise do not interact with each other. In such a
gas, all the internal energy is in the form of kinetic energy and any change in internal energy is
accompanied by a change in temperature.
(http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/idegas.html)
Some help for units and calculations can be found at http://www.chemguide.co.uk/physical/kt/idealgases.html
14
FY2 materiaali.nb
Ideal Gas Law
pV =nRT
(16)
This means that the product of pressure and volume is always a constant number (= n R T) at a
fixed temperature. Pressure is inversely proportional to volume.
The quantities in the equation are:
† pressure p, @pD = Pa =
N
m2
† volume V, @VD = m2
† temperature T [T] = K, This is important, only Kelvins work here
† number of moles n. One mole of substance has Avogadro number of particles, NA = Avogadro's
constant is approximately 6.02214 µ 1023 Mole-1 (Also one mole is the amount of substance
that contains as many elementary entities as there are atoms in 0.012 kg of carbon-12)
† One mole of an ideal gas occupies a volume of 22.4 dm3 .
† R is universal gas constant 8.31447
J
K mol
Generalization of Ideal Gas Law
pV
T
= n R , which is always a constant number.
(17)
if the same amount (same number of moles) of gas changes volume, temperature and pressure,
then
FY2 materiaali.nb
p1 V1
T1
=
p2 V2
15
(18)
T2
To find out the number of moles
n =
m
(19)
M
† m = mass in grams
† M = molar mass. For example molar mass of O2 = 32, H2 = 2, CO2 = 12 + 16 + 16 = 44.
Examples
1. A mass of oxygen occupies 0.0200 m3 at atmospheric pressure, 101 kPa, and 5°C.
Determine its volume if its pressure is increased to 108 kPa while its temperature is
changes to 30°C.
T1 = 5 + 273 = 278 K
T2 = 30 + 273 = 303 K
p1 V1
p2 V2
=
T1
T2
Solve for V2
V2 =
p1 V1 T2
T 1 p2
=
101 kPa µ 0.0200 m3 µ 303 K
278 K µ 108 kPa
= 0.0204 m3
2. An ideal gas has volume 1.0 liter at 1.0 atm and - 20.0°C. To how many atmospheres
pressure must it be subjected to be compressed to 0.50 liter when the temperature is 40.0
°C? Express your answer also in kPa.
† Note : You may use a nonstandard unit (atm) if it cancels away or if the question asks for it.
T1 = 273 - 20 = 253 K
T2 = 40 + 273 = 313 K
V1 = 1 L
V2 = 0.5 L
p1 = 1 atm
p1 V1
p2 V2
=
T1
T2
Solve for V2
p1 V1 T2 1 atm µ 1 L µ 313 K
p2 =
=
= 2.474 atm º 2.5 atm
T1 V2
253 K µ 0.5 L
Convert to kPa :
2.474 atm = 101.325 µ 2.474 kPa = 250.67805` kPa º 250 kPa
3. A pressurized container has 30.0 liters of nitrogen. Pressure of nitrogen is 15.0 MPa and
temperature 20.0°C. Find a) the number of moles of nitrogen b) mass of nitrogen c)
density of nitrogen.
16
FY2 materiaali.nb
g
M = 28
mol
pV=nRT
n=
n=
pV
RT
=
; p = 15 µ 106 Pa, T = 273 + 20 = 293 K; V = 0.030 m3 ;
15 µ 106 Pa µ 0.030 m3
8.31447
J
K Mol
µ 293 K
m
M
m = n µ M = 184.718 mol µ 28
r=
= 184.718 mol º 185 mol
m
V
=
5.17210 kg
0.030 m3
=
g
mol
= 5172.10 g º 5.17 kg
172.403 kg
m3
º 172
kg
m3
Chapter 11
Heat (Lämpö)
In 1948 scientists agreed that because heat (like work) is a measure of transfer of energy, its SI unit
should be JOULE.
† heat Q, [Q] = J
Internal Energy (Sisäenergia) – This is needed mainly in chapter 12
Internal energy is the energy (mainly kinetic) associated with the atoms and molecules of the
system.
† internal energy U, [U] = J
Specific Heat (Ominaislämpö tai ominaislämpökapasiteetti)
† specific heat c, [c] =
c=
J
kg °C
Q
(20)
m DT
Q = c m DT
This formula can be used to find how many joules are needed to raise the temperature of certain
object.
It also gives the amount of energy released when the object cools down.
Example: specific heat for water is 4186
J
,
kg °C
which means that 1 kg of water needs 4186 joules to
heat up by 1°C (or 1K),
So the amount needed to heat 6 kg of water by 50°C is 4186
J
×
kg °C
6 kg × 50°C =1255800 J º
1.3MJ
This is the same energy as in 60 grams of chocolate. Lisää energiatietoutta http://www.fineli.fi/
This formula can be used to find how many joules are needed to raise the temperature of certain
object.
It also gives the amount of energy released when the object cools down.
FY2 materiaali.nb
Example: specific heat for water is 4186
J
,
kg °C
17
which means that 1 kg of water needs 4186 joules to
heat up by 1°C (or 1K),
So the amount needed to heat 6 kg of water by 50°C is 4186
J
×
kg °C
6 kg × 50°C =1255800 J º
1.3MJ
This is the same energy as in 60 grams of chocolate. Lisää energiatietoutta http://www.fineli.fi/
Calorimetry
Means experiments made so that heat is not lost to surrounding materials or air. In these cases
changes of temperature can be calculated by using equations. For example if hot water is poured to
cold water, hot water gives heat to cold water (let’s say the amount given is Q) and cold water takes
heat (amount taken must be also Q).
Qgiven = Qtaken
c1 m1 HTi - Tf L = c2 m2 HTf - Ti L
(21)
Can you explain the subscripts in temperatures?
Examples
1. 20 grams of milk at 7°C is poured into 180 grams of coffee at 80°C. Find final temperature
Tf . Take the specific heat of both milk and coffee to be same as for water.
c1 m1 HTi - Tf L = c2 m2 HTf - Ti L
c µ 180 µ H80 - Tf L = c µ 20 µ HTf - 7L
cancel specific heats Hthey are now the sameL
multiply out the brackets
180 µ 80 - 180 Tf = 20 Tf - 20 µ 7
collect unknowns HTf L to left side
-20 Tf - 180 Tf = -14 400 - 140
-200 Tf = -14 540
-14 540
Tf =
º 73 °C
-200
Phase Changes (Olomuodon muutokset)
18
FY2 materiaali.nb
This paragraph involves answering questions like “how much heat is needed to melt 500 g of ice at
0°C?” or “how much heat is released when 2 kg of water at 0°C freezes?”.
† NOTE that temperature stays constant while a phase change happens.
substance
heat of
fusion
aluminum
iron
tin
ethanol
water HiceL
heat of
Lf J kJ
N
kg
vaporization Lv J MJ
N
kg
397
276
57
102
333
10.9
6.8
2.45
0.841
2.26
Q = m Lf Hfusion, sulaminenL = m s
(22)
Q = m Lv Hvaporization,
höyrystyminenL = m h
(23)
Examples
1. How much heat is needed to melt 2 kg of iron at its melting point 1808°C?
Q = m Lf = 2 kg µ 276
kJ
kg
= 552 kJ
2. How much heat is needed to heat 2 kg of iron from 20°C to 1808°C
FY2 materiaali.nb
c = 0.448
kJ
kg °C
19
; m = 2.00 kg; DT = 1788 °C
Q = c m DT
Q = 0.448
kJ
µ 2.00 kg µ 1788 °C =
kg °C
Q = 1602 kJ º 1600 kJ = 1.6 MJ
3. How much heat is needed to melt 2 kg of iron originally at 20°C?
552 kJ + 1600 kJ = 2152 kJ º 2.2 MJ
Energy Transfer
Conduction (johtuminen)
Heat transfer through collisions of adjacent molecules.
The rate of heat transfer in solid substances like glass etc. can be calculated:
P=k
A DT
(24)
L
† power P, [P] = W (watt)
† cross sectional area A, [A] = m2
† thickness L, [L] = m
† thermal conductivity k, [k] =
J
s m °C
† In house building common term is conductivity U, U =
k
L
L
k
(The corresponding term in US is resistivity R = )
† more in http://fi.wikipedia.org/wiki/Lämpöresistanssi
Examples
1. Find the rate of heat transfer through a single glass window having measures 100.0cm x
140.0cm, thickness of the glass 5.00 mm when temperature inside is +20°C and outside –
5°C.
A = 1.40; DT = 25; L = 0.005; k = 0.9;
A DT
1.4 µ 25
P=k
= 0.9
= 6300 W
L
0.005
Heat tranfer rate is very big. For this window the U - value U =
k
L
=
0.84
0.005
= 168.
This is the reason why windows have triple glass panes and air or special gases between glass
panes as an extra insulator. This way the U-value of the whole window can be much less than 1.
20
FY2 materiaali.nb
http://www.skaala.com/weboost.php?sivu=tiedosto&t=69&url=skaala_kaytto__asennus__ja_huolto_ohjeetjojp&type=pdf
2. According to building regulations the U-value for the outer walls of a house should be less
W
than 0.25 2 .
m K
At what rate is heat transferred through the walls of a room if the temperature inside the
house is 20°C and outside the house -20°C? The area of the walls is 10 m2 .
P = U A DT
U = 0.25
W
m2 °C
; A = 10 m2 ; DT = 40 °C;
P = U A DT = 0.25
W
2
* 10 m2 * 40 °C = 100 W
m °C
3. (Problem 34) A window has a glass surface of 1.6µ103 cm2 and a thickness of 3.0 mm. (a)
Find the rate of energy transfer (=power) by conduction through this pane when the
temperature of the inside surface of the glass is 21°C and the outside temperature is -18°C.
A = 0.16; k = 0.84; T1 = 21; T2 = - 18; L = 0.003;
T1 - T2
P = kA
= 1700 W
L
Convection (kuljetus)
Convection means taking heat energy from one place to another using a flowing substance, like
water or air.
Radiation (säteily)
Heat energy can move through radiation (which is electromagnetic wave) at certain infrared wavelengths. This way heat energy from the Sun reaches the Earth without any medium. Radiation can
move through vacuum.
Stefan’s law gives the rate of radiation in watts.
P = s A e T4
† s =the Stefan-Boltzmann constant, 5.6696µ10-8
† A = surface area of the object
(25)
W
m2 K 4
FY2 materiaali.nb
21
† e = constant called the emissivity, may have values from 0 to 1.
for white and shiny objects º 0 and for black objects º 1. Its given in problems!
† T = temperature of the object in kelvins.
P = s A e IT4 - T04M
(26)
This form is more useful because it accounts for the temperature of the surroundings, T0 .
Examples
1. (Problem 34 continued) A window has a glass surface of 1.6 x 103 cm2 and a thickness of
3.0 mm.
(b) At what rate is the energy radiating through the same window as the one above.
e = 1;
s = 5.6696 * 10-8
W
m 2 K4
;
A = 0.16 m2 ;
T1 = H-18 + 273L K = 255 K
T2 = H21 + 273L K = 294 K
P = s A e I T2 4 - T1 4 M = 29 W
Chapter 12
The First law of Thermodynamics (Lämpöopin ensimmäinen pääsääntö)
The change in the internal energy of a gas is the sum of heat input and work input.
DU = Q + W
(27)
This is one way of stating the law of conservation of energy
Work input means that someone compresses the gas. When volume decreases by DV, the work
done on the gas is
W = p DV
(28)
Note that if gas expands because of being heated, the work done is negative, see example 12.3 in
the book.
Skip paragraph 12.3 in the book.
Heat Engines and the Second Law of Thermodynamics
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Work done by heat engine from the figure above
W = Qh - Qc
(29)
Efficiency (hyötysuhde)
work done by engine
e=
Weng
heat taken from hot reservoir
= e Qh
=
Weng
Qh
=
Qh - Qc
Qh
(30)
Example 12.10 is a good one but skip example 12.11.
The efficiency of an ideal (best possible) heat engine was defined by Sadi Carnot:
ec =
Th - Tc
(31)
Th
Notice that temperatures have to be still expressed in Kelvins.
Examples
1. (Example 12.10)
During one cycle and engine extracts 2.00µ103 J of energy from a hot reservoir and
transfers 1.50µ103 J to a cold reservoir.
a) Find the total thermal efficiency of the engine
b) How much work does this engine do in one cycle?
c) How much power does the engine generate if it goes through four cycles in 2.50 s?
Qh = 2.00 µ 103 ; Qc = 1.50 µ 103 ;
e=
Qh - Qc
Qh
=
2.00 µ 103 - 1.50 µ 103
2.00 µ 103
W = Qh - Qc = 5.00 µ 102 J
= 25 %
FY2 materiaali.nb
P=
W
t
=
4 µ 5.00 µ 102 J
2.5 s
=
800.` J
s
23
= 800 W
2. (Example 12.13)
A steam engine has a boiler that operates at 5.00µ102 K. The energy from the boiler
changes water to steam, which drives the piston. The temperature of the exhaust is that of
the outside air, 3.00µ102 K.
a) What is the engine’s efficiency if it is an ideal engine?
b) If the 3.50µ103 J of energy is supplied from the boiler, find the energy transferred to the
cold reservoir and the work done by the engine on its environment.
ec =
Th - Tc
Th
=
500 - 300
500
= 40 %
Second Law of Thermodynamics
† No heat engine operating in a cycle can absorb energy from a reservoir and use it entirely for the
performance of an equal amount of work.
† Lämpöä on mahdotonta muuttaa kokonaan työksi
This law can be stated in a variety of ways:
† Luonnon tapahtumat etenevät kohti systeemin todennäköisintä tilaa, suurempaa epäjärjestystä.
Eristetyssä systeemissä entropia (= epäjärjestys) kasvaa.
Refrigerators and Heat Pumps
A heat pump or refrigerator is a heat engine running in reverse.
Because with refrigerators it is not important how much heat goes to hot container (outside air) but
rather how much heat is taken from the cold reservoir and how much work (electrical energy) is
used on that.
Efficiency formula is replaced by coefficient of performace (käyttökerroin), in cooling mode:
COP =
Qc
W
in heating mode:
(32)
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FY2 materiaali.nb
COP =
Qh
(33)
W
† The larger COP is the better cooling engine we have. Usually COP = 3 - 6.
Examples
1. (Example 12.12 modified)
A 2.00 liter container of leftover soup at temperature of 50°C is places in a refrigerator.
Assume the specific heat of the soup is the same as that of water and the density is
1.25µ103
kg
m3
. The refrigerator cools the soup to 10°C.
a) If the COP of the refrigerator is 5.00, find the energy needed to cool the soup.
b) If the compressor has a power rating of 200W, for what minimum length of time
must it operate to cool the soup to 10°C?
FY2 materiaali.nb
m = r V = 1.25 µ 103
kg
m
3
µ 0.002 m3 = 2.50 kg
Q = m c DT = 2.50 kg µ 4186
COP =
Qc
418 600 J
kg °C
=5 Ø W=
µ 40 °C = 418 600 J
418 600 J
= 83 720 J
W
5
W
W 83 720 J
P= Ø t= =
= 418.6` s º 6.98 min º 7.0 min
t
P
200 Js
W
=
J
Alla erään ilmalämpöpumpun teknisiä tietoja:
Laske alla olevasta mittauskuvasta COP lämpötilan 0°C kohdalta.
Mikä on alla olevan laitteen COP lämpötilan 0°C kohdalla?.
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