02. Hukum I Termodinamika
Transcript of 02. Hukum I Termodinamika
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02. Hukum I Termodinamika
Zulfiadi Zulhan
Teknik MetalurgiFakultas Teknik Pertambangan dan PerminyakanInstitut Teknologi BandungINDONESIA
Termodinamika Metalurgi (MG-2112)
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Jangan Mengunggah
Materi Kuliah ini di
INTERNET!
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NO TALKING
NO SLEEPING
NO MOBILE PHONEhttp://www.longestlife.com
https://www.pinterest.com
https://www.pinterest.se
http://clipart-library.com
https://www.dreamstime.com
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Hukum Pertama TermodinamikaKekekalan Energi
Bentuk energi yang umum dijumpai:
▪ Energi panas
▪ Energi kerja atau energi mekanik
▪ Energi listrik
▪ Energi kimia
• Energi tidak dapat diciptakan dan tidak dapat dimusnahkan.
• Energi dapat ditransportasi atau dikonversi dari satu bentuk
ke bentuk lainnya, tetapi energi tidak dapat diciptakan dan
tidak dapat dimusnahkan.
• Perubahan kimia dan atau fisika selalu diikuti oleh
perubahan energi.
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Energi
Energi fosil (Batubara, minyak, gas alam)
Pembangkit listrik termal
Pembangkit listrik tenaga air
Pembangkit listrik panas bumi
http://www.broadstarwindsystems.com/fossil-fuels/
https://greentumble.com/why-do-we-use-fossil-fuels-instead-of-other-fuels/
https://internationalfinance.com/indonesia-replace-coal-
power-plants-renewable-energy-sources/
https://www.beritadaerah.co.id/2021/02/17/hydropower-batang-toru-
harapan-listrik-sumatera-utara/
https://www.thinkgeoenergy.com/video-overview-on-the-sarulla-
geothermal-power-plants-in-indonesia/
https://economy.okezone.com/read/2018/11/14/320/1977586/g
as-alam-jadi-sumber-energi-terbesar-kedua-dunia-pada-2030
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Energi
http://kabar6.com/chef-michael-dibalik-lezatnya-aneka-menu-hotel-santika-ice-bsd/
https://www.99.co/blog/indonesia/informasi-lengkap-colokan-listrik/
https://www.idntimes.com/tech/gadget/izza-namira-1/cara-charging-
hp-yang-cepat/10https://news.ddtc.co.id/pemprov-dki-resmi-naikkan-tarif-bea-balik-nama-kendaraan-bermotor-17778
https://nextren.grid.id/read/01216080
2/5-tips-memilih-laptop-untuk-
belajar-dan-bekerja-pemula-wajib-
tahu-nih?page=all
ILUSTRASI menonton televisi.* /PIXABAY /
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Energi
https://www.lenntech.com/greenhouse-effect/fossil-fuels.htm
https://biology-for-all.weebly.com/carbon-cycle.html
http://www.fossilfuelconnections.org/coal
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Eksotermik dan Endotermik
H < 0 → Reaksi eksotermik (menghasilkan panas)
H > 0 → Reaksi endotermik (membutuhkan panas)
https://teachsciencewithfergy.com
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https://bifrostonline.org/global-carbon-cycle/
Sumber: http://berkeleyearth.org/
https://www.iea.org/reports/net-zero-by-2050
Balance between the amount of greenhouse gas produced
and the amount removed from the atmosphere
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The Sun
our main Source of Energy
http://igbiologyy.blogspot.com/2014/03/106-energy-flow-energy-loss.html
https://www.slideserve.com/nerice/the-sun-the-primary-source-of-energy-for-all-living-things
https://slideplayer.com/slide/14761278/
https://www.bitlanders.com/blogs/energy-flow-in-ecosystem/235900
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https://www.sciencealert.com/this-awesome-periodic-table-shows-the-origins-of-every-atom-in-your-body
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http://new-universe.org/zenphoto/albums/Chapter3/Illustrations/Abrams24.jpg
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Mesin Uap (Sumber Energi = Panas + Kerja)
https://www.youtube.com/watch?v=9mhYnQGZJuM
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1401 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Panas dan Kerja
U
Q
UUUU
Q
W
UUU
• If heat (Q) is supplied to the system, the internal energy of the system (U) will increase.
• If the surroundings does work (W) to the system, the internal energy (U) of the system will
increase.
𝛅𝐐𝐥𝐢𝐧𝐠𝐤𝐮𝐧𝐠𝐚𝐧 ≅ +𝐝U𝐬𝐢𝐬𝐭𝐞𝐦 𝛅𝐖𝐥𝐢𝐧𝐠𝐤𝐮𝐧𝐠𝐚𝐧 ≅ +𝐝U𝐬𝐢𝐬𝐭𝐞𝐦
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1501 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Panas, Kerja dan Energi Dalam
Panas (Q) mengalir karena ada perbedaan temperatur.
Panas mengalir hingga perbedaan temperatur tidak ada.
Pada saat panas mengalir, energi berpindah.
(+): dari Lingkungan ke Sistem
Kerja (W) adalah perpindahan energi akibat interaksi
antara sistem dengan lingkungan.
(+): lingkungan melakukan kerja kepada sistem
https://www.tokopedia.com/
https://www.qraved.com
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Panas Kerja dan Energi Dalam
Internal Energy (U) is the energy contained in the system
• If heat (Q) is supplied to the system, the internal
energy of the system (U) will increase.
• If the surroundings does work (W) to the system,
the internal energy (U) of the system will
increase.
https://www.tokopedia.com/
https://www.qraved.com
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1701 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Heat, Work and Internal Energy (First Low Energy: Closed System)
The relationship between Heat, Work and Energy in a „closed system“:
UdWQ =+
Notation is used in front of Q and W Q and W are not state function (function of the path)
Internal energy U is a state function, its differential is indicated by d
To make it more complete, Kinetic and Potensial energies can be added to the system, so
„first law of thermodynamic for closed system“:
d(KE) d(PE) UdWQ ++=+
„Closed system“ : no matter enters or leaves the system. The boundaries of the
system may expand or contract as work is done by or on the system, and the thermal
energy may flow into or out the system as heat.
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Panas dan Kerja: Konvensi Tanda
Positif (+) : Kerja dilakukan oleh lingkungan kepada sistem
Negative (-) : Sistem melakukan kerja ke lingkungan
SISTEM
Lingkungan
Lingkungan
Q+ W+
Tanda positif untuk Panas dan Keja Energi sistem dalam
bentuk Panas dan Kerja bertambah (ber +).
LingkunganLingkungan
𝛅𝐐 + 𝛅𝐖 = 𝐝U
𝐁𝐞𝐛𝐞𝐫𝐚𝐩𝐚 𝐛𝐮𝐤𝐮𝐦𝐞𝐧𝐮𝐥𝐢𝐬:𝛅𝐐 − 𝛅𝐖 = 𝐝U
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1901 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Case 1:
One kg mass will be raised from the ground level to a height of 10 m.
d(KE) d(PE) UdWQ ++=+
Solution:
Q = 0, because there was no heat transfer involved in the process.
dU = 0, there was no change in internal energy.
d(KE) = 0
h g m W
d(PE) W
=
=
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Sifat-Sifat Intensif dan Ekstensif
Sifat Ekstensif bergantung pada besar atau massa atau ukuran dari sistem
Contoh Sifat Ekstensif: Volume
Sifat Intensif tidak bergantung pada massa atau ukuran dari sistem.
Contoh Sifat Intensif: Tekanan, Temperatur, Densitas, Volume Spesifik (Volume per
Satuan Massa), Volume Molar (Volume per Mol).
/kgm dalam Spesifik VolumeV
m
V V
3=
=
Densitas dari sebuah material adalah kebalikan
dari volume spesifik
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Heat, Work and Internal Energy (First Low Energy: Open System)
Vd P workflowV
0
ii =„i“ is designated for entering material, and
„o“ for leaving material.
„ Open System“ :matter may enter or leave the system.
Consider:
• internal energy (U) of the materials entering and leaving of the system
• work (W) done on the system when the material is pushed into the system and the work (W)
done by the system, when material is pushed out of the system. This work is called „flow work“.
System
boundary
Flow work in an open system
P
mi
iiiii dm V dV or m VV ==
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Hukum I Termodinamika: Sistem Terbuka
(OPEN SYSTEM)
Vd P workflowV
0
ii =„i“ is designated for entering material,
and „o“ for leaving material.
System
boundary
Flow work in an open system
P
mi
iiiii dm V dV or m VV ==
http://www.bintang.comhttp://www.startribune.com
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2301 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Heat, Work and Internal Energy (First Low Energy: Open System) cont.
Vd P workflowV
0
ii =„i“ is designated for entering material, and
„o“ for leaving material.
Omitting kinetic and potential energy, first law of thermdynamic for „ Open System“:
iiiii dm V dV or m VV ==
ii
m
0
iii m V P md V P workflow ==
m V P work)flow( iii =In differential form:
dU W Q work)flow( work)flow(m U -m U oiooii =++−+
( ) ( ) dU W Q m V P U -m V P U ooooiiii =++++
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2401 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Heat, Work and Internal Energy (First Low Energy: Open System) cont.
Ui mi and Uo mo represent the internal energy carried into and out of
the system by the matter entering and leaving of the system.
Pi Vi mi and Po Vo mo are the flow work into and out of the system by
the matter entering and leaving of the system
Ui + Pi Vi δmi− Uo + Po Vo δmo + δQ + δW = dU
( ) ( ) dU W Q m V P U -m V P Uoi m
oooo
m
iiii =++++
For all streams entering or leaving the system:
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Entalpi
U + PV didefinisikan sebagai Entalpi (H)
PV U H +
https://webmuda.files.wordpress.comhttps://webmuda.files.wordpress.com
𝐔𝐢 + P𝐢 𝐕𝐢 δmi− 𝐔𝐨 + P𝐨 𝐕𝐨 δmo + δQ + δW = dU
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2601 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Enthalpy
U + PV is defined as Enthalpy (H)
PV U H +
Specific enthalpy is enthalpy per unit mass
VP U H +=
( ) ( ) )KEPEU(d W Q m PE KE H -m PE KE Hoi m
oooo
m
iiii ++=++++++
Consider potential and kinetic energy, First Law of Thermodynamic for
open system:
For closed system: mi and mo = 0
Enthalpy is a measure of the total energy of
a thermodynamic system. It includes
the internal energy, which is the energy
required to create a system, and the amount of
energy required to make room for it by
displacing its environment and establishing
its volume and pressure. wikipedia
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2701 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Steady State
(„Bahasa Indonesia = Keadaan Tunak)
Steady State is defined as one in which the system does not change with time.
▪ Every quantity or property of the system is time invariant.
▪ Material may enter and leave the system, but the system itself remains unchanged.
Steady State principle can be used to analyse processing apparatus, such as chemical
reaction vessel, smelters, blast furnace, pumps, turbines, etc.
In these apparatus, once steady operations have been achieved, material enters and
material leaves, but the system itself remains basically unchanged with time.
First law of thermodynamic (steady state): dU = 0
( ) ( ) 0 W Q m PE KE H -m PE KE Hoi m
oooo
m
iiii =++++++
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2801 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Steady State, cont.
First law of thermodynamic:
( ) ( ) 0 W Q m PE KE H -m PE KE Hoi m
oooo
m
iiii =+++++
For a finite process, it can be expressed as follows:
( ) ( ) m PE KE H -m PE KE H W Q io m
iiii
m
oooo +++=+
When no changes in kinetic or potential energy:
m H -m H W Q io m
ii
m
oo =+
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2901 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Heat Capacity at Constant Volume
Heat capacity of a material is the amount of thermal energy required to change the
temperature of the material.
For a closed system at constant volume (dV=0) no mechanical work is done (W=0),
First Law Thermodynamics:
𝛅𝐐 + 𝜹𝐖 = 𝐝𝐔
dU dT C m Q ==
U d m
dU dT C ==
where C is the Heat capacity
Heat capacity at constant volume is given by
notation Cv:
vv
vT
Q
T
U C
=
=
Cv: is a function of temperature and of the specific
volume of the material
dT C m dU v= dT C Ud v=
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3001 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Heat Capacity at Constant PressureA material heated at constant pressure usually expands.
The internal energy added to the material is accounted for by the increase in the
internal energy of the material plus the work done by the material as it expands against
the constant pressure imposed on it.
Considering only mechanical work:
dUW Q =+At constant pressure (dP =0),
P dV = d(PV), so:
dUdV P - Q =
dV P dU Q +=
( ) dH VP Ud Q =+=
dT C m dH p=
dT C Hd p=
PP
pT
Q
T
H C
=
=
d(PV) = PdV + VdP
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3101 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Heat Capacity at Constant Pressure
Heat capacity are usually tabulated at constant pressure heat capacities, because most
heating or cooling of materials takes place at constant pressure ambient conditions.
Cp is a function of temperature and pressure.
The Cp values tabulated in the following slides are at one atmosphere (1 atm) pressure.
J/mol.K cT T b a C -2
p ++=
https://www.dream.co.id/
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3201 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Buku yang berisi data-data termodinamika
Kubaschewski, Alcock, Spencer, Materials
Thermochemistry edisi 6, 1993
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3301 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Heat Capacity at Constant Pressure
Kubaschewski, Alcock, Spencer, Materials
Thermochemistry edisi 6, 1993
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3401 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Heat Capacity at Constant Pressure
Kubaschewski, Alcock, Spencer, Materials
Thermochemistry edisi 6, 1993
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3501 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Heat Capacity at Constant Pressure
Kubaschewski, Alcock, Spencer, Materials
Thermochemistry edisi 6, 1993
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3601 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Heat Capacity at Constant Pressure
Kubaschewski, Alcock, Spencer, Materials
Thermochemistry edisi 6, 1993
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3701 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Panas Kerja dan Energi Dalam
Kubaschewski, Evans, Metallurgical
Thermochemistry, edisi 2, 1956
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Kubaschewski, Evans, Metallurgical
Thermochemistry, edisi 2, 1956
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3901 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Cara membaca konstanta di Tabel, misal untuk MgO: a = 42,59 b = 7,28 x 10-3 c = -6,19 x 105
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Cp
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FACTSAGE
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4201 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
FACTSAGEhttps://www.factsage.com/
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4301 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Differential Scanning Calorimetry (DSC)
https://www.netzsch-thermal-analysis.com/en/contract-testing/methods/differential-scanning-calorimetry/
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4401 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Differential Scanning Calorimetry (DSC)
https://www.mt.com/id/en/home/applications/Application_Browse_Laboratory_Analytics/Application_Browse_thermal_analysis/specific-heat-capacity-measurement.html
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4501 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Example of the use of heat capacities
Calculate the energy required and the cost of heating a slab of aluminium of mass one metric
ton (1000 kg) from 300 K to 800 K, a temperature that might be used to reduce the thickness of
the aluminium through rolling.
The aluminium will be heated by passing it though a furnace that uses electricity as its source
of energy. The cost of electrical energy is assumed to be Rp. 1000 per kWh (kilowatt-hour).
Assume that there are no extraneous heat losses from the furnace, all the electrical energy
entering the furnace is used to heat the aluminium.
Select the furnace as a system. Note that it is
an open system. Assume, that it is at steady
state.
U W Q m H - m H ooii =++
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4601 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Example of the use of heat capacities, cont.Note that the heat flow term is zero because the energy did not flow into the system
because of a temperature difference. Energy entered the system as work, because of an
electrical difference.
U W Q m H - m H ooii =++
( ) H - H m W ioAl=
( ) ( ) +==
800
300
3-
800
300
pio dT T10 x 12.38 20.67dT C H - H
( ) )300800(2
10 x 12.38 300)-20.67(800 H - H 22
-3
io −+=
Material a b x 103 c x 10-5 Range (K)
Al(s) 20.67 12.38 298 – 933
Al (l) 29.3 933 - 1273
𝐇𝐓 −𝐇T1 = න
T1
𝐓
(a + bT + cT−2 + dT𝟐) dT
𝐇𝐓 −𝐇T1 = a (T−T1) +𝐛
𝟐(𝑻𝟐 − 𝑻𝟏𝟐) − 𝒄
𝟏
𝑻−
𝟏
𝑻𝟏+𝐝
𝟑(𝐓𝟑−T1𝟑)
𝑪𝒑 = a + bT + cT−2 + dT𝟐
Melting temperature
Al = 660 C = 933 K
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4701 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Example of the use of heat capacities, cont.
( ) )300800(2
10 x 12.38 300)-20.67(800 H - H 22
-3
io −+=
( ) J/mol 3,7301 H - H io =
( ) J/kg 508,560 27
1000 x 3,7301 H - H io ==
kWh 141.4 kg 1000 x J/kg 508,560 W ==
-141,400. Rp. kWh / 1000 Rp. x kWh 141.4 Cost ==
1 J = 1 W.s
1 J = 1/3600 Wh
1 J = 1/(3600*1000) kWh
Ho − Hi = න
298
Tm
Cp,s dT+ Hm + න
Tm
T
Cp,l dT + …
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4801 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Perhitungan Menggunakan FACTSAGE
141.1 kWh
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4901 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Example of the use of heat capacities, cont.
Another way to approach the same problem is to select an aluminium slab as a system
(closed system).
12 U - U U W Q ==+
The work term is not zero, because aluminium expands upon heating.
)VV(PU - U Q 1212 −+=
)HH(mH - H Q 1212 −==
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5001 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
T. Rosenqvist, Principles of Extractive Metallurgy, 2004
Dihitung dengan
FACTSAGE
Ho − Hi = න
298
Tm
Cp,s dT + Hm + න
Tm
T
Cp,l dT + …
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5101 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Enthalpies of Formation
The same as internal energy, the enthalpy change of a material
between states 1 and 2 is given as:
Neither internal energy nor enthalpy can be defined in absolute terms.
Even at a temperature of absolute zero, a material may posses energy.
For convenience in tabulation, it is useful to define a reference state for a
material and to assign a value of zero to enthalpy for certain materials in
that reference state.
HHH 12 −=
Reference conditions: the elements in their equilibrium states at 298
K and one atmosphere pressure.
Example: enthalpy of diatomic oxygen at 298K and 1 atm pressure = 0.
Enthalpy of monoatomic oxygen is not zero, because the monoatomic
form is not the equilibrium form at 298 K and one atmosphere.
https://www.idntimes.com/travel
https://www.instagram.com/p/BvYwSXBg5lt/
A
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5201 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Enthalpies of Formation (cont.)
Another example: the enthalpy of carbon as graphite (not diamond) at 298K and one
atmosphere is taken as zero.
If the elements are assumed to have zero enthalpy when in their reference states, then
compounds must have some other value of enthalpy at the reference conditions.
System
boundary
C
O2
CO2
Q = -393.5 kJ/mol
P = 1 atm, T =298 K P = 1 atm, T =298 K
To illustrate, consider a system in which carbon (as graphite) and oxygen are
introduced into a steady state system to form carbon dioxide at 298 K and 1 atm.
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5301 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Enthalpies of Formation (cont.)
Because no work is done:
m H m H W Q io m
ii
m
oo −=+
m H m H Q io m
ii
m
oo −=
For the reaction: C + O2 = CO2
n H -n H n H Q 2222 OOCCCOCO −=
In such an experiment, the heat
transfered from the system would be
393.5 kJ per mole of carbon
introduced or mole of CO2 leaving the
system
H - H HkJ 5.393 Q 22 OCCO −=−=
HC = 0 and HO2 = 0, because
these elements are in their
reference state, therefore:
atm) 1 K, 298 (at kJ 5.393 H2CO −=
This is called the heat formation
of carbon dioxide (Hf).
Enthalpy change is negative, the
heat is evolved = exothermic
Enthalpy change is positive, heat
must be added to the process,
endothermic
n is the stoichiometric coefficient
of the reaction
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5401 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
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5501 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Enthalpy Changes in Chemical Reaction
Enthalpy change for a chemical reaction can be calculated from the heats of
formation of the compounds and element involved in the reaction.
Example: calculate the enthalpy change (at 298K) for the oxidation of
methane (CH4). The chemical equation:
CH4 (g) + 2 O2 (g) = CO2 (g) + 2 H2O (g)
The equation above can be written as the sum of three equations:
1. C + O2 = CO2 H = Hf (CO2)
2. 2H2 + O2 = 2 H2O H = 2Hf (H2O)
3. C + 2 H2 = CH4 H = Hf (CH4)
Hreaction = Hf (CO2) + 2 Hf (H2O) - Hf (CH4)
= -393.5 + 2(-241.8) – (-62.3) = -814.8 kJ at 298 K
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5601 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Eksotermik dan Endotermik
H < 0 → Reaksi eksotermik (menghasilkan panas)
H > 0 → Reaksi endotermik (membutuhkan panas)
https://teachsciencewithfergy.com
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5701 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Adiabatic Temperature Calculation
Nitrogen and carbon dioxide do not react.
Combustion reaction for the carbon monoxide:
CO + ½ O2 = CO2
Gas Moles in Moles outExit gas
composition (%)
CO 0.20 0
CO2 0.30 0.5 36
O2 0.10 0
N2 0.50 + (79/21)*0.1 0.88 64
Total moles in exit gas 1.38 100
Gas contains 20% CO, 30% CO2 and 50% N2. Caculate adibatic
temperature from gas burning!
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5801 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Adiabatic Temperature Calculation
For a steady state burner, ther first law
n H -n H W Q io m
ii
m
oo =+
W = 0 (no work is done on or by the burner.
Assume, no heat losses to the surroundings (adiabatic), aim to find the
highest temperature that can be reached.
n H n Hoi m
oo
m
ii =
H 88.0H 5.0n Hom
T,2NT,2COoo +=
H 88.0H 0.1 H 3.0H 2.0n H298,2N298,2O
m298,2CO298,COii
i
+++=
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5901 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Adiabatic Temperature Calculation
state) (reference 0 H and 0 H 298,2O298,2N ==
HN2,T= න
298
T
Cp,N2dT
HCO2,T= ΔHf,CO2,298
+ න
298
T
Cp,CO2dT
HCO2,298= ΔHf,CO2,298
HCO,298= ΔHf,CO,298
T flame = 1368K
= 1095 C
mo
Ho no = 0.5 HCO2,T+ 0.88 HN2,T
n H n Hoi m
oo
m
ii =
mi
Hi ni = 0.2 HCO,298 + 0.3 HCO2,298
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6001 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Perhitungan Menggunakan FACTSAGE
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6101 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Perhitungan Menggunakan FACTSAGE
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6201 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Flame Temperature
https://www.messergroup.com/ironandsteel/foundries/electricarcfurnace
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6301 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
LatihanGas metana (CH4) dibakar pada temperatur kamar (298 K dan 1 atm) dengan:
a. Oksigen (100% O2)
b. Udara (21% O2 dan 71% N2)
Hitung temperatur nyala api pada kondisi adiabatik!
Gas Masuk (mol) Keluar (mol)
CH4 1 0
O2 2 0
CO2 0 1
H2O 0 2
Total mol gas keluar 3
Kasus a: Reaksi: CH4 + 2O2 = CO2 + 2H2O
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6401 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
1. Pendahuluan, istilah-istilah dan notasi
2. Hukum I Termodinamika
3. Hukum II Termodinamika
4. Hubungan Besaran-Besaran Termodinamika
5. Kesetimbangan
6. Kesetimbangan Kimia dan Diagram Ellingham
7. Proses Elektrokimia dan Diagram Potensial - pH (Pourbaix)
8. Ujian Tengah Semester
9. Aktivitas Ion
10. Termodinamika Larutan
11. Penggunaan Persamaan Gibbs - Duhem
12. Penggunaan Metoda Elektrokimia untuk menentukan Sifat-Sifat / Besaran-Besaran
Termodinamika
13. Keadaan Standar Alternatif
14. Koefisien Aktivitas dalam Larutan Encer Multi-Komponen
15. Diagram Fasa
16. Ujian Akhir Semester
Materi Perkuliahan
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6501 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Terima kasih!
Zulfiadi Zulhan
Program Studi Teknik Metalurgi
Fakultas Teknik Pertambangan dan Perminyakan
Institut Teknologi Bandung
Jl. Ganesa No. 10
Bandung 40132
INDONESIA
www.metallurgy.itb.ac.id