02. Hukum I Termodinamika

02. Hukum I Termodinamika

Zulfiadi Zulhan

Teknik MetalurgiFakultas Teknik Pertambangan dan PerminyakanInstitut Teknologi BandungINDONESIA

Termodinamika Metalurgi (MG-2112)

201 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021

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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.


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


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 /


Energi

https://www.lenntech.com/greenhouse-effect/fossil-fuels.htm

https://biology-for-all.weebly.com/carbon-cycle.html

http://www.fossilfuelconnections.org/coal


Eksotermik dan Endotermik

H < 0 → Reaksi eksotermik (menghasilkan panas)

H > 0 → Reaksi endotermik (membutuhkan panas)

https://teachsciencewithfergy.com


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


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


https://www.sciencealert.com/this-awesome-periodic-table-shows-the-origins-of-every-atom-in-your-body


http://new-universe.org/zenphoto/albums/Chapter3/Illustrations/Abrams24.jpg


Mesin Uap (Sumber Energi = Panas + Kerja)

https://www.youtube.com/watch?v=9mhYnQGZJuM


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


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


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


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.


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


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

=

=


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


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 ==


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


http://www.bintang.comhttp://www.startribune.com


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“:


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 =++++


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:


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


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

http://en.wikipedia.org/wiki/Energy

http://en.wikipedia.org/wiki/Thermodynamic_system

http://en.wikipedia.org/wiki/Internal_energy

http://en.wikipedia.org/wiki/Environment_(systems)

http://en.wikipedia.org/wiki/Volume

http://en.wikipedia.org/wiki/Pressure


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 =++++++


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 =+


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=


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


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/

https://www.dream.co.id/


Buku yang berisi data-data termodinamika

Kubaschewski, Alcock, Spencer, Materials

Thermochemistry edisi 6, 1993


















Panas Kerja dan Energi Dalam

Kubaschewski, Evans, Metallurgical

Thermochemistry, edisi 2, 1956


Kubaschewski, Evans, Metallurgical

Thermochemistry, edisi 2, 1956


Cara membaca konstanta di Tabel, misal untuk MgO: a = 42,59 b = 7,28 x 10-3 c = -6,19 x 105


Cp


FACTSAGE


FACTSAGEhttps://www.factsage.com/


Differential Scanning Calorimetry (DSC)

https://www.netzsch-thermal-analysis.com/en/contract-testing/methods/differential-scanning-calorimetry/


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


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 =++


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


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 + …


Perhitungan Menggunakan FACTSAGE

141.1 kWh


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 −==


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 + …


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.

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A

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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.


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



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


Eksotermik dan Endotermik

H < 0 → Reaksi eksotermik (menghasilkan panas)

H > 0 → Reaksi endotermik (membutuhkan panas)

https://teachsciencewithfergy.com


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!



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

+++=



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






Flame Temperature

https://www.messergroup.com/ironandsteel/foundries/electricarcfurnace

https://www.messergroup.com/ironandsteel/foundries/electricarcfurnace


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


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


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

[email protected]

Documents

02. Hukum I Termodinamika