02. Hukum I Termodinamika

02. Hukum I Termodinamika

Zulfiadi Zulhan

Teknik Metalurgi

Fakultas Teknik Pertambangan dan Perminyakan Institut Teknologi Bandung

INDONESIA

Termodinamika Metalurgi (MG-2112)

Jangan Mengunggah

Materi Kuliah ini di

INTERNET!

NO TALKING

NO SLEEPING

NO MOBILE PHONE

http://www.longestlife.com https://www.pinterest.com https://www.pinterest.se http://clipart-library.com

Hukum Pertama Termodinamika

Kekekalan 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.thinkgeoenergy.com/video-overview-on-the-sarulla- geothermal-power-plants-in-indonesia/

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/10

https://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

Eksotermik dan Endotermik

H < 0 → Reaksi eksotermik (menghasilkan panas)

H > 0 → Reaksi endotermik (membutuhkan panas)

https://teachsciencewithfergy.com

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)

Panas dan Kerja

U Q

U

U

U U

Q

W

U U

U

• 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.

𝛅𝐐

_{𝐥𝐢𝐧𝐠𝐤𝐮𝐧𝐠𝐚𝐧}

≅ +𝐝U

_{𝐬𝐢𝐬𝐭𝐞𝐦}

𝛅𝐖

_{𝐥𝐢𝐧𝐠𝐤𝐮𝐧𝐠𝐚𝐧}

≅ +𝐝U

_{𝐬𝐢𝐬𝐭𝐞𝐦}

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.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“:

U d W

Q +  =



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)

U d W

Q +  = + +



„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

+

^).

Lingkungan Lingkungan

𝛅𝐐 + 𝛅𝐖 = 𝐝U

𝐁𝐞𝐛𝐞𝐫𝐚𝐩𝐚 𝐛𝐮𝐤𝐮 𝐦𝐞𝐧𝐮𝐥𝐢𝐬:

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)

U d W

Q +  = + +



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).

/kg m

dalam Spesifik

Volume V

m V V

=

3

=

Densitas dari sebuah material adalah kebalikan

dari volume spesifik

Heat, Work and Internal Energy (First Low Energy:

Open System)

V d P work

flow

V

0

i

i

⁼ 

„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

m_i

V

ⁱ

= V

ⁱ

m

ⁱ

or dV = V

ⁱ

dm

ⁱ

Hukum I Termodinamika: Sistem Terbuka (OPEN SYSTEM)

V d P work

flow

V

0

i

i

⁼ 

„i“ is designated for entering material, and „o“ for leaving material.

System boundary

Flow work in an open system P

m_i

V

ⁱ

= V

ⁱ

m

ⁱ

or dV = V

ⁱ

dm

ⁱ

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

Heat, Work and Internal Energy (First Low Energy:

Open System) cont.

V d P work

flow

V

0

i

i

⁼ 

„i“ is designated for entering material, and

„o“ for leaving material.

Omitting kinetic and potential energy, first law of thermdynamic for „ Open System“:

i i

i i

i

V m or dV V dm

V = =

i i m

0

i i

i

P V d m P V m

work

flow =  =

m V

P work)

flow

(

_i

=

_i



_i

In differential form:



dU W

Q work)

flow (

work) flow

( m

U - m

U

_i



_{i o}



_o

+ 

_i

− 

_o

+  +  =

( ^U

_i

^{+ P}

_i

^V

_i

) ^{ m}

_i

^- ( ^U

_o

^{+ P}

_o

^V

_o

) ^{ m}

_o

^{+  Q +  W = dU}

Heat, Work and Internal Energy (First Low Energy:

Open System) cont.

U_i m_i and U_o m_o represent the internal energy carried into and out of the system by the matter entering and leaving of the system.

P_i V_i m_i and P_o V_o m_o are the flow work into and out of the system by the matter entering and leaving of the system

U

_i

+ P

_i

V

_i

δm

_i

− U

_o

+ P

_o

V

_o

δm

_o

+ δQ + δW = dU

( ^{U P V} ) ^{m -} ( ^{U P V} ) ^{m Q W dU}

o

i m

o o o o

m

i i i

i

+   +  +  +  =



For all streams entering or leaving the system:

Entalpi

U + PV didefinisikan sebagai Entalpi (H)

PV U

H  +

https://webmuda.files.wordpress.com https://webmuda.files.wordpress.com

𝐔

_𝐢

+ P

_𝐢

𝐕

_𝐢

δm

_i

− 𝐔

_𝐨

+ P

_𝐨

𝐕

_𝐨

δm

_o

+ δQ + δW = dU

Enthalpy

U + PV is defined as Enthalpy (H)

PV U

H  +

Specific enthalpy is enthalpy per unit mass

V P U

H = +

(

^{H KE PE}

)

^{m -}

(

^{H KE PE}

)

^{m Q W d(U PE KE)}

o

i m

o o o

o m

i i i

i + + 



+ +  +  +  = + +



Consider potential and kinetic energy, First Law of Thermodynamic for open system:

For closed system: m_i and  m_o = 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

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

(

^{H KE PE}

)

^{m -}

(

^{H KE PE}

)

^{m Q W 0}

o

i m

o o

o o

m

i i

i

i + + 



+ +  +  +  =



Steady State, cont.

First law of thermodynamic:

(

^{H KE PE}

)

^{m -}

(

^{H KE PE}

)

^{m Q W 0}

o

i m

o o

o o

m

i i

i

i + + 



+  +  +  =



For a finite process, it can be expressed as follows:

(

^{H KE PE}

)

^{m -}

(

^{H KE PE}

)

^m

W Q

i

o m

i i i

i m

o o o

o





^{+ + +}

= +

When no changes in kinetic or potential energy:

m H -

m H W

Q

i

o m

i i m

o o





= +

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 m d

dT dU

C = =

where C is the Heat capacity

Heat capacity at constant volume is given by notation C_v:

v v

v

T

Q T

C U 



 







= 

 

 







= 

C_v: is a function of temperature and of the specific volume of the material

dT C

m

dU =

_v

d U = C

_v

dT

Heat Capacity at Constant Pressure

A 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:

dU W

Q +  =



At constant pressure (dP =0),

P dV = d(PV), so:

dU dV

P -

Q =



dV P

dU

Q = +



( ^{U P V} ) ^dH

d

Q = + =



dT C

m dH =

_p

dT C

H

d =

_p

P P

p

T

Q T

C H 



 







= 

 

 







= 

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.

C_p is a function of temperature and pressure.

The C_p values tabulated in the following slides are at one atmosphere (1 atm) pressure.

J/mol.K cT

T b a

C

_p

= + +

^-2

Buku yang berisi data-data termodinamika

Kubaschewski, Alcock, Spencer, Materials Thermochemistry edisi 6, 1993

Heat Capacity at Constant Pressure

Kubaschewski, Alcock, Spencer, Materials Thermochemistry edisi 6, 1993

Heat Capacity at Constant Pressure

Kubaschewski, Alcock, Spencer, Materials Thermochemistry edisi 6, 1993

Heat Capacity at Constant Pressure

Kubaschewski, Alcock, Spencer, Materials Thermochemistry edisi 6, 1993

Heat Capacity at Constant Pressure

Kubaschewski, Alcock, Spencer, Materials Thermochemistry edisi 6, 1993

Panas Kerja dan Energi Dalam

Kubaschewski, Evans, Metallurgical

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

FACTSAGE

https://www.factsage.com/

Differential Scanning Calorimetry (DSC)

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

_{i i o o}

+ + = 

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

_{i i o o}

+ + = 

( ^{H - H} )

m

W =

_{Al o i}

( ) ⁼ _ ⁼

⁸⁰⁰

_ ( ⁺ )

300

3 - 800

300 p i

o

- H C dT 20.67 12.38 x 10 T dT H

( ) ^{( 800 300 )}

2

10 x 12.38 300)

- 20.67(800 H

-

H

^{2 2}

-3 i

o

= + −

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

𝐇_𝐓 − 𝐇T1 = න T1

𝐓

(a + bT + cT−2 + dT^𝟐) dT

𝐇_𝐓 − 𝐇T1 = a (T−T1) + 𝐛

𝟐(𝑻^𝟐 − 𝑻𝟏^𝟐) − 𝒄 𝟏

𝑻 − 𝟏

𝑻𝟏 +𝐝

𝟑(𝐓^𝟑−T1^𝟑)

𝑪_𝒑 = a + bT + cT−2 + dT^𝟐

Melting temperature Al = 660 C = 933 K

Example of the use of heat capacities, cont.

( ) ^{( 800 300 )}

2

10 x 12.38 300)

- 20.67(800 H

-

H

^{2 2}

-3 i

o

= + −

( ^H

_o

^{- H}

_i

) ^{= 1 3,730 J/mol}

( ) ^{508,560 J/kg}

27 x 1000 3,730

1 H

-

H

_{o i}

= =

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

H_o − H_i = න 298

T_m

C_p,s dT + H_m + න

T_m

T

C_p,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).

1 2

- U U

U W

Q + =  =

The work term is not zero, because aluminium expands upon heating.

) V V

( P U

- U

Q =

_{2 1}

+

₂

−

₁

) H H

( m H

- H

Q =

_{2 1}

=

₂

−

₁

T. Rosenqvist, Principles of Extractive Metallurgy, 2004

Dihitung dengan FACTSAGE H_o − H_i = න

298

T_m

C_p,s dT + H_m + න

T_m

T

C_p,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.

H H

H =

₂

−

₁



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

A

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

i

o m

i i m

o

o





⁻

= +

m H

m H Q

i

o m

i i m

o

o





⁻

=

For the reaction: C + O₂ = CO₂

n H

- n H n

H

Q =

CO2 CO2

−

C C O2 O2

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 H

kJ 5 . 393

Q = − =

CO2

−

C O2

H_C = 0 and H_O2 = 0, because these elements are in their reference state, therefore:

atm) 1

K, 298 (at

kJ 5 . 393 HCO2 = −

This is called the heat formation of carbon dioxide (H_f).

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 = H_f (CO₂)

2. 2H₂ + O₂ = 2 H₂O H = 2H_f (H₂O)

3. C + 2 H₂ = CH₄ H = H_f (CH₄)

H_reaction = H_f (CO₂) + 2 H_f (H₂O) - H_f (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 out Exit 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!

Adiabatic Temperature Calculation

For a steady state burner, ther first law n

H -

n H W

Q

i

o m

i i m

o

o





= +

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

H

o

i m

o o m

i

i





⁼

H 88 . 0 H

5 . 0 n

H

mo o o CO2,T N2,T



^{= +}

H

88 . 0 H

0.1 H

3 . 0 H

2 . 0 n

H O2,298 N2,298

m i i CO,298 CO2,298

i

+ +

+



=

Adiabatic Temperature Calculation

state) (reference

0 H

and 0

H

N2,298

=

O2,298

=

H_N2,

T = න

298

T

Cp,N₂ dT H_CO2,

T = ΔH_f,CO2,

298 + න 298

T

Cp,CO₂ dT

H_CO2,

298 = ΔH_f,CO2,

298

H_CO,

298 = ΔH_f,CO,

298

T _flame = 1368K

= 1095 C

෍

m_o

H_o n_o = 0.5 H_CO2,

T + 0.88 H_N2,

T

n H n

H

o

i m

o o m

i i





⁼

෍

m_i

H_i n_i = 0.2 H_CO,298 + 0.3 H_CO2,

298

Perhitungan Menggunakan FACTSAGE

Perhitungan Menggunakan FACTSAGE

Flame Temperature

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

Latihan

Gas 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 zulfiadi.zulhan@gmail.com