first law of thermodynamics

1. Ahmed Medhat - Thermodynamics 1 - ahmedhatfa@hotmail.com
First law of
thermodynamics

2. Ahmed Medhat - Thermodynamics 1 - ahmedhatfa@hotmail.com
First law of thermodynamics ,also known as the conservation of energy principle, studying the relationships
among the various forms of energy and energy interactions.
‫ألخري‬ ‫صورة‬ ‫من‬ ‫تتحول‬ ‫لكن‬ ‫و‬ ‫العدم‬ ‫من‬ ‫تستحدث‬ ‫ال‬ ‫و‬ ‫تفني‬ ‫ال‬ ‫الطاقة‬
(mgΔz)
𝑬𝟏 + 𝑬𝒊𝒏
=
𝑬𝟐 + 𝑬𝒐𝒖𝒕
State
1
State
2
‫خالل‬ ‫الطاقة‬ ‫محصلة‬
‫خالل‬
‫المسار‬
=
‫الثانية‬ ‫و‬ ‫األولي‬ ‫الحالة‬ ‫بين‬ ‫خالل‬ ‫الطاقة‬

3. Ahmed Medhat - Thermodynamics 1 - ahmedhatfa@hotmail.com
Energy (E)
Heat (Q)
𝐻2 − 𝐻1
𝐻𝑔 − 𝐻𝑓
= 𝐻𝑓𝑔
m*C*ΔT
m*C*ΔT
m*C*ΔT
ሶ
𝑚 * H.V
Work (W)
T * 𝜔
1
2
𝑘(x2
2
− 𝑥1
2
)
ሶ
𝑊 = 𝑉 ∗ 𝐼
Internal
Energy (U)
Kinetic Energy
(KE)
1
2
𝑚𝑉2
Potential
Energy (PE)
M*g*z
Moving Flow
Work
P*V
▪Polytropic process
▪Isentropic process
▪Isothermal process
▪Isobaric process
▪Isochoric process

4. Ahmed Medhat - Thermodynamics 1 - ahmedhatfa@hotmail.com
𝐻5 − 𝐻4
Enthalpy

5. Ahmed Medhat - Thermodynamics 1 - ahmedhatfa@hotmail.com
Specific Heat ( C )
𝒌𝑱
𝑲𝒈 ∗𝑲
‫طاقة‬ ‫مخزن‬
𝒅𝒒
𝒅𝑻
𝑸 = 𝒎 ∗ 𝑪 ∗ ΔT
Cv
constant Volume
Cp
constant Pressure
Cv = Cp=C
solids & liquids

6. Ahmed Medhat - Thermodynamics 1 - ahmedhatfa@hotmail.com
Energy (E)
Heat (Q)
𝐻2 − 𝐻1
𝐻𝑔 − 𝐻𝑓
= 𝐻𝑓𝑔
m*C*ΔT
m*Cv*ΔT
m*Cp*ΔT
ሶ
𝑚 * H.V
Work (W)
T * 𝜔
1
2
𝑘(x2
2
− 𝑥1
2
)
ሶ
𝑊 = 𝑉 ∗ 𝐼
Internal
Energy (U)
Kinetic Energy
(KE)
1
2
𝑚𝑉2
Potential
Energy (PE)
M*g*z
Moving Flow
Work
P*V
▪Polytropic process
▪Isentropic process
▪Isothermal process
▪Isobaric process
▪Isochoric process

7. Ahmed Medhat - Thermodynamics 1 - ahmedhatfa@hotmail.com
First law of thermodynamics ,also known as the conservation of energy principle, studying the relationships
among the various forms of energy and energy interactions.
‫ألخري‬ ‫صورة‬ ‫من‬ ‫تتحول‬ ‫لكن‬ ‫و‬ ‫العدم‬ ‫من‬ ‫تستحدث‬ ‫ال‬ ‫و‬ ‫تفني‬ ‫ال‬ ‫الطاقة‬
(mgΔz)
𝑬𝟏 + 𝑬𝒊𝒏
=
𝑬𝟐 + 𝑬𝒐𝒖𝒕
State
1
State
2
Heat (Q)
Work (W)

8. Energy (E)
Heat (Q)
𝐻2 − 𝐻1
𝐻𝑔 − 𝐻𝑓
= 𝐻𝑓𝑔
m*C*ΔT
m*Cv*ΔT
m*Cp*ΔT
ሶ
𝑚 * H.V
Work (W)
T * 𝜔
1
2
𝑘(x2
2
− 𝑥1
2
)
ሶ
𝑊 = 𝑉 ∗ 𝐼
Internal
Energy (U)
Kinetic Energy
(KE)
1
2
𝑚𝑉2
Potential
Energy (PE)
M*g*z
Moving Flow
Work
P*V
Ahmed Medhat - Thermodynamics 1 - ahmedhatfa@hotmail.com
𝑬𝟏 + 𝑸 = 𝑬𝟐 + 𝑾
𝑸 − 𝑾 = 𝑬𝟐 − 𝑬𝟏
𝑬𝟏 + 𝑬𝒊𝒏= 𝑬𝟐 + 𝑬𝒐𝒖𝒕
Heat (Q)
Work (W)

9. Ahmed Medhat - Thermodynamics 1 - ahmedhatfa@hotmail.com
A rigid tank contains a hot fluid that is cooled while being stirred by a paddle wheel. Initially the internal
energy of the fluid is 800 KJ. During the cooling process, the fluid losses 500 KJ of heat, and the paddle
wheel does 100 KJ of work on the fluid. Determine the final internal energy of the fluid.
Neglect the energy stored in the paddle wheel.
100KJ – 500KJ = 𝑼𝟐 – 800 KJ
Solution
𝑸 − 𝑾 = (𝑼𝟐 − 𝑼𝟏)
(-value)
(-value)
𝑼𝟐 = 400 KJ
(+value)
(+value)
(-value)
(-value)