Fysik - Orbit

Opgave O4.4

V = 250L =0,25m^3

V = 250 L = 0, 25 m ​ 3 ​ ​

T_{luft-ind}=65^{\circ}=338K

T ​ l u f t - i n d ​ ​ = 6 5 ​ \circ ​ ​ = 338 K

T_{luft-ud}=10^{\circ}=283K

T ​ l u f t - u d ​ ​ = 1 0 ​ \circ ​ ​ = 283 K

p=102,4kPa=102400Pa

p = 102, 4 k P a = 102400 P a

Opgave O4.4

\rho=\frac{M}{R}\cdot\frac{p}{T}

ρ = \frac{​ M ​ ​}{​ R ​} \cdot \frac{​ p ​ ​}{​ T ​}

M=\frac{\rho\cdot R\cdot T}{p}

M = \frac{​ ρ \cdot R \cdot T ​ ​}{​ p ​}

\rho_{luft}=1,29\frac{kg}{m^3}=densitet-ved-0-grader

ρ ​ l u f t ​ ​ = 1, 29 \frac{​ k g ​ ​}{​ m ​ 3 ​ ​ ​} = d e n s i t e t - v e d - 0 - g r a d e r

Opgave O4.4

F_{op}=\rho_{luft}\cdot V \cdot g

F ​ o p ​ ​ = ρ ​ l u f t ​ ​ \cdot V \cdot g

F_{t}=m\cdot g=\rho_{365K}\cdot V \cdot g

F ​ t ​ ​ = m \cdot g = ρ ​ 365 K ​ ​ \cdot V \cdot g

m=\frac{F_t}{g}

m = \frac{​ F ​ t ​ ​ ​ ​}{​ g ​}

Opgave O4.9

T_{start} = 10^{\circ}=283K

T ​ s t a r t ​ ​ = 1 0 ​ \circ ​ ​ = 283 K

T_{slut} = 45^{\circ}=318K

T ​ s l u t ​ ​ = 4 5 ​ \circ ​ ​ = 318 K

p = 255kPa=255000Pa

p = 255 k P a = 255000 P a

Beregn ny dæktryk

Opgave O4.9

\frac {p_1}{T_1}=\frac {p_2}{T_2}

\frac{​ p ​ 1 ​ ​ ​ ​}{​ T ​ 1 ​ ​ ​} = \frac{​ p ​ 2 ​ ​ ​ ​}{​ T ​ 2 ​ ​ ​}

Gay-Lussacs lov

\frac {255000Pa}{283K}=\frac {p}{318K}

\frac{​ 2 5 5 0 0 0 P a ​ ​}{​ 2 8 3 K ​} = \frac{​ p ​ ​}{​ 3 1 8 K ​}

p = 2,865371025\cdot 10^5

p = 2, 865371025 \cdot 1 0 ​ 5 ​ ​

Opgave O4.10

T_{start} = 16,1^{\circ}=289,1K

T ​ s t a r t ​ ​ = 16, 1 ​ \circ ​ ​ = 289, 1 K

T_{slut} = 78,2^{\circ}=351,2K

T ​ s l u t ​ ​ = 78, 2 ​ \circ ​ ​ = 351, 2 K

p = 101kPa=101000Pa

p = 101 k P a = 101000 P a

V=1,0L=0,001m^3

V = 1, 0 L = 0, 001 m ​ 3 ​ ​

Opgave O4.10

Beregn omgivelsernes stempelarbejde

A=-p\cdot \Delta V

A = - p \cdot Δ V

Charles' lov

\frac {V_1}{T_1}=\frac {V_2}{T_2}

\frac{​ V ​ 1 ​ ​ ​ ​}{​ T ​ 1 ​ ​ ​} = \frac{​ V ​ 2 ​ ​ ​ ​}{​ T ​ 2 ​ ​ ​}

\frac {0,002m^3}{289,1K}=\frac {V}{351,2K}

\frac{​ 0 , 0 0 2 m ​ 3 ​ ​ ​ ​}{​ 2 8 9 , 1 K ​} = \frac{​ V ​ ​}{​ 3 5 1 , 2 K ​}

V=0,001214804566m^3

V = 0, 001214804566 m ​ 3 ​ ​

Opgave O4.10

A=-p\cdot \Delta V

A = - p \cdot Δ V

\Delta V=0,001-0,001214804566

Δ V = 0, 001 - 0, 001214804566

A=-101000Pa\cdot -0,000214804566

A = - 101000 P a \cdot - 0, 000214804566

\Delta V=-0,000214804566m^3

Δ V = - 0, 000214804566 m ​ 3 ​ ​

A=21,7J

A = 21, 7 J

Opgave O4.10

Beregn Q, den varme som luften tilføres

\rho_{luft}=1,29\frac {kg}{m^3}

ρ ​ l u f t ​ ​ = 1, 29 \frac{​ k g ​ ​}{​ m ​ 3 ​ ​ ​}

Q=m\cdot c\cdot \Delta T

Q = m \cdot c \cdot Δ T

m=\rho_{luft}\cdot V=1,29\frac {kg}{m^3}\cdot 0,001m^3=0,00129kg

m = ρ ​ l u f t ​ ​ \cdot V = 1, 29 \frac{​ k g ​ ​}{​ m ​ 3 ​ ​ ​} \cdot 0, 001 m ​ 3 ​ ​ = 0, 00129 k g

\Delta T=351,2K-289,1K=62,1K

Δ T = 351, 2 K - 289, 1 K = 62, 1 K

c=1,00\frac{kJ}{kg\cdot K}=1000\frac{J}{kg\cdot K}

c = 1, 00 \frac{​ k J ​ ​}{​ k g \cdot K ​} = 1000 \frac{​ J ​ ​}{​ k g \cdot K ​}

Q=0,00129kg\cdot 1000\frac{J}{kg\cdot K}\cdot 62,1K=80,1J

Q = 0, 00129 k g \cdot 1000 \frac{​ J ​ ​}{​ k g \cdot K ​} \cdot 62, 1 K = 80, 1 J

Opgave O4.10

Beregn tilvæksten af den indre energi

\Delta E_{indre}=Q+A

Δ E ​ i n d r e ​ ​ = Q + A

\Delta E_{indre}=80,1J+21,7J=102J

Δ E ​ i n d r e ​ ​ = 80, 1 J + 21, 7 J = 102 J