Fysik - Orbit
Opgave O4.4
V = 250L =0,25m^3
V=250L=0,25m3
T_{luft-ind}=65^{\circ}=338K
Tluft−ind=65∘=338K
T_{luft-ud}=10^{\circ}=283K
Tluft−ud=10∘=283K
p=102,4kPa=102400Pa
p=102,4kPa=102400Pa
Opgave O4.4
\rho=\frac{M}{R}\cdot\frac{p}{T}
ρ=RM⋅Tp
M=\frac{\rho\cdot R\cdot T}{p}
M=pρ⋅R⋅T
\rho_{luft}=1,29\frac{kg}{m^3}=densitet-ved-0-grader
ρluft=1,29m3kg=densitet−ved−0−grader
Opgave O4.4
F_{op}=\rho_{luft}\cdot V \cdot g
Fop=ρluft⋅V⋅g
F_{t}=m\cdot g=\rho_{365K}\cdot V \cdot g
Ft=m⋅g=ρ365K⋅V⋅g
m=\frac{F_t}{g}
m=gFt
Opgave O4.9
T_{start} = 10^{\circ}=283K
Tstart=10∘=283K
T_{slut} = 45^{\circ}=318K
Tslut=45∘=318K
p = 255kPa=255000Pa
p=255kPa=255000Pa
Beregn ny dæktryk
Opgave O4.9
\frac {p_1}{T_1}=\frac {p_2}{T_2}
T1p1=T2p2
Gay-Lussacs lov
\frac {255000Pa}{283K}=\frac {p}{318K}
283K255000Pa=318Kp
p = 2,865371025\cdot 10^5
p=2,865371025⋅105
Opgave O4.10
T_{start} = 16,1^{\circ}=289,1K
Tstart=16,1∘=289,1K
T_{slut} = 78,2^{\circ}=351,2K
Tslut=78,2∘=351,2K
p = 101kPa=101000Pa
p=101kPa=101000Pa
V=1,0L=0,001m^3
V=1,0L=0,001m3
Opgave O4.10
Beregn omgivelsernes stempelarbejde
A=-p\cdot \Delta V
A=−p⋅ΔV
Charles' lov
\frac {V_1}{T_1}=\frac {V_2}{T_2}
T1V1=T2V2
\frac {0,002m^3}{289,1K}=\frac {V}{351,2K}
289,1K0,002m3=351,2KV
V=0,001214804566m^3
V=0,001214804566m3
Opgave O4.10
A=-p\cdot \Delta V
A=−p⋅ΔV
\Delta V=0,001-0,001214804566
ΔV=0,001−0,001214804566
A=-101000Pa\cdot -0,000214804566
A=−101000Pa⋅−0,000214804566
\Delta V=-0,000214804566m^3
ΔV=−0,000214804566m3
A=21,7J
A=21,7J
Opgave O4.10
Beregn Q, den varme som luften tilføres
\rho_{luft}=1,29\frac {kg}{m^3}
ρluft=1,29m3kg
Q=m\cdot c\cdot \Delta T
Q=m⋅c⋅ΔT
m=\rho_{luft}\cdot V=1,29\frac {kg}{m^3}\cdot 0,001m^3=0,00129kg
m=ρluft⋅V=1,29m3kg⋅0,001m3=0,00129kg
\Delta T=351,2K-289,1K=62,1K
ΔT=351,2K−289,1K=62,1K
c=1,00\frac{kJ}{kg\cdot K}=1000\frac{J}{kg\cdot K}
c=1,00kg⋅KkJ=1000kg⋅KJ
Q=0,00129kg\cdot 1000\frac{J}{kg\cdot K}\cdot 62,1K=80,1J
Q=0,00129kg⋅1000kg⋅KJ⋅62,1K=80,1J
Opgave O4.10
Beregn tilvæksten af den indre energi
\Delta E_{indre}=Q+A
ΔEindre=Q+A
\Delta E_{indre}=80,1J+21,7J=102J
ΔEindre=80,1J+21,7J=102J
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