Tres exempla quaestionum de expansione areae
1. Lamina ferrea temperatura 20 ° C longitudinem 50 cm et latitudinem 30 cm habet. Si coefficiens expansionis linearis ferri 10⁻⁵ ° C⁻¹ est , tum incrementum areae et areae totalis temperatura 60 ° C est…
Disputatio
Constat:
Temperatura initialis (T1 ) = 20 ° C
Temperatura finalis (T² ) = 60 ° C
Mutatio temperaturae (ΔT ) = 60 ° C – 20 ° C = 40 ° C
Area initialis (A1 ) = Longitudo × latitudo = 50 cm × 30 cm = 1500 cm²
Coefficiens expansionis linearis chalybis ( α ) = 10⁻⁵ ° C⁻¹
Coefficiens expansionis chalybis (β ) = 2a = 2 × 10⁻⁵ ° C⁻¹
Quaestio: Incrementum areae ( ΔA )
Incrementum areae (ΔA ) :
ΔA = βA 1ΔT
ΔA = ( 2 × 10⁻⁵ ° C⁻¹ ) (1500 cm² )(40 ° C )
A = (80 x 10 -5 ) (1500 cm 2 )
ΔA = 1 × 20 000 × 10⁻⁵ cm²
ΔA = 1.2 × 10⁻⁵ × 10⁻⁵ cm²
ΔA = 1.2 cm²
Area totalis ( A2 ) :
A² = A¹ + ΔA
A² = 1500 cm² + 1.2 cm²
A² = 1501 , 2 cm²
2. Lamina aluminii cum coefficiente expansionis linearis 24 × 10⁻⁶ / ° C aream 40 cm⁻² habet ad temperaturam 30 ° C. Determina temperaturam finalem si area laminae aluminii ad 40,2 cm⁻² crescit.
Disputatio
Constat:
Temperatura initialis (T1 ) = 30 ° C
Coefficiens expansionis linearis aluminii (α ) = 24 × 10⁻⁶ ° C⁻¹
Coefficiens expansionis areae aluminii (β ) = 2α = 2 × 24 × 10⁻⁶ ° C⁻¹ = 48 × 10⁻⁶ ° C⁻¹
Area initialis (A1 ) = 40 cm2
Area finalis (A² ) = 40,2 cm²
Mutatio areae (ΔA ) = 40,2 cm² – 40 cm² = 0,2 cm²
Quaestio: Temperaturam finalem (T² ) determina .
Responsum:
Formula mutationis areae (ΔA ) :
ΔA = βA1ΔT
Temperatura finalis (T² ) :
ΔA = βA₁ ( T₂ – T₁ )
0,2 cm2 = (48 × 10-6 oC-1)(40 cm2)(T2 - 30oC)
0,2 = (1920 × 10⁻⁶ ) ( T² – 30 )
0,2 = (1,920 × 10⁻³ ) (T² – 30 )
0,2 = (2 × 10⁻³ ) (T² – 30 )
0,2 / (2 × 10⁻³ ) = T² – 30
0,1 × 10³ = T² – 30
1 × 10² = T² – 30
100 = T2 – 30
100 + 30 = T²
T² = 130
Temperatura finalis = 130 ° C
3. Res circularis, temperatura 20 ° C, radium 20 cm habet. Si temperatura 100 ° C radius ad 20,5 cm crescit, tum coefficiens expansionis linearis rei est…
Disputatio
Constat:
Temperatura initialis (T1 ) = 30 ° C
Temperatura finalis (T² ) = 100 ° C
Mutatio temperaturae (ΔT ) = 100 ° C – 30 ° C = 70 ° C
Radius initialis (r₁ ) = 20 cm
Radius finalis (r² ) = 20,5 cm
Quaestio: Coefficiens expansionis areae obiecti ( b )
Responsum:
Area initialis circuli (A₁ ) = π r₁² = ( 3,14 )(20 cm) ² = (3,14)(400 cm² ) = 1256 cm²
Area finalis circuli (A² ) = π /r²² = ( 3,14 )(20,5 cm²) ² = (3,14)(420,25 cm² ) = 1319,585 cm²
Incrementum areae circuli (Δ A) = 1319,585 cm² – 1256 cm² = 63,585 cm²
Formula mutationis areae (ΔA ) :
ΔA = βA 1 ΔT
Coefficiens expansionis areae:
ΔA = βA 1 ΔT
63,585 cm 2 = β (1 256 cm 2 )(70 o C)
63,585 = β ( 87920 ° C)
β = 63,585 / 87920 ° C
β = 0,00072 / ° C
β = 7,2 × 10⁻⁴ / ° C
β = 7,2 × 10⁻⁴ ° C⁻¹
Coefficiens expansionis linearis (α ):
β = 2α
α = β / 2
α = (7,2 × 10-4 ) / 2
α = 3,6 × 10⁻⁴ ° C⁻¹
