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Linear Thermal Expansion (in one dimension)
1) The change in length ΔL is proportional to the original length L, and the change in temperature ΔT : ΔL = αLΔT, where ΔL is the change in length , and α is the coefficient of linear expansion.
a) The main span of San Francisco’s Golden Gate Bridge is 1275 m long at its coldest (–15ºC). The coefficient of linear expansion, α , for steel is 12×10−6 /ºC. When the temperatures rises to 25 °C, what is its change in length in meters?
2) The change in volume ΔV is very nearly ΔV ≈ 3αVΔT . This equation is usually written as ΔV = βVΔT, where β is the coefficient of volume expansion and β ≈ 3α . V is the original volume. ΔT is the change in temperature. Suppose your 60.0-L (15.9-gal) steel gasoline tank is full of gasoline, and both the tank and the gasoline have a temperature of 15.0ºC . The coefficients of volume expansion, for gasoline is βgas = 950×10−6 /ºC , for the steel tank is βsteel = 35×10−6 /ºC .
a) What is the change in volume (in liters) of the gasoline when the temperature rises to 25 °C in L?
b) What is the change in volume (in liters) of the tank when the temperature rises to 25 °C in L?
c) How much gasoline would be spilled in L?

Respuesta :

Answer:

1) [tex]\Delta L= 0.612\ m[/tex]

2) a. [tex]\Delta V_G=0.57\ L[/tex]

   b. [tex]\Delta V_S=0.021\ L[/tex]

   c. [tex]V_0=0.549\ L[/tex]

Explanation:

1)

  • given initial length, [tex]L=1275\ m[/tex]
  • initial temperature, [tex]T_i=-15^{\circ}C[/tex]
  • final temperature, [tex]T_f=25^{\circ}C[/tex]
  • coefficient of linear expansion, [tex]\alpha=12\times 10^{-6}\ ^{\circ}C^{-1}[/tex]

∴Change in temperature:

[tex]\Delta T=T_f-T_i[/tex]

[tex]\Delta T=25-(-15)[/tex]

  • [tex]\Delta T=40^{\circ}C[/tex]

We have the equation for change in length as:

[tex]\Delta L= L.\alpha. \Delta T[/tex]

[tex]\Delta L= 1275\times 12\times 10^{-6}\times 40[/tex]

[tex]\Delta L= 0.612\ m[/tex]

2)

Given relation:

[tex]\Delta V=V.\beta.\Delta T[/tex]

where:

[tex]\Delta V[/tex]= change in volume

V= initial volume

[tex]\Delta T[/tex]=change in temperature

  • initial volume of tank, [tex]V_{Si}=60\ L[/tex]
  • initial volume of gasoline, [tex]V_{Gi}=60\ L[/tex]
  • initial temperature of steel tank, [tex]T_{Si}=15^{\circ}C[/tex]
  • initial temperature of gasoline, [tex]T_{Gi}=15^{\circ}C[/tex]
  • coefficients of volumetric expansion for gasoline, [tex]\beta_G=950\times 10^{-6}\ ^{\circ}C[/tex]
  • coefficients of volumetric expansion for gasoline, [tex]\beta_S=35\times 10^{-6}\ ^{\circ}C[/tex]

a)

final temperature of gasoline, [tex]T_{Gf}=25^{\circ}C[/tex]

∴Change in temperature of gasoline,

[tex]\Delta T_G=T_{Gf}-T_{Gi}[/tex]

[tex]\Delta T_G=25-15[/tex]

[tex]\Delta T_G=10^{\circ}C[/tex]

Now,

[tex]\Delta V_G= V_G.\beta_G.\Delta T_G[/tex]

[tex]\Delta V_G=60\times 950\times 10^{-6}\times 10[/tex]

[tex]\Delta V_G=0.57\ L[/tex]

b)

final temperature of tank, [tex]T_{Sf}=25^{\circ}C[/tex]

∴Change in temperature of tank,

[tex]\Delta T_S=T_{Sf}-T_{Si}[/tex]

[tex]\Delta T_S=25-15[/tex]

[tex]\Delta T_S=10^{\circ}C[/tex]

Now,

[tex]\Delta V_S= V_S.\beta_S.\Delta T_S[/tex]

[tex]\Delta V_S=60\times 35\times 10^{-6}\times 10[/tex]

[tex]\Delta V_S=0.021\ L[/tex]

c)

Quantity of gasoline spilled after the given temperature change:

[tex]V_0=\Delta V_G-\Delta V_S[/tex]

[tex]V_0=0.57-0.021[/tex]

[tex]V_0=0.549\ L[/tex]

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