201k views
5 votes
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?

User Kadir
by
5.0k points

1 Answer

7 votes

Answer:

1)
\Delta L= 0.612\ m

2) a.
\Delta V_G=0.57\ L

b.
\Delta V_S=0.021\ L

c.
V_0=0.549\ L

Step-by-step explanation:

1)

  • given initial length,
    L=1275\ m
  • initial temperature,
    T_i=-15^(\circ)C
  • final temperature,
    T_f=25^(\circ)C
  • coefficient of linear expansion,
    \alpha=12* 10^(-6)\ ^(\circ)C^(-1)

∴Change in temperature:


\Delta T=T_f-T_i


\Delta T=25-(-15)


  • \Delta T=40^(\circ)C

We have the equation for change in length as:


\Delta L= L.\alpha. \Delta T


\Delta L= 1275* 12* 10^(-6)* 40


\Delta L= 0.612\ m

2)

Given relation:


\Delta V=V.\beta.\Delta T

where:


\Delta V= change in volume

V= initial volume


\Delta T=change in temperature

  • initial volume of tank,
    V_(Si)=60\ L
  • initial volume of gasoline,
    V_(Gi)=60\ L
  • initial temperature of steel tank,
    T_(Si)=15^(\circ)C
  • initial temperature of gasoline,
    T_(Gi)=15^(\circ)C
  • coefficients of volumetric expansion for gasoline,
    \beta_G=950* 10^(-6)\ ^(\circ)C
  • coefficients of volumetric expansion for gasoline,
    \beta_S=35* 10^(-6)\ ^(\circ)C

a)

final temperature of gasoline,
T_(Gf)=25^(\circ)C

∴Change in temperature of gasoline,


\Delta T_G=T_(Gf)-T_(Gi)


\Delta T_G=25-15


\Delta T_G=10^(\circ)C

Now,


\Delta V_G= V_G.\beta_G.\Delta T_G


\Delta V_G=60* 950* 10^(-6)* 10


\Delta V_G=0.57\ L

b)

final temperature of tank,
T_(Sf)=25^(\circ)C

∴Change in temperature of tank,


\Delta T_S=T_(Sf)-T_(Si)


\Delta T_S=25-15


\Delta T_S=10^(\circ)C

Now,


\Delta V_S= V_S.\beta_S.\Delta T_S


\Delta V_S=60* 35* 10^(-6)* 10


\Delta V_S=0.021\ L

c)

Quantity of gasoline spilled after the given temperature change:


V_0=\Delta V_G-\Delta V_S


V_0=0.57-0.021


V_0=0.549\ L

User Kino
by
4.7k points