Question

Linear Thermal Expansion (in one dimension)

Answer 1

Answer:

1) $\Delta L= 0.612\ m$

2) a. $\Delta V_G=0.57\ L$

   b. $\Delta V_S=0.021\ L$

   c. $V_0=0.549\ L$

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, $\alpha=12\times 10^{-6}\ ^{\circ}C^{-1}$

∴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:

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

$\Delta L= 1275\times 12\times 10^{-6}\times 40$

$\Delta L= 0.612\ m$

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, $\beta_G=950\times 10^{-6}\ ^{\circ}C$

• coefficients of volumetric expansion for gasoline, $\beta_S=35\times 10^{-6}\ ^{\circ}C$

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,

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

$\Delta V_G=60\times 950\times 10^{-6}\times 10$

$\Delta V_G=0.57\ L$

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,

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

$\Delta V_S=60\times 35\times 10^{-6}\times 10$

$\Delta V_S=0.021\ L$

c)

Quantity of gasoline spilled after the given temperature change:

$V_0=\Delta V_G-\Delta V_S$

$V_0=0.57-0.021$

$V_0=0.549\ L$