All Solutions
Page 357: Lesson Check
$$
Delta L = L_ialpha Delta T
$$
Where $L_i$ is the initial length of the rod,
$alpha$ is the coefficient of thermal expansion,
$Delta T$ is the change in temperature.
Therefore, the final length of the rod becomes:
$$
begin{align*}
L &= L_i + Delta L\
\
L &= L_i + L_ialpha Delta T\
\
L &= L_ileft( {1 + alpha Delta T} right)
end{align*}
$$
$$
L = L_ileft[ {1 + alpha left( {T + 10} right)} right]
$$
Where $T$ is the initial temperature of the rod.
Final length of the rod which is cooled by $10^circ text{C}$ is:
$$
L’ = L_ileft[ {1 + alpha left( {T – 10} right)} right]
$$
The ratio of the two lengths will be:
$$
begin{align*}
dfrac{L}{L’} &= dfrac{L_ileft[ {1 + alpha left( {T + 10} right)} right]}{{L_i}left[ {1 + alpha left( {T – 10} right)} right]}\
\
dfrac{L}{L’} &= dfrac{left[ {1 + alpha left( {T + 10} right)} right]}{left[ {1 + alpha left( {T – 10} right)} right]}\
end{align*}
$$
Conduction occurs as high energy particles collide with and jostle neighbouring lower energy particles, thereby transferring kinetic energy from one particle to the next. Eventually, the conduction process transfers energy from particle to particle.
Therefore the process of conduction occurs through collision of particles.
Convection transfers thermal energy through the physical movement of particles from one place to another, therefore the process that carries thermal energy through boiling water as it heats on top of a stove is convection.
$$
begin{equation*}
Delta L = alpha , L_i , Delta T
end{equation*}
$$
for the four systems we have:
$$
begin{align*}
Delta L_A &= alpha_A , L_{iA} , Delta T_A = 24 cdot 10^{-6} frac{1}{text{textdegree}text{C}} cdot 2text{m} cdot 40text{textdegree}text{C} = 1.92 cdot 10^{-3}text{m} \
Delta L_B &= alpha_B , L_{i B} , Delta T_B = 12 cdot 10^{-6} frac{1}{text{textdegree}text{C}} cdot 2text{m} cdot 20text{textdegree}text{C} = 4.8 cdot 10^{-4}text{m} \
Delta L_C &= alpha_C , L_{i C} , Delta T_C = 12 cdot 10^{-6} frac{1}{text{textdegree}text{C}} cdot 1text{m} cdot 30text{textdegree}text{C} = 3.6 cdot 10^{-4}text{m} \
Delta L_D &= alpha_D , L_{i D} , Delta T_D = 24 cdot 10^{-6} frac{1}{text{textdegree}text{C}} cdot 1text{m} cdot 10text{textdegree}text{C} = 2.4 cdot 10^{-4}text{m} \
end{align*}
$$
Ranking the systems in order increasing length:
$$
begin{equation*}
Delta L_A > Delta L_B > Delta L_C > Delta L_D
end{equation*}
$$
Delta L_A > Delta L_B > Delta L_C > Delta L_D
$$
$$
begin{equation*}
Delta L = alpha L_i , , T
end{equation*}
$$
Where $alpha = 17 cdot 10^{-6} frac{1}{text{textdegree}text{C}}$ for copper, $L_i = 1.325text{cm}$
The change in temperature is:
$$
begin{align*}
Delta T &= T_f – T_i \
Delta T &= 224.0text{textdegree}text{C} – 21.0text{textdegree}text{C} \
Delta T &= 203text{textdegree}text{C}
end{align*}
$$
We have:
$$
begin{equation*}
Delta L = 17 cdot 10^{-6}frac{1}{text{textdegree}text{C}} cdot 1.325text{cm} cdot 203text{textdegree}text{C} approx 0.0045text{cm}
end{equation*}
$$
The diameter is
$$
begin{equation*}
L_f = L_i + Delta L = 1.325text{cm} + 0.0045text{cm} approx 1.33text{cm}
end{equation*}
$$
L_f approx 1.33text{cm}
$$
$$
begin{align*}
Delta L &= alpha L_iDelta T\
Delta T&=frac{Delta L }{alpha L_i}\
T_f-T_i&=frac{Delta L }{alpha L_i}\
T_f&=frac{Delta L }{alpha L_i}+T_i\
&=frac{1.164-1.166}{12*10^{-6}*1.166}+23\
&=boxed{color{#4257b2}{-119.93^circ c}}
end{align*}
$$
– The diameter of the hole $L_i = 1.166text{cm}$ at the initial temperature $T_i = 23.00text{textdegree}text{C}$
– The diameter of the hole $L_f = 1.164text{cm}$ at the final temperature $T_f$.
As the steel plate expands, so does the hole.
It expands by the following amount:
$$
begin{equation*}
Delta L = alpha , L_i , Delta T
end{equation*}
$$
Where $alpha = 12 cdot 10^{-6}frac{1}{text{textdegree}text{C}}$ is the coefficient of thermal expansion for steel.
The changes in length and temperature are expressed as follows:
$$
begin{align*}
Delta L &= L_f – L_i = 1.164text{cm} – 1.166text{cm} = -0.002text{cm} \
Delta T &= T_f – T_i
end{align*}
$$
Inserting this into our formula we have:
$$
begin{align*}
& Delta L = alpha , L_i(T_f – T_i) \
& T_f – T_i = frac{Delta L}{alpha , L_i} \
& T_f = T_i + frac{Delta L}{alpha , L_i}
end{align*}
$$
Pluging in the values the final temperature is found:
$$
begin{align*}
T_f &= 23.00text{textdegree}text{C} + frac{-(0.002text{cm})}{12 cdot 10^{-6}frac{1}{text{textdegree}text{C}} cdot 1.166text{cm}} \
T_f &= 23.00 – 142.94 = – 119.94 \
T_f & approx -120text{textdegree}text{C}
end{align*}
$$
T_f approx – 120text{textdegree}text{C}
$$
We use the formula:
$$
begin{equation*}
Delta L = alpha cdot L_I , Delta T
end{equation*}
$$
Where $Delta L = 0.0033text{cm}$, and $Delta T = 120text{textdegree}text{C}$
We find:
$$
begin{equation*}
L_i = frac{Delta L}{alpha , Delta T} = frac{0.0032text{cm}}{24 cdot 10^{-6}frac{1}{text{textdegree}text{C}}} cdot 120text{textdegree}text{C} = 1.11text{cm}
end{equation*}
$$
L_i = 1.11text{cm}
$$
$$
begin{equation*}
Delta L = alpha cdot L_i , Delta T
end{equation*}
$$
Where $Delta L = 0.36text{cm}$, $Delta T = 85text{textdegree}text{C}$ and $L_i = 2.5text{m}$
We find the coefficient of the thermal expansion
$$
begin{equation*}
alpha = frac{Delta L}{L_i , Delta T} = frac{0.36text{cm}}{2.5text{m} cdot 85text{textdegree}text{C}} = 1.69 cdot 10^{-5}frac{1}{text{textdegree}text{C}} approx 17 cdot 10^{-6}frac{1}{text{textdegree}text{C}}
end{equation*}
$$
This is most likely copper
begin{align*}
alpha = 17 cdot 10^{-6}frac{1}{text{textdegree}text{C}} \
text{This is copper}
end{align*}
$$
