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\begin{center}
{\Large {Solution Derivations for Capa \#14\\[20pt]} }
\end{center}
1) A hollow sphere of inner radius $r_i = 6.00\ cm$ and outer radius
$r_o = 8.60\ cm$ floats half submerged in a liquid of density
$\rho_{liq} = 850\ kg/m^3$. What is the mass $m$ of the sphere?
\begin{quote}
$r_{i}$ = Given in $cm$.\\
$r_{o}$ = Given in $cm$.\\
$\rho _{liq}$ = Given\\
Per Archemedes, the force on an object is equal to the weight of displaced
fluid. If the sphere is half-submerged, we can find the displaced volume
of fluid and then its weight. From the weight we can extract the mass.
\[
v=\frac{4}{3}\pi r_{o}^{3}
\]
Since the sphere is half-submerged,
\[
\frac{1}{2}v=\frac{2}{3}\pi r_{o}^{3}
\]
is what we want. Since density is mass per unit volume,
\[
\rho _{liq}=\frac{m}{v}
\]
or
\[
v\rho _{liq}=m
\]
where $v$ in our case is half the volume of the sphere, or
\[
\frac{2}{3}\pi r_{o}^{3}.
\]
\[
\frac{2}{3}\pi r_{o}^{3}\rho _{liq}=m.
\]
Make sure to convert $r_{o}$ to meters.
\end{quote}
2) Calculate the density $\rho$ of the material of which the sphere is made.
\begin{quote}
$r_{i}$ = Given in $cm$.\\
$r_{o}$ = Given in $cm$.\\
Again, density is mass per unit volume. The mass of this sphere was
calculated in problem (1), so we can use it for this one. We just need to
find the volume of the sphere. Actually, we need to find the volume of the
{\em mass} in the sphere which we will use to find the density. The
volume of this hollow sphere is the volume of the large, outer sphere minus
the volume of the smaller, inner sphere. Or,
\begin{eqnarray*}
v_{tot} & = & v_{out}-v_{in}\\
v_{tot} & = & \frac{4}{3}\pi r_{o}^{3}-\frac{4}{3}\pi r_{i}^{3}=\frac{4}{3}\pi
\left( r_{o}^{3}-r_{i}^{3}\right)
\end{eqnarray*}
Using the mass from problem (1),
\[
\rho =\frac{m}{v}=\frac{m}{\frac{4}{3}\pi \left( r_{o}^{3}-r_{i}^{3}\right)
}=\frac{3m}{4\pi \left( r_{o}^{3}-r_{i}^{3}\right) }
\]
Remember to convert both $r_{i}$ and $r_{o}$ to meters. Density has units
of $kg/m^{3}$.
\end{quote}
3) Find the initial downward acceleration $a$ of an iron ball of density
$\rho = 8.40\ g/cm^3$ when placed in water. (Use $\rho_{water} = 1000\ kg/m^3$).
\begin{quote}
$\rho $ = Given in $g/cm^{3}$\\
$\rho _{water}$ = Given\\
Assume you have one $cm^{3}$ of iron of density $\rho $. So you have
whatever magnitude of $\frac{\rho }{1000}$ kilograms of iron (the mass,
since $\rho $ was given in $g/cm^{3}$) and 1 $cm^{3}$ of volume. The
buoyant force exerted on this object is equal to the weight of the displaced
water. The force is simple the weight of 1 $cm^{3}$ of water.
\[
F=\rho _{water}\ast 1\ cm^{3}\ast \left( \frac{1\ m}{100\ cm}\right) ^{3}\ast g
=\frac{g\rho _{water}}{1000000}
\]
(which has units of newtons). That is the upward force due to buoyancy. The
downward force is due to the weight of the ball. The weight is simply $W=mg$
where the mass is defined above. The forces are acting in opposite
directions, so pick a coordinate system and find the difference. Here, we
assume that up is the positive direction.
\[
F_{net}=ma=F-W
\]
To find the resulting initial acceleration, simply divide the total force by
the mass of the object.
\begin{eqnarray*}
ma & = & \frac{g\rho _{water}}{1000000}-mg\\
a & = & \frac{g\rho _{water}}{1000000\ast m}-g
\end{eqnarray*}
To help solve this, just think of $m$ as equaling $\frac{\rho }{1000}$.
This quantity will have units of $kg/cm^{3}$ but then you are assuming you
have 1 $cm^{3}$ of the substance, so the mass in kilograms is all that's
left. This method of solving the problem will probably produce a negative
acceleration, meaning downward. The problem asks for the initial downward
acceleration, so take the absolute value of the answer.
\end{quote}
4) One of the problems with debris in the ocean is that it is often difficult
to see, because much of the object is under the surface of the water. An
object with a density of $\rho = 751\ kg/m^3$ and a mass $m = 2443\ kg$ is
thrown into the ocean. Find the fraction $f$ of the object that is sticking
out of the water (use $\rho_{seawater} = 1024\ kg/m^3$).
\begin{quote}
$\rho $ = Given\\
$m$ = Given\\
$\rho _{seawater}$ = Given\\
To find the volume of the object thrown in the ocean,
\begin{eqnarray*}
\rho & = & \frac{m}{v},\\
v & = & \frac{m}{\rho }
\end{eqnarray*}
This problem is basically asking for what percentage of the total volume of
the object is sticking out of the ocean. This can be found when the object
is floating; that is, when the buoyant force equals the weight of the
object. The buoyant force is the weight of displaced water which equals
\[
F=\rho _{seawater}v_{displaced}g
\]
The weight of the object is simply $W=mg$. Setting these equal and solving
for $v_{displaced}$,
\begin{eqnarray*}
\rho _{seawater}v_{displaced}g & = & mg\\
v_{displaced} & = & \frac{m}{\rho _{seawater}}
\end{eqnarray*}
Computing the ratio $\frac{v_{displaced}}{v}$ will give you the percentage
of the object {\em below }the surface. Take this number and subtract it
from 1 to find the percentage above the surface.
\end{quote}
5) A piston with a tight-fitting, but movable lid contains an ideal gas at a
pressure $p$, an initial temperature $T_i$ and an initial volume $V_i$. The
temperature of the piston is slowly increased by heating the outside while the
pressure is maintained constant. The temperature is raised to a final
temperature of $T_f = 3.5\,T_i$. As the temperature increases, the volume
changes until the final volume is $V_f = f\,V_i$, where $f$ is a dimensionless
factor. What is the factor $f$ by which the volume changes?
\begin{quote}
$p$ = Given\\
$T_{i}$ = Given\\
$V_{i}$ = Given\\
$T_{f}$ = Given in terms of $T_{i}$\\
$V_{f}=f\, V_{i}$\\
The problem asks for the scalar $f$ by which the volume changes. To solve
this, you must use the ideal gas law.
\[
PV=nRT
\]
to relate the initial and final conditions.
\[
P_{i}V_{i}=n_{i}RT_{i}
\]
and
\[
P_{f}V_{f}=n_{f}RT_{f}
\]
Since the pressure does not change, we can relate the two equations by
solving for the pressure and then substitute it back into the first
equation. The container is sealed, so the number of gas particles ($n$)
also does not change.
\begin{eqnarray*}
P & = & \frac{nRT_{f}}{V_{f}}\\
\frac{nRT_{f}}{V_{f}}V_{i} & = & nRT_{i}
\end{eqnarray*}
Simplifying,
\[
T_{f}V_{i}=T_{i}V_{f}
\]
Or
\[
\frac{T_{f}}{T_{i}}V_{i}=V_{f}
\]
Which is similar to
\[
V_{f}=f\ast V_{i}
\]
So,
\[
f=\frac{T_{f}}{T_{i}}
\]
\end{quote}
6) A container of fixed volume $V$ has a tiny hole through which gas can enter
or leave. The container is filled with an ideal gas at the same temperature
and pressure $p$ as the surrounding room air. The initial temperature is
$T_i$ and the initial number of molecules of ideal gas in the container is
$N_i$. The temperature of the container is increased by heating the outside
until the final temperature is $T_f = 2.50\, T_i$. As the temperature
increases, the number of molecules remaining in the container decreases until
the final number is $N_f = f\, N_i$, where $f$ is a dimensionless factor.
What is the factor $f$ by which the number of molecules changes?
\begin{quote}
$V$ = constant\\
$T_{i}$ = Given\\
$n_{i}$ = Given\\
$T_{f}$ = Given in terms of $T_{i}$\\
$n_{f}=f\, n_{i}$\\
The problem asks for the scalar $f$ by which the number of gas particles
changes. To solve this, you must use the ideal gas law.
\[
P_{i}V_{i}=n_{i}RT_{i}
\]
and
\[
P_{f}V_{f}=n_{f}RT_{f}
\]
Dividing the two equations,
\[
\frac{P_{i}V_{i}}{P_{f}V_{f}}=\frac{n_{i}RT_{i}}{n_{f}RT_{f}}
\]
Since the volume does not change, the pressure is allowed to equalize with
the air outside the container, and $R$ is a constant,
\begin{eqnarray*}
1 & = & \frac{n_{i}T_{i}}{n_{f}T_{f}}\\
n_{f}T_{f} & = & n_{i}T_{i}\\
n_{f} & = & \frac{n_{i}T_{i}}{T_{f}}
\end{eqnarray*}
Which is similar to
\[
n_{f}=f\ast n_{i},
\]
so
\[
f=\frac{T_{i}}{T_{f}}
\]
\end{quote}
7) A piston with a tight-fitting, but movable lid contains an ideal gas at an
initial temperature $T_i = 280\ K$, an initial pressure $p_i = 1.00\ atm$ and
an initial volume $V_i = 250\ cm^3$. The volume of the piston is increased by
pulling out the lid until the final volume is $V_f = 3.70\, V_i$. At the same
time the volume is increased, the temperature is decreased by $50.0\ K$ with
an external temperature controller. What is the final pressure $p_f$
{\em in atmospheres}? Give the answer in the form ``X atm'', where X is the
number of atmosphere.
\begin{quote}
$T_{i}$ = Given\\
$P_{i}$ = Given\\
$V_{i}$ = Given\\
$V_{f}$ = Given\\
$T_{f}$ = Given\\
$P_{f}$ = ?\\
Using a calculation from problem (6),
\[
\frac{P_{i}V_{i}}{P_{f}V_{f}}=\frac{n_{i}RT_{i}}{n_{f}RT_{f}}
\]
The number of particles does not change and $R$ is a constant,
\begin{eqnarray*}
P_{i}V_{i}T_{f} & = & P_{f}V_{f}T_{i}\\
\frac{P_{i}V_{i}T_{f}}{V_{f}T_{i}} & = & P_{f}
\end{eqnarray*}
\end{quote}
8) A mass $m_{Pb} = 4.00\ kg$ of lead shot at a temperature
$T_i = 97.9^{\circ}C$ is poured into $m_w = 4.00\ kg$ of water at
$T_w = 20.5^{\circ}C$. Find the final temperature $T_f$ of the mixture. Give
the answer in degrees Celsius, using the unit ``C''. (CAPA can't read a
superscripted degree zero.) Use $c_{water} = 4187\ \frac{J}{kg^{\circ}C}$ and
$c_{lead} = 128\ \frac{J}{kg^{\circ}C}$.
\begin{quote}
$m_{Pb}$ = Given\\
$T_{i}$ = Given\\
$m_{w}$ = Given\\
$T_{w}$ = Given\\
$T_{f}$ = ?\\
We can use the equation $\Delta Q=mc\Delta T$ to relate the heat gained by
the water and lost by the lead.
\[
-m_{Pb}c_{lead}\Delta T_{lead}=m_{water}c_{water}\Delta T_{w}
\]
So,
\begin{eqnarray*}
m_{water}c_{water}\left( T_{f}-T_{w}\right) +m_{Pb}c_{lead}\left(
T_{f}-T_{i}\right) & = & 0\\
m_{water}c_{water}T_{f}+m_{Pb}c_{lead}T_{f} & = & m_{water}
c_{water}T_{w}+m_{Pb}c_{lead} T_{i}\\
T_{f} & = & \frac{m_{water}c_{water}T_{w}+m_{Pb}c_{lead} T_{i}}{m_{water}c_{water}+m_{Pb}c_{lead}}
\end{eqnarray*}
\end{quote}
9) A brass cube has an edge length of $L = 35.0\ cm$. What is the increase in
its surface area $\Delta A$ when it is heated from $T_i = 21.0^{\circ}C$ to
$T_f = 82.0^{\circ}C$? Data: $\alpha_{Brass} = 19 \times 10^{-6}\ ^{\circ}C^{-1}$
(at $20^{\circ}C$).
\begin{quote}
$L$ = Given in centimeters\\
$T_{i}$ = Given\\
$T_{f}$ = Given\\
$\alpha _{Brass}$ = Given\\
$\Delta A$ = ?\\
The initial surface area can easily be calculated.
\[
A_{i}=6L^{2}
\]
To find the new length of each side (and the new surface area), use the
equation
\begin{eqnarray*}
\alpha & = & \frac{\Delta L}{L\Delta T}\\
\alpha L\Delta T & = & \Delta L
\end{eqnarray*}
Then find the difference in surface area
\[
A_{f}-A_{i}=6\left( L+\Delta L\right) ^{2}-6L^{2}
\]
Remember to convert $L$ to meters. The area difference will be in square
meters.
\end{quote}
10) At $T_i = 20^{\circ}C$, an iron bar has length $L_{Fe} = 1.5000\ m$ long
and a brass bar has length $L_{brass} = 1.4965\ m$ long. Find the temperature
$T$ at which they will have the same length. Data:
$\alpha_{Iron} = 12 \times 10^{-6}\ ^{\circ}C^{-1}$,
$\alpha_{Brass} = 19 \times 10^{-6}\ ^{\circ}C^{-1}$ (at $20^{\circ}C$).
\begin{quote}
$T_{i}$ = Given\\
$L_{Fe}$ = Given\\
$L_{brass}$ = Given\\
$\alpha _{Iron}$ = Given\\
$\alpha _{brass}$ = Given\\
$T_{f}$ = ?\\
We want to know when the two rods have equal length. This will happen when
\[
L_{brass}+\Delta L_{brass}=L_{Fe}+\Delta L_{Iron}
\]
We solved for $\Delta L$ in the previous problem. Substituting in,
\begin{eqnarray*}
L_{brass}+\alpha _{brass}L_{brass}\Delta T & = & L_{Fe}+\alpha_{Iron}L_{Fe}\Delta T\\
\alpha _{brass}L_{brass}\Delta T-\alpha _{Iron}L_{Fe}\Delta T & = & L_{Fe}-L_{brass}\\
\Delta T\left( \alpha _{brass}L_{brass}-\alpha _{Iron}L_{Fe}\right) & = & L_{Fe}-L_{brass}\\
\Delta T=T_{f}-T_{i} & = & \frac{L_{Fe}-L_{brass}}{\left( \alpha_{brass}L_{brass}
-\alpha _{Iron}L_{Fe}\right) }\\
T_{f} & = & \frac{L_{Fe}-L_{brass}}{\left( \alpha _{brass}L_{brass}-\alpha
_{Iron}L_{Fe}\right) }+T_{i}
\end{eqnarray*}
\end{quote}
11) Choose all correct statements, e.g. B, AC, BCD, etc. NOTE!!! You may need
to know the coefficient of expansion of copper, which is NOT listed in the
text. (slightly less than brass).
\begin{quote}
QUESTION:\\
A) When the Celsius temperature doubles, the Fahrenheit temperature doubles.\\
B) A copper plate has a hole cut in its center. When the plate is
heated, the hole gets smaller.\\
C) When heat is added to a system, the temperature must rise.\\
D) When the pendulum of a grandfather clock is heated, the clock runs more
slowly.\\
E) Cold objects do not radiate heat energy.\\
F) In a bimetallic strip of aluminum and brass which curls when heated, the
aluminum is on the inside of the curve.\\
ANSWER:\\
A) FALSE: A Fahrenheit degree is not the same size as a Celsius degree.\\
B) FALSE: When heated, all linear dimensions expand (the hole gets larger).\\
C) FALSE: could use up heat.\\
D) TRUE: The pendulum will increase in length, increasing the moment of
inertia, and slowing the angular movement.\\
E) FALSE: All objects with an absolute temperature above absolute zero (all
objects) radiate heat energy.\\
F) FALSE: Aluminum has a higher expansion coefficient than does brass so it
will expand further. It will be on the outside of the curve.\\
NOTE: CAPA is looking for the letters of the correct answers. In this
case, D.
\end{quote}
12) How much energy is lost (that is, what is the decrease in energy
$-\Delta E$) when $m = 136.0\ g$ of steam, at a temperature
$T_i = 181.0\ ^{\circ}C$, is cooled and frozen into $m = 136.0\ g$ of ice
at $T_f = 0\ ^{\circ}C$? take the specific heat of steam to be
$c_{steam} = 2.01\ kJ/kg \cdot K$, the heat of vaporization of water is
$L_v = 539\ cal/g$, the heat of fusion of ice is $L_F = 79.7\ cal/g$. use the
known value of the specific heat of water $c_{water}$.
\begin{quote}
$-\Delta E$ = ?\\
$m_{steam}=m_{ice}=m_{water}=m$ = Given in grams.\\
$T_{i}$ = Given\\
$m_{ice}$ = Given\\
$T_{f}$ = Given\\
$c_{steam}$ = Given\\
$L_{v}$ = Given in cal/g\\
$\vspace{1pt}L_{f}$ = Given in cal/g\\
$c_{water}=4.184$ $\frac{kJ}{kg \cdot K}$\\
This problem has four parts. Cool the steam to $100^{\circ}C$, condense into
water, cool the water to $0^{\circ}C$, and freeze the water.
\begin{eqnarray*}
\Delta Q_{1} & = & mc_{steam}\Delta T\\
\Delta Q_{2} & = & mL_{v}\\
\Delta Q_{3} & = & mc_{water}\Delta T\\
\Delta Q_{4} & = & mL_{f}
\end{eqnarray*}
$\Delta Q_{1}$ is the energy that must be removed from the steam
in order to bring it to $100^{\circ}C$. $\Delta Q_{2}$ is the
energy that must be removed to condense the steam into water. $\Delta
Q_{3} $ is the energy that must be removed to cool the water to
$0^{\circ}C$. $\Delta Q_{4}$ is the energy that must be removed to
freeze the water. In all, the total energy that is removed from the
system,
\[
Q_{tot}=\Delta Q_{1}+\Delta Q_{2}+\Delta Q_{3}+\Delta Q_{4}
\]
Before calculating anything, make sure to convert your mass to kilograms as
well as $L_{f}$ and $L_{v}$ to $kJ/kg$ since that is what $c_{steam}$ and
$c_{water}$ are given in. You could also convert them the other way around.
\ To do this, use the conversion factors $\frac{1\ Cal}{1000\ cal},\
\frac{4184\ J}{1\ Cal},\ \frac{1\ kJ}{1000\ J}$ and, of course,
$\frac{1000\ g}{1\ kg}.$ You do not need to convert the temperature to Kelvin
since you are concerned about the {\em change} in temperature. The change in
Celsius temperature is the same magnitude as the change in Kelvin. If you
converted $L_{f}$ and $L_{v}$ then your answer will have units of kJ.
\end{quote}
13) It is possible to warm your hands on a cold day by rubbing them together.
Assume that the coefficient of sliding friction between your hands is $0.5$
and that the normal force between the hands is $44.0\ N$, and that you rub
them together with an average speed of $33.0\ cm/s$. Assume that the mass of
each of your hands is $323.0\ g$, that the specific heat of your hands is
$4.2\ kJ/kg \cdot K$, and that all the heat generated goes into raising the
temperature of your hands. How long must you rub your hands together to
produce a $6.0\ ^{\circ}C$ increase in temperature?
\begin{quote}
$\mu $ = Given (coefficient of sliding friction)\\
$F_{N}$ = Given (Normal force)\\
$s$ = Given (speed) in cm/s\\
$m$ = Given in grams\\
$c_{hands}$ = Given\\
$\Delta T$ = Given\\
$t$ = ?\\
This is also a multipart problem. First, calculate the amount of heat that
must be generated to raise the temperature of your hands by $\Delta T$.
\[
\Delta Q=2mc_{hands}\Delta T
\]
Where the $2$ is introduced because you have two hands. To find the time
necessary for your hands to heat up by this amount, we must relate the work
you must do to heat your hands to the time it takes to heat them. This is
precisely the definition of power.
\[
P=\frac{W}{t}
\]
But another, obscure formula for power is that
\[
P=Fv
\]
By having two formulae for power, you can set them equal and solve for the
unknown (time in this case).
\begin{eqnarray}
Fv & = & \frac{W}{t} \nonumber \\
t & = & \frac{W}{Fv} \label{eqn:13-1}
\end{eqnarray}
Where
\[
F=\mu F_{N}
\]
is the force of friction,
\[
v=s
\]
is the speed at which you rub your hands together (convert to m/s), and
\[
W=\Delta Q=2mc_{hands}\Delta T
\]
is the work which must be done to heat your hands (convert mass to kg).
Another, less obvious, step is to convert $c_{hands}$ to $J$ (since it is
in $kJ$). Remember that $\frac{1000\ J}{1\ kJ}$. Looking at the formula for
time below, we will end up with
\[
\frac{kg*\frac{kJ}{kg K}* K}{N * m/s}.
\]
A joule is defined to be a Newton * meter. But the units will not cancel if the
value of $c_{hands}$ is left in $kJ$. Substituting this back in to equation
(\ref{eqn:13-1}),
\[
t=\frac{2mc_{hands}\Delta T}{\mu F_{N}\ast s}
\]
\end{quote}
\end{document}