All Solutions
Page 378: Assessment
On the other hand, temperature is proportional to the average kinetic energy of the particles in a given substance.
Thermal energy is internal energy of the body, expressed in Joules.
Let $T_{C1}$ be the first temperature in degrees Celsius.
Then temperature in kelvin will be:
$$
T_{K1} = T_{C1} + 273.15 – text{equation }1
$$
And $T_{C2}$ be the second temperature in degrees Celsius.
Then temperature in kelvin will be:
${T_{K2}} = {T_{C2}} + 273.15 – {text{equation 2}}$
Now subtracting equation $1$ from equation $2$ we get:
$$
T_{K2} – T_{K1} = T_{C2} – T_{C1}
$$
Therefore, a change in temperature in degree Celsius is the same in Kelvin.
Thus, a change in temperature of $20^circ text{C}$ is equal to a change in temperature of $20text{ K}$
$$
begin{align*}
T_F = frac{9}{5} , T_c + 32
end{align*}
$$
we can see that the changes in temperature are related as follows:
$$
begin{align*}
Delta T_F = frac{9}{5} , Delta T_c
end{align*}
$$
We can see that a change of $20text{textdegree}text{C}$ is actually a change of:
$$
begin{align*}
Delta T_F = frac{9}{5}Delta T_c = frac{9 cdot 20}{5} = 36text{textdegree}text{F}
end{align*}
$$
Obviously a change of $20text{textdegree}text{C}(36text{textdegree}text{F})$ is greater than a change of $36text{textdegree}text{F}$
Next, it is in no way obvious that boiling and freezing are opposites, though one can understand the analogy.
And thirdly the fact the number of 180 appears is irrelevant, in Celsius and Kelvin scale that number would have been 100.
To conclude: Though the comment is amusing, it is scientifically absurd for more than one reason.
Hence it is invalid to say a certain body “contains” heat.
It would be valid to state that the hotter body has higher temperature.
For the analysis of the quantities of thermal energies, further information about that system is needed (such as the masses and compositions of the objects)
To convert to Celsius we use:
$$
begin{align*}
T_C &= frac{5}{9} left(T_F – 32 right) \
T_C &= frac{5}{9} left(98.6 – 32 right) = 37text{textdegree}text{C}
end{align*}
$$
To convert to Kelvin we use:
$$
begin{align*}
T &= T_C + 273.15 \
T &= 37 + 273.15 = 310.15text{textdegree}text{K}
end{align*}
$$
item
$T_c = 37textdegreetext{C}$
item
$T = 310.15textdegreetext{K}$
end{enumerate}
$$
begin{align*}
T &= T_C + 273.15 \
T_C &= T – 273.15 \
T_C &= 1.0 – 273.15 = – 272.15text{textdegree}text{C}
end{align*}
$$
begin{align*}
T_C = – 272.15text{textdegree}text{C}
end{align*}
$$
$$
begin{align*}
T_C &= frac{5}{9} left(T_F – 32 right) \
T_C &= frac{5}{9} left(30 – 32 right) = frac{5}{9} cdot left(-2 right) \
T_C &= -frac{10}{9} text{textdegree}text{C} = – 1.11text{textdegree}text{C}
end{align*}
$$
begin{align*}
T_C = – 1.11text{textdegree}text{C}
end{align*}
$$
$$
begin{align*}
T &= T_C + 273.15 \
T &= 1200 + 273.15 \
T &= 1473.15text{ K}
end{align*}
$$
To convert this to the Fahrenheit scale we use:
$$
begin{align*}
T_F &= frac{9}{5} T_C + 32 \
T_F &= frac{9}{5} cdot 1200 + 32 \
T_F &= 2192text{textdegree}text{F}
end{align*}
$$
item
$T = 1473.15text{K}$
item
$T_F = 2192textdegreetext{F}$
end{enumerate}
$$
begin{equation*}
T_F = frac{9}{5} T_C + 32
end{equation*}
$$
With the condition $T_F = 2 , T_C$
Using this we get
$$
begin{align*}
& 2T_C = frac{9}{5} T_C + 32 \
& T_C left(2 – frac{9}{5} right) = 32 \
& T_C cdot frac{1}{5} = 32 \
& T_C = 160text{textdegree}text{C}
end{align*}
$$
Or in Fahrenheit
$$
begin{align*}
T_F = 2 T_C = 320text{textdegree}text{F}
end{align*}
$$
begin{align*}
& T_C = 160text{textdegree}text{C} \
& T_F = 320text{textdegree}text{F}
end{align*}
$$
– The initial temperature $T_1 = -4.0text{textdegree}text{F}$
– The final temperature $T_2 = 45text{textdegree}text{F}$
– The elapsed time interval $Delta t = 2 text{ min}$
Firstly we will find the change in temperature in Fahrenheit
$$
begin{align*}
Delta T_F &= T_2 – T_1 \
Delta T_F &= 45text{textdegree}text{F} – left(-4.0text{textdegree}text{F} right) \
Delta T_F &= 49text{textdegree}text{F}
end{align*}
$$
To convert the interval to Kelvin, we will convert it to Celsius since intervals in Kelvin and Celsius are the same thing. We will use:
$$
begin{equation*}
T_C = frac{5}{9} left(T_F – 32 right)
end{equation*}
$$
That is:
$$
begin{align*}
Delta T_C = frac{5}{9} Delta T_F = frac{5}{9} cdot 49 = 27.2text{textdegree}text{C}
end{align*}
$$
in Kelvin this is the same:
$$
begin{align*}
Delta T = Delta T_C = 27.2text{ K}
end{align*}
$$
The time interval in seconds it:
$$
begin{align*}
t = 2text{ min} = 120text{ s}
end{align*}
$$
Now we can calculate the rate of temperature change in kelvins per second:
$$
begin{align*}
& frac{Delta T}{Delta t} = frac{27.2text{ K}}{120text{ s}} = 0.2267 , frac{text{K}}{text{s}} \
& frac{Delta T}{Delta t} approx 0.23 frac{text{K}}{text{s}}
end{align*}
$$
begin{align*}
frac{Delta T}{Delta t} approx 0.23 , frac{text{K}}{text{s}}
end{align*}
$$
Therefore, the lid loosens due to thermal expansion when it is held under hot running water.
For a three dimensional object the increase of length along each dimension proportional to its initial length along that dimension.
So even though the relative changes in length are the same for all dimensions, in the general case the absolute changes aren’t the same.
Since the typical car has a greater length than height, it follows that its change in length is greater than its change in height.
For copper
$$
begin{align*}
alpha_{text{c}} = 17 cdot 10^{-6}text{ K}^{-1}
end{align*}
$$
For steel:
$$
begin{align*}
alpha_{text{s}} = 12 cdot 10^{-6}text{ K}^{-1}
end{align*}
$$
For aluminium:
$$
begin{align*}
alpha_{text{a}} = 24 cdot 10^{-6}text{ K}^{-1}
end{align*}
$$
So for strip A made out of copper and steel we see the difference in thermal expansion coefficients is:
$$
begin{align*}
Delta alpha_{text{a}} = alpha_{text{c}} – alpha_{text{s}} = left(17 – 12 right) cdot 10^{-6}text{ K}^{-1} = 5 cdot 10^{-6} cdot text{ K}^{-1}
end{align*}
$$
For strip B made out of aluminium and steel we see the difference in thermal expansion coefficient is:
$$
begin{align*}
Delta alpha_{text{B}} = alpha_{text{a}} – alpha_{text{s}} = left(24 – 12 right) cdot 10^{-6} cdot text{K}^{-1} = 12 cdot 10^{-6} cdot text{K}^{-1}
end{align*}
$$
Since $Deltaalpha_{text{B}}$ is greater than $Deltaalpha_{text{A}}$ it follows that the bimetallic strip B bend more than the strip A
begin{equation*}
Delta L = alpha Delta T L_{text{i}} = text{const} cdot L_{text{i}}
end{equation*}
where $alphaDelta T$ is constant in our problem (not the general case) for all the plates. \
Knowing all this we basically only need to rank the lengths in increasing order along the vertical and horizontal. \
The results are:
begin{enumerate}[a)]
item
the vertical: $B = C < E = D < A$
item
the horizontal: $A = C = E < B < D$
end{enumerate}
item
the vertical: $B = C < E = D < A$
item
the horizontal: $A = C = E < B < D$
end{enumerate}
The expansion coefficient of aluminum is higher than brass. Since the size of the aluminum ball is more, we have to decrease the temperature, so that the aluminum can pass through the hole.
Explanation (A) is the correct explanation.
– The diameter of a hole in an aluminium plate $L_{text{i}} = 1.18text{ cm}$
– The initial temperature $T_{text{i}} = 23.0text{textdegree}text{C}$
– The final temperature $T_{text{f}} = 199.0text{textdegree}text{C}$
The diameter of a hole changes during thermal expansion just as any other length. So we can write:
$$
begin{equation*}
Delta L = alpha , Delta T , L_{text{i}}
end{equation*}
$$
Where the thermal expansion coefficient for aluminium is $alpha = 24 cdot 10^{-6} text{ K}^{-1}$, and the change in temperature is found as follows:
$$
begin{align*}
Delta T &= T_{text{f}} – T_{text{i}} = 199.0text{textdegree}text{C} – 23.0text{textdegree}text{C} \
Delta T &= 176text{textdegree}text{C} = 176text{ K}
end{align*}
$$
So we can find the change in length:
$$
begin{align*}
Delta L &= 24 cdot 10^{-6}text{ K}^{-1} cdot 176text{ K} cdot 1.178text{ cm} \
Delta L &= 0.00498text{ cm}
end{align*}
$$
So the final length is:
$$
begin{align*}
L_{text{f}} = L_{text{i}} + Delta L = 1.178text{ cm} + 0.005text{ cm} = 1.183text{ cm}
end{align*}
$$
begin{align*}
L_{text{f}} = 1.183text{ cm}
end{align*}
$$
– The initial length of the steel beam $L_{text{i}} = 5.5text{ m}$
– The change in length of the beam $Delta L = 0.0012text{ m}$
– The thermal expansion coefficient for steel $alpha = 12 cdot 10^{-6}text{ K}^{-1}$
The required change in temperature can be found from the well known formula for change in length during thermal expansion:
$$
begin{equation*}
Delta L = alpha , Delta T , L_{text{i}}
end{equation*}
$$
rearranging for temperature:
$$
begin{align*}
Delta T = frac{Delta L}{alpha , L_{text{i}}}
end{align*}
$$
Plugging in the values:
$$
begin{align*}
Delta T & = frac{0.0012text{ m}}{12 cdot 10^{-6}text{ K}^{-1} cdot 5.5text{ m}} = 18.1919text{ K} \
Delta T & approx 18.2text{textdegree}text{C}
end{align*}
$$
begin{align*}
Delta T & approx 18.2text{textdegree}text{C}
end{align*}
$$
– The initial diameter of the ball $L_{text{i}} = 1.2text{ m}$
– The change in diameter of the ball $Delta L = 2.2text{ mm}$
– The change in temperature $Delta T = 95text{textdegree}text{C}$
begin{enumerate}[a)]
item $\$
The required coefficient of thermal expansion can be found from the well known formula for change in length during thermal expansion:
begin{equation*}
Delta L = alpha , Delta T , L_{text{i}}
end{equation*}
rearranging for $alpha$
begin{align*}
alpha = frac{Delta L}{Delta T , L_{text{i}}}
end{align*}
Plugging in the values:
begin{align*}
alpha & = frac{2.2text{ mm}}{95textdegreetext{C} cdot 1.2text{ m}} = frac{2.2 cdot 10^{-3}text{ m}}{95textdegreetext{C} cdot 1.2text{ m}} = 19.29 cdot 10^{-6}text{ K}^{-1} \
alpha & approx 19.3 cdot 10^{-6}text{ K}^{-1}
end{align*}
item
Looking at the table, we conclude that the substance is most likely to be brass.
end{enumerate}
item
$alpha = 19.3 cdot 10^{-6}text{ K}^{-1}$
item
brass
end{enumerate}
– The initial diameters of the ring $L_1 = 4.000text{ cm}$ and rod $L_2 = 4.040text{ cm}$
– The initial temperature $T_{text{i}} = 10text{textdegree}text{C}$
– The coefficient of thermal expansion of aluminium $alpha = 24 cdot 10^{-6}text{ K}^{-1}$
item $\$
Since we need to slip the ring over the rod, we need to increase its diameter. This means we need to heat it (the ring).
end{enumerate}
item $\$
The required change in temperature of the ring can be found from the well known formula for change in length during thermal expnasion:
begin{equation*}
Delta L = alpha , Delta T , L_{text{i}}
end{equation*}
Where $L_{text{i}}$ is the initial diameter of the ring $L_{text{i}} = L_1$, and the change in length is the difference between the rod’s and ring’s diameters:
begin{align*}
Delta L = L_2 – L_1 = 4.040text{ cm} – 4.000text{ cm} = 0.04text{ cm}
end{align*}
rearranging we get:
begin{align*}
Delta T = frac{Delta L}{alpha , L_1} = frac{0.04text{ cm}}{24 cdot 10^{-6}text{ K}^{-1} cdot 4.000text{ cm}} approx 416.67text{ K}
end{align*}
So the final temperature of the ring, at which it fits over the rod is
begin{align*}
T_{text{f}} = T_{text{i}} + Delta T = 10textdegreetext{C} + 416.67textdegreetext{C} = 426.67textdegreetext{C}
end{align*}
end{enumerate}
begin{align*}
T_{text{f}} = 426.67text{textdegree}text{C}
end{align*}
$$
– The thermal expansion coefficient of the steel interior $alpha_1 = 17.3 cdot 10^{-6}text{ K}^{-1}$
– The thermal expansion coefficient of the copper bottom $alpha_2 = 17.0 cdot 10^{-6}text{ K}^{-1}$
– The initial temperature of the pot $T_{text{i}} = 22text{textdegree}text{C}$
– The final temperature of the pot $T_{text{f}} = 610text{textdegree}text{C}$
– The initial diameter of the pot $L_{text{i}} = 21.0text{ cm}$
As the pot is heated it expands. Since steel and copper have different coefficients of thermal expansion, the interior and bottom will expand for different amounts. Those are found separately using the well known thermal expansion formula:
$$
begin{equation*}
Delta L = alpha , Delta T , L_{text{i}}
end{equation*}
$$
where:
$$
begin{align*}
Delta T = T_{text{f}} – T_{text{i}} = 610text{textdegree}text{C} – 22text{textdegree}text{C} = 588text{textdegree}text{C}
end{align*}
$$
First we find the change for the steel interior:
$$
begin{align*}
Delta L_1 &= alpha_1 , Delta T , L_{text{i}} \
Delta L_1 &= 17.3 cdot 10^{-6}text{ K}^{-1} cdot 588text{textdegree}text{C} cdot 21.0text{ cm} \
Delta L_1 &= 0.21362text{ cm}
end{align*}
$$
Now for the copper bottom:
$$
begin{align*}
Delta L_2 &= alpha_2 , Delta T , L_{text{i}} \
Delta L_2 &= 17.0 cdot 10^{-6}text{ K}^{-1} cdot 588text{textdegree}text{C} cdot 21.0text{ cm} \
Delta L_2 &= 0.20992text{ cm}
end{align*}
$$
Now we find the difference in diameter change using the previous results:
$$
begin{align*}
Delta x &= Delta L_1 – Delta L_2 \
Delta x &= (0.21362 – 0.20992)text{ cm} \
Delta x &= 0.0037text{ cm}
end{align*}
$$
begin{align*}
Delta x = 0.0037text{ cm}
end{align*}
$$
$$
begin{equation*}
Q = m cdot c cdot Delta T
end{equation*}
$$
Since in our problem the thermal energy $Q$ is fixed, we can see that the change in temperature $Delta T$ will depend on the values of mass $m$ and specific heat capacity $c$.
Hence the mentioned two reasons for difference in change of temperature of two bodies that receive the same thermal energy are:
– They have different mass
– The have different specific heat capacities (are made out of different materials, i.e. steel and wood)
The have different specific heat capacities
When the balloon is filled with water, this water can quickly absorb and redistribute a vast amount of thermal energy, due to its large mass and specific heat capacity compared to that of air. This movement of thermal energy doesn’t allow the rapid heating of the piece of balloon closest to the match, and hence stops the bursting that follows it.
Filling the balloon with water increases the total heat capacity of the system, making it much harder to rapidly raise its temperature.
item $\$
Looking at the table we can see that the specific heat capacities of aluminium and ice are:
begin{align*}
c_{text{A}} &= 900 , frac{text{J}}{text{kg}textdegreetext{C}} \
c_{text{i}} &= 2090 , frac{text{J}}{text{kg}textdegreetext{C}}
end{align*}
we also know that the mass of aluminium is twice the mass of ice:
begin{align*}
m_{text{A}} = 2 m_{text{i}} = 2text{ kg}
end{align*}
The increase in temperature while heating is governed by the formula
begin{equation*}
Q = m , c , Delta T
end{equation*}
So comparing $Delta T_{text{A}}$ and $Delta T_{text{i}}$ for the same thermal energy $Q$ gives us
begin{align*}
Delta T_{text{A}} &= frac{Q}{m_{text{A}} , c_{text{A}}} = frac{Q}{2text{ kg} cdot 900 frac{text{J}}{text{kg}textdegreetext{C}}} = frac{Q}{1800frac{text{J}}{textdegreetext{C}}} \
Delta T_{text{i}} &= frac{Q}{m_{text{i}} , c_{text{i}}} = frac{Q}{1text{ kg} cdot 2090frac{text{J}}{text{kg}textdegreetext{C}}} = frac{Q}{2090frac{text{J}}{textdegreetext{C}}}
end{align*}
it follows that
begin{align*}
Delta T_{text{A}} > Delta T_{text{i}}
end{align*}
since $2090 > 1800$
item
The correct explanation is A
end{enumerate}
item
Greater
item
the correct explanation is A
end{enumerate}
item $\$
When fluids are poured in the same container heat is exchanged between them until a final equilibrium temperature is reached. Since they are of the same mass, only the specific heat capacity determines their change in temperature. Since the specific heat capacity of water is higher than that of alcohol its change in temperature will be smaller than that of alcohol. This can be seen from the formula:
begin{equation*}
Q = m cdot c cdot Delta T
end{equation*}
Which in our problem takes the form:
begin{align*}
c cdot Delta T = text{const}
end{align*}
\
Hence the final temperature is closer to starting temperature of water, than to that of alcohol. Meaning it is above $25textdegreetext{C}$
\
item
the correct explanation is B
end{enumerate}
item
greater than $25textdegreetext{C}$
item
the correct explanation is B
end{enumerate}
– The mass of the piece of iron $m = 0.141text{ kg}$
– The initial temperature of the piece of iron $T_{text{i}} = 26.5text{textdegree}text{C}$
– The amount of energy removed $Q = 66.2text{ J}$
– The specific heat capacity of iron $c = 448 , frac{text{J}}{text{kg}text{textdegree}text{C}}$
When we remove the energy $Q$ from the iron it’s temperature drops. This can be found using the formula:
$$
begin{equation*}
Q = m , c , Delta T
end{equation*}
$$
rearranging for the change in temperature:
$$
begin{equation*}
Delta T = frac{Q}{m , c}
end{equation*}
$$
and plugging in the values:
$$
begin{align*}
Delta T = frac{66.2text{ J}}{0.141text{ kg} cdot 448 , frac{text{J}}{text{kg}text{textdegree}text{C}}} = 1.0479text{textdegree}text{C} approx 1.05text{textdegree}text{C}
end{align*}
$$
the final temperature is found:
$$
begin{align*}
T_{text{f}} &= T_{text{i}} – Delta T = 26.5text{textdegree}text{C} – 1.05text{textdegree}text{C} \
T_{text{f}} &= 25.45text{textdegree}text{C}
end{align*}
$$
begin{align*}
T_{text{f}} = 25.45text{textdegree}text{C}
end{align*}
$$
– The mass of the chunk of ice $m = 0.21text{ kg}$
– The change of temperature of the chunk of ice $Delta T = 7.5text{textdegree}text{C}$
– The specific heat capacity of ice $c = 2090 , frac{text{J}}{text{kg}text{textdegree}text{C}}$
When we remove the energy $Q$ from the ice it’s temperature drops. This can be found using the formula:
$$
begin{equation*}
Q = m , c , Delta T
end{equation*}
$$
Plugging in the values we find:
$$
begin{align*}
Q & = 0.21text{ kg} cdot 2090 , frac{text{J}}{text{kg}text{textdegree}text{C}} cdot 7.5text{textdegree}text{C} \
Q & approx 3292text{ J}
end{align*}
$$
We have found how much thermal energy must be removed from the chunk of ice
begin{align*}
Q = 3292text{ J}
end{align*}
$$
– The mass of the apple $m = 1.5text{ kg}$
– The initial temperature of the apple $T_{text{i}} = 12text{textdegree}text{C}$
– The final temperature of the apple $T_{text{f}} = 36text{textdegree}text{C}$
– The specific heat capacity of water $c = 4186 , frac{text{J}}{text{kg}text{textdegree}text{C}}$
When we add energy $Q$ to the apple it’s temperature increases. This can be found using the formula:
$$
begin{equation*}
Q = m , c , Delta T
end{equation*}
$$
Where the change in temperature is:
$$
begin{equation*}
Delta T = T_{text{f}} – T_{text{i}}
end{equation*}
$$
So our formula becomes:
$$
begin{align*}
Q = m , c (T_{text{f}} – T_{text{i}})
end{align*}
$$
Plugging in the values we get:
$$
begin{align*}
Q &= 0.15text{ kg} cdot 4186 , frac{text{J}}{text{kg}text{textdegree}text{C}} left(36 – 12 right)text{textdegree}text{C} \
Q &= 15069.6text{ J} approx 15070text{ J}
end{align*}
$$
begin{align*}
Q = 15070text{ J}
end{align*}
$$
– The metabolic rate when sleeping $A = 2.6 cdot 10^{-4}frac{text{C}}{text{s} cdot text{kg}}$
– The mass of our person $m = 75text{ kg}$
– The duration of sleep $t = 8text{ hr}$
First let’s clarify what the metabolic rate is. Looking at it’s dimensions $frac{text{C}}{text{s} cdot text{kg}}$ we see that it amounts to a certain amount of energy divided by unit of time, divided by unit of mass. Hence we conclude that it’s the specific (meaning per unit mass) power of burning of energy. We can then write:
$$
begin{equation*}
A = frac{Delta Q}{Delta m , Delta t}
end{equation*}
$$
To find the total energy a person metabolites we use
$$
begin{align*}
Delta m &= m \
Delta t &= t
end{align*}
$$
Now rearranging for $Q$:
$$
begin{align*}
Q = A , m , t
end{align*}
$$
Plugging in the values we get:
$$
begin{align*}
Q &= 2.6 cdot 10^{-4}frac{text{C}}{text{s} cdot text{kg}} cdot 75text{ kg} cdot 8 cdot 3600text{ s}
end{align*}
$$
Where we used $1text{ hr} = 3600text{ s}$. Finally we have:
$$
begin{align*}
Q = 561.6text{ C}
end{align*}
$$
– The total amount of work done $W = 2.5 text{ C}$
– The duration of the workout $t = 1frac{1}{2}text{ min} = 90text{ s}$
We need to find the power output which is defined as the rate of work done, meaning:
$$
begin{equation*}
P = frac{W}{t}
end{equation*}
$$
to find this in Wats we first need to convert the nutritional calories (C) to Joules. Remember $1text{C} = 1text{ kcal} = 4186text{ J}$
Now plugging in the values we get:
$$
begin{align*}
P = frac{2.5text{ C}}{90text{ s}} = frac{2.5 cdot 4186text{ J}}{90text{ s}} = 116.28text{ W}
end{align*}
$$
We will now convert this to Horse powers, using: $1text{HP} = 746text{ W}$
it follows $1text{W} = frac{1}{746}text{HP}$. Now we find the power output:
$$
begin{align*}
P = 116.28 , frac{1}{746}text{HP} = 0.156text{ HP}
end{align*}
$$
begin{align*}
P = 116.28text{ W} = 0.156text{ HP}
end{align*}
$$
– The mass of the iron horseshoe $m_{text{i}} = 0.50text{ kg}$
– The mass of the water $m_{text{w}} = 25.0text{ kg}$
– The initial temperature of the iron $T_{text{i}} = 450text{textdegree}text{C}$
– The initial temperature of the water $T_{text{w}} = 23text{textdegree}text{C}$
– The specific heat capacity of iron $c_{text{i}} = 448 , frac{text{J}}{text{kg}text{textdegree}text{C}}$
– The specific heat capacity of water $c_{text{w}} = 4186 , frac{text{J}}{text{kg}text{textdegree}text{C}}$
We know that no thermal energy is exchanged with the surroundings.
So we know that thermal energy is transferred from the iron horseshoe to the bucked of water only:
$$
begin{gather*}
Q_{text{i}} + Q_{text{w}} = 0 \
m_{text{i}} , c_{text{i}} , Delta T_{text{i}} + m_{text{w}} , c_{text{w}} , Delta T_{text{w}} = 0 \
m_{text{i}} , c_{text{i}} (T_{text{f}} – T_{text{i}}) + m_{text{w}} , c_{text{w}} (T_{text{f}} – T_{text{w}}) = 0 \
end{gather*}
$$
Where $T_{text{f}}$ is the final equilibrium temperature.
Rearranging it follows:
$$
begin{align*}
& T_{text{f}} (m_{text{i}} , c_{text{i}} + m_{text{w}} , c_{text{w}}) = m_{text{i}} , c_{text{i}} , T_{text{i}} + m_{text{w}} , c_{text{w}} , T_{text{w}} \
& T_{text{f}} = frac{m_{text{i}} , c_{text{i}} , T_{text{i}} + m_{text{w}} , c_{text{w}} , T_{text{w}}}{m_{text{i}} , c_{text{i}} + m_{text{w}} , c_{text{w}}}
end{align*}
$$
Plugging in the values:
$$
begin{align*}
T_{text{f}} &= frac{0.50text{ kg} cdot 448frac{text{J}}{text{kg}text{textdegree}text{C}} cdot 450text{textdegree}text{C} + 25.0text{ kg} cdot 4186frac{text{J}}{text{kg}text{textdegree}text{C}} cdot 23text{textdegree}text{C}}{0.50text{ kg} cdot 448frac{text{J}}{text{kg}text{textdegree}text{C}} + 25.0text{ kg} cdot 4186frac{text{J}}{text{kg}text{textdegree}text{C}}} \
T_{text{f}} &= 23.91text{textdegree}text{C}
end{align*}
$$
begin{align*}
T_{text{f}} = 23.91text{textdegree}text{C}
end{align*}
$$
$$
Q=m_bc_l(T_{iL}-T_e);quad Q=m_wc_w(T_e-T_{iw}).
$$
Where $T_{iL} = 84.2^circ$, $T_{iw} = 21.5^circ$. Equating the right sides and solving for $T_e$ we get
$$
T_e = frac{m_bc_lT_{iL}+m_wc_wT_{iw}}{m_bc_l+m_wc_w} =24.5^circ.
$$
$$
Q=Nm_0c_s(T_{is}-T_e);quad Q= m_wc_w(T_e-T_{iw})
$$
which, solved for $N$ gives
$$
N= frac{m_wc_w(T_e-T_{iw})}{m_0c_s(T_{is}-T_e)} approx 706.
$$
b)Since specific heat of copper is higher, it carries more thermal energy at the same temperature so we will need less copper pellets to achieve the same effect.
c) By analogous procedure we did in part a) we get
$$
N= frac{m_wc_w(T_e-T_{iw})}{m_0c_c(T_{is}-T_e)} approx 440.
$$
– The mass of the object $m_{text{o}} = 38text{ g}$
– The mass of water $m_{text{w}} = 103text{ g}$
– The mass of the aluminium calorimeter $m_{text{A}} = 155text{ g}$
– The initial temperature of the aluminium and water $T_{text{A}} = T_{text{w}} = 20text{textdegree}text{C}$
– The initial temperature of object $T_{text{o}} = 100text{textdegree}text{C}$
– The final temperature $T_{text{f}} = 22text{textdegree}text{C}$
– The specific heat capacity of water $c_{text{w}} = 4186 , frac{text{J}}{text{kg}text{textdegree}text{C}}$
– The specific heat capacity of aluminium $c_{text{A}} = 900 , frac{text{J}}{text{kg}text{textdegree}text{C}}$
We know that no thermal energy is exchanged with the surroundings.
So we know that thermal energy is transferred from the object to the calorimeter, and the water inside. We can write:
$$
begin{gather*}
Q_{text{o}} + Q_{text{w}} + Q_{text{A}} = 0 \
m_{text{o}} , c_{text{o}} , Delta T_{text{o}} + m_{text{w}} , c_{text{w}} , Delta T_{text{w}} + m_{text{A}} , c_{text{A}} , Delta T_{text{A}} = 0 \
m_{text{o}} , c_{text{o}} (T_{text{f}} – T_{text{o}}) + m_{text{w}} , c_{text{w}} (T_{text{f}} – T_{text{w}}) + m_{text{A}} , c_{text{A}} (T_{text{f}} – T_{text{A}}) = 0 \
end{gather*}
$$
Where $T_{text{f}}$ is the final equilibrium temperature.
Rearranging it follows:
$$
begin{align*}
c_{text{o}} = frac{- m_{text{w}} , c_{text{w}} (T_{text{f}} – T_{text{w}}) – m_{text{A}} , c_{text{A}}(T_{text{f}} – T_{text{A}}) }{m_{text{o}}(T_{text{f}} – T_{text{o}})}
end{align*}
$$
Plugging in the values we get:
$$
begin{align*}
c_{text{o}} &= frac{- 0.103text{ kg} cdot 4186frac{text{J}}{text{kg}text{textdegree}text{C}} left(22 – 20 right)text{textdegree}text{C} – 0.155text{ kg} cdot 900frac{text{J}}{text{kg}text{textdegree}text{C}} left(22 – 20 right)text{textdegree}text{C} }{0.038text{ kg} (22 – 100)text{textdegree}text{C} } \
c_{text{o}} &= 385.06 , frac{text{J}}{text{kg}text{textdegree}text{C}}
end{align*}
$$
This material is copper.
begin{gather*}
c_{text{o}} approx 385 , frac{text{J}}{text{kg}text{textdegree}text{C}} \
text{This material is copper}
end{gather*}
$$
– The mass of the cup $m_{text{cu}} = 116text{ g}$
– The mass of the coffee $m_{text{co}} = 225text{ g}$
– The mass of the cream $m_{text{cr}} = 12.2text{ g}$
– The initial temperature of the cup $T_{text{cu}} = 24text{textdegree}text{C}$
– The initial temperature of coffee $T_{text{co}} = 80.3text{textdegree}text{C}$
– The initial temperature of the cream $T_{text{cr}} = 5.0text{textdegree}text{C}$
– The specific heat capacity of the cup $c_{text{cu}} = 1090 , frac{text{J}}{text{kg}text{textdegree}text{C}}$
– The specific heat capacity of the water $c_{text{w}} = 4186 , frac{text{J}}{text{kg}text{textdegree}text{C}}$
We know that no thermal energy is exchanged with the surroundings.
So we know that thermal energy is transferred from the coffee to the cup and the cream, we can write:
$$
begin{gather*}
Q_{text{co}} + Q_{text{cu}} + Q_{text{cr}} = 0 \
m_{text{co}} , c_{text{co}} , Delta T_{text{co}} + m_{text{cu}} , c_{text{cu}} , Delta T_{text{cu}} + m_{text{cr}} , c_{text{cr}} , Delta T_{text{cr}} = 0 \
m_{text{co}} , c_{text{co}} (T_{text{f}} – T_{text{co}}) + m_{text{cu}} , c_{text{cu}} (T_{text{f}} – T_{text{cu}}) + m_{text{cr}} , c_{text{cr}} (T_{text{f}} – T_{text{cr}}) = 0
end{gather*}
$$
Where $T_{text{f}}$ is the final equilibrium temperature.
Rearranging it follows:
$$
begin{align*}
T_{text{f}} = frac{m_{text{co}} , c_{text{co}} , Delta T_{text{co}} + m_{text{cu}} , c_{text{cu}} , Delta T_{text{cu}} + m_{text{cr}} , c_{text{cr}} , Delta T_{text{cr}}}{m_{text{co}} , c_{text{co}} + m_{text{cu}} , c_{text{cu}} + m_{text{cr}} , c_{text{cr}} }
end{align*}
$$
Plugging in the values, along with $c_{text{co}} = c_{text{cr}} = c_{text{w}}$ we get:
$$
begin{align*}
T_{text{f}} &= frac{0.225text{ kg} cdot 4186 frac{text{J}}{text{kg}text{textdegree}text{C}} cdot 80.3text{textdegree}text{C} + 0.116text{ kg} cdot 1090 frac{text{J}}{text{kg}text{textdegree}text{C}} cdot 24text{textdegree}text{C} + 0.0122text{ kg} cdot 4186frac{text{J}}{text{kg}text{textdegree}text{C}} cdot 5.0text{textdegree}text{C}}{0.225text{ kg} cdot 4186frac{text{J}}{text{kg}text{textdegree}text{C}} + 0.116text{ kg} cdot 1090frac{text{J}}{text{kg}text{textdegree}text{C}} cdot 0.0122text{ kg} cdot 4186frac{text{J}}{text{kg}text{textdegree}text{C}} } \
T_{text{f}} &= 70.5text{textdegree}text{C}
end{align*}
$$
begin{align*}
T_{text{f}} = 70.5text{textdegree}text{C}
end{align*}
$$
They hit the droplets of perspiration on our skin, giving the water molecules enough speed to leave the liquid phase and evaporate.
This lowers the average temperature of the remaining molecules of water, and the new temperature gradient channels heat from the body to the droplet.
Hence the body loses thermal energy, and cools down.
Look at the solution of the previous problem for further explanation of said mechanism.
Keeping this in mind, we know that the answers containing changes in temperature of the system are invalid. \
All that remains is to analyze the sign of the latent heat in the phase transition process. \
As we know from everyday life, melting of ice requires an addition of thermal energy, so it follows that the freezing of water requires a removal of thermal energy. \ \
Hence the final answer is:
begin{enumerate}[d)]
item
a removal of thermal energy from the water
end{enumerate}
item
a removal of thermal energy from the water
end{enumerate}
As was mentioned in the solution of the previous problem, during a phase transition the temperature cannot change. As long as there is ice in the bucket containing water, heating the bucket will not raise its temperature, only melt the ice.
We conclude that the resulting temperature is $0text{textdegree}text{C}$
From the table we read the removed heats:
$$
begin{align*}
Q_{text{A}} &= 16,600text{ J} \
Q_{text{B}} &= 3150text{ J} \
Q_{text{C}} &= 3350text{ J} \
Q_{text{D}} &= 5400text{ J}
end{align*}
$$
and the masses
$$
begin{align*}
m_{text{A}} &= 0.05text{ kg} \
m_{text{B}} &= 0.025text{ kg} \
m_{text{C}} &= 0.01text{ kg} \
m_{text{D}} &= 0.05text{ kg}
end{align*}
$$
For all four liquids we find the latent heat of fusion using the formula that governs phase transiitons:
$$
begin{equation*}
Q = lambda , m
end{equation*}
$$
rearranging
$$
begin{equation*}
lambda = frac{Q}{m}
end{equation*}
$$
Plugging in the numbers for all liquids (A,B,C,D) we get:
$$
begin{align*}
lambda_{text{A}} &= 332 cdot 10^{3} frac{text{J}}{text{kg}} \
lambda_{text{B}} &= 126 cdot 10^{3} frac{text{J}}{text{kg}} \
lambda_{text{C}} &= 335 cdot 10^{3} frac{text{J}}{text{kg}} \
lambda_{text{D}} &= 108 cdot 10^{3} frac{text{J}}{text{kg}}
end{align*}
$$
Now to rank them in increasing order:
$$
begin{align*}
lambda_{text{D}} < lambda_{text{B}} < lambda_{text{A}} < lambda_{text{C}}
end{align*}
$$
begin{align*}
lambda_{text{D}} < lambda_{text{B}} < lambda_{text{A}} < lambda_{text{C}}
end{align*}
$$
Using the given graph, we can read the point at which the vapor pressure curve intersects with a vertical, constant temperature.
Looking at the temperature $30text{textdegree}text{C}$ we see that the intersection point is $4.25text{ kPa}$, which we will take for our final estimate.
4.25text{ kPa}
$$
For pressure $1.5text{ kPa}$, we will estimate the temperature of the intersection point to be $14text{textdegree}text{C}$
14text{textdegree}text{C}
$$
– The mass of water $m = 1.7text{ kg}$
– The initial and final temperatures $T_{text{i}} = T_{text{f}} = 0 text{textdegree}text{C}$
– The latent of heat of freezing water $lambda = 335 cdot 10^{3} frac{text{J}}{text{kg}}$
The amount of thermal energy that needs to be removed can be found using the formula that governs phase transitions:
$$
begin{equation*}
Q = m , lambda
end{equation*}
$$
plugging in the values we get:
$$
begin{align*}
Q &= 1.7text{ kg} cdot 335 cdot 10^{3} frac{text{J}}{text{kg}} \
Q &= 569500text{ J} \
Q &approx 570text{ kJ}
end{align*}
$$
Q approx 570text{ kJ}
$$
– The mass of lead $m = 0.95text{ kg}$
– The initial and final temperatures $T_{text{i}} = T_{text{f}} = 327.5text{textdegree}text{C}$
– The latent heat of fusion of lead $lambda = 2.32 cdot 10^{4} frac{text{J}}{text{kg}}$
The amount of thermal energy that needs to be added can be found using the formula that governs phase transitions:
$$
begin{equation*}
Q = m , lambda
end{equation*}
$$
plugging in the values we get:
$$
begin{align*}
Q &= 0.95text{ kg} cdot 2.32 cdot 10^{4} frac{text{J}}{text{kg}} \
Q &= 22040text{ J} \
Q &approx 22text{ kJ}
end{align*}
$$
Q approx 22text{ kJ}
$$
– The initial temperature $T_{text{i}} = – 12text{textdegree}text{C}$
– The final temperature $T_{text{f}} = 24text{textdegree}text{C}$
– The amount of thermal energy added $Q = 8.8 cdot 10^{5}text{ J}$
– The latent heat of fusion of water $lambda = 335 cdot 10^{3} , frac{text{J}}{text{kg}}$
– The specific heat capacity of ice $c_{text{i}} = 2090 , frac{text{J}}{text{kg}text{textdegree}text{C}}$
– The specific heat capacity of water $c_{text{w}} = 4186 , frac{text{J}}{text{kg}text{textdegree}text{C}}$
When heat is added to our system, first the temperature of the ice increases to its melting point ($0text{textdegree}text{C}$), then ice melts into water, and then the water is heated from $0text{textdegree}text{C}$ to $24text{textdegree}text{C}$. Hence the total amount of added thermal energy is:
$$
begin{align*}
Q = m , c_{text{i}} (0text{textdegree}text{C} – T_{text{i}}) + m , lambda + m , c_{text{w}} (T_{text{f}} – 0text{textdegree}text{C})
end{align*}
$$
Where we used the formulas that govern heating of bodies, and phase transitions:
$$
begin{align*}
Q &= m , c , Delta T \
Q &= m , lambda
end{align*}
$$
rearranging for mass we get:
$$
begin{align*}
m = frac{Q}{c_{text{i}}(0text{textdegree}text{C} – T_{text{i}}) + lambda + c_{text{w}}(T_{text{f}} – 0text{textdegree}text{C}) }
end{align*}
$$
Plugging in the values we get:
$$
begin{align*}
m &= frac{8.8 cdot 10^{5}text{ J} }{2090 frac{text{J}}{text{kg}text{textdegree}text{C}} cdot left(0 – (-12) right)text{textdegree}text{C} + 335 cdot 10^{3}frac{text{J}}{text{kg}} + 4186 frac{text{J}}{text{kg}text{textdegree}text{C}} cdot left(24 – 0 right)text{textdegree}text{C} } \
m &= 1.91text{ kg}
end{align*}
$$
m = 1.91text{ kg}
$$
– The initial temperature $T_{text{i}} = – 5 text{textdegree}text{C}$
– The mass of the block of ice $m = 1.1text{ kg}$
– The amount of thermal energy added $Q = 5.2 cdot 10^{5}text{ J} = 520text{ kJ}$
– The latent heat of fusion of water $lambda = 335 cdot 10^{3}frac{text{J}}{text{kg}}$
– The specific heat capacity of ice $c_{text{i}} = 2090 , frac{text{J}}{text{kg}text{textdegree}text{C}}$
– The specific heat capacity of water $c_{text{w}} = 4186 , frac{text{J}}{text{kg}text{textdegree}text{C}}$
When heat is added to our system, first the temperature of the ice increases to its melting point ($0text{textdegree}text{C}$), then the ice melts into water, and if there is enough energy, the whole ice will melt. Then the ice is heated from $0text{textdegree}text{C}$ to $T_{text{f}}$.
For the heating we will write:
$$
begin{align*}
Q_1 &= m , c_{text{i}} , Delta T = m , c_{text{i}} (0text{textdegree}text{C} – T_{text{i}}) \
Q_1 &= 11.496text{ kJ}
end{align*}
$$
then for the meling:
$$
begin{align*}
Q_2 &= m , lambda = 1.1text{ kg} cdot 335 cdot 10^{3} , frac{text{J}}{text{kg}} \
Q_2 &= 368.5text{ kJ}
end{align*}
$$
We see that $Q_1 + Q_2 < Q$ which means that the ice is totally melted.
For the heating of water we write:
$$
begin{align*}
Q_3 = m , c_{text{w}} (T_{text{f}} – 0text{textdegree}text{C})
end{align*}
$$
the total added thermal energy is the sum of the ones previously found:
$$
begin{align*}
Q = Q_1 + Q_2 + Q_3
end{align*}
$$
we find $Q_3$:
$$
begin{align*}
Q_3 &= Q – Q_1 – Q_2 \
Q_3 &= 520text{ kJ} – 11.495text{ kJ} – 368.5text{ kJ} \
Q_3 &approx 140text{ kJ}
end{align*}
$$
Now we find the final temperature using what we already found for $Q_3$
$$
begin{align*}
Q_3 &= m , c_{text{w}} , T_{text{f}} \
T_{text{f}} &= frac{Q_3}{m , c_{text{w}}} = frac{140text{ kJ}}{1.1text{ kg} cdot 4186 , frac{text{J}}{text{kg}text{textdegree}text{C}}} \
T_{text{f}} &= 30.4text{textdegree}text{C}
end{align*}
$$
We have found the final temperature to equal $T_{text{f}} = 30.4text{textdegree}text{C}$, obviously there is no ice in the end, it has all melted.
begin{align*}
T_{text{f}} = 30.4text{textdegree}text{C}
text{ , all the ice has melted.}
end{align*}
$$
– The mass of water $m = 865text{ kg}$
– The power of the heater $P = 200text{ kW}$
– The initial temperature of the water $T_{text{i}} = 20text{textdegree}text{C}$
– The final temperature of the water $T_{text{f}} = 0text{textdegree}text{C}$
– The latent heat of fusion of water $lambda = 335 cdot 10^{3} , frac{text{J}}{text{kg}}$
– The specific heat capacity of water $c_{text{i}} = 4186 , frac{text{J}}{text{kg} text{textdegree}text{C}}$
To find the total heat released by the water during its cooling and freezing, we use the following Laws, for the cooling of objects:
$$
begin{equation*}
Q_1 = m , c , Delta T = m , c (T_{text{f}} – T_{text{i}})
end{equation*}
$$
Notice that the sign of $Q_1$ is negative. By convention this means that body releases thermal energy. We can safely remove the minus and continue with $Q_1 > 0$, bearing in mind that it is released thermal energy.
And for phase transitions:
$$
begin{equation*}
Q_2 = m , lambda
end{equation*}
$$
For the total process we have:
$$
begin{align*}
Q &= Q_1 + Q_2 \
Q &= m , c (T_{text{i}} – T_{text{f}}) + m , lambda
end{align*}
$$
Plugging in the numbers we get:
$$
begin{align*}
Q &= 865text{ kg} left(4186 , frac{text{J}}{text{kg}text{textdegree}text{C}} (20text{textdegree}text{C} – 0text{textdegree}text{C}) + 335 cdot 10^{3} , frac{text{J}}{text{kg}} right) \
Q &= 362192800text{ J} \
Q &approx 3.622 cdot 10^{8}text{ J}
end{align*}
$$
Our heater of power $P = 200text{ kW}$ would need to operate for:
$$
begin{align*}
P &= frac{Q}{t} Rightarrow t= frac{Q}{P} \
t &= frac{3.622 cdot 10^{8}text{ J}}{2000text{ W}} = 181100text{ s} approx 50.5text{ h}
end{align*}
$$
The amount of thermal energy stored in the water is equivalent to $50.5text{ h}$ of the heater working.
t = 50.5text{ h}
$$
– The mass of water $m = 1text{ kg}$
– The rate of thermal energy addition $P = 12,250 , frac{text{J}}{text{s}}$
– The initial temperature of the water $T_{text{i}} = 0text{textdegree}text{C}$
– The final temperature of the water $T_{text{f}} = 15text{textdegree}text{C}$
– The latent heat of fusion of water $lambda = 335 cdot 10^{3} , frac{text{J}}{text{kg}}$
– The specific heat capacity of water $c = 4186 , frac{text{J}}{text{kg}text{textdegree}text{C}}$
To find the total thermal energy needed to raise the temperature of water we use the following Laws, for phase transitions:
$$
begin{equation*}
Q_1 = m , lambda
end{equation*}
$$
and for the heating of objects:
$$
begin{equation*}
Q_2 = m , c (T_{text{f}} – T_{text{i}})
end{equation*}
$$
For the total process we have:
$$
begin{align*}
Q &= Q_1 + Q_2 \
Q &= m , lambda + m , c (T_{text{f}} – T_{text{i}})
end{align*}
$$
Plugging in the numbers we get:
$$
begin{align*}
Q &= 1text{ kg} left(335 cdot 10^{3} , frac{text{J}}{text{kg}} + 4186 , frac{text{J}}{text{kg}text{textdegree}text{C}} (15text{textdegree}text{C} – 0text{textdegree}text{C}) right) \
Q &= 397790 \
Q &approx 398text{ kJ}
end{align*}
$$
With the given rate of addition of thermal energy the required time is
$$
begin{align*}
t &= frac{Q}{P} = frac{398text{ kJ}}{12,250 , frac{text{J}}{text{s}}} = 31.81text{ s} \
t &approx 32text{ s}
end{align*}
$$
The required time is found to be $t = 32text{ s}$
Note that the total thermal energy $Q$ could have been read from the graph, but this way is more accurate.
t = 32text{ s}
$$
– The mass of lemonade(water) $m_{text{w}} = 3.99text{ kg}$
– The mass of ice $m_{text{i}} = 0.0550text{ kg}$
– The initial temperature of water $T_{text{w}} = 20.5text{textdegree}text{C}$
– The initial temperature of ice $T_{text{i}} = – 10.2text{textdegree}text{C}$
– The latent heat of fusion of water $lambda = 335 cdot 10^{3} , frac{text{J}}{text{kg}}$
– The specific heat capacity of water $c_{text{w}} = 4186 , frac{text{J}}{text{kg}text{textdegree}text{C}}$
– The specific heat capacity of ice $c_{text{i}} = 2090 , frac{text{J}}{text{kg}text{textdegree}text{C}}$
Comparing the total thermal energy needed to heat and melt all the ice:
$$
begin{align*}
Q_1 &= m_{text{i}} , c_{text{i}} (0text{textdegree}text{C} – T_{text{i}}) + m_{text{i}} , lambda \
Q_1 &approx 19600text{ J}
end{align*}
$$
and the thermal energy the water can supply via cooling only:
$$
begin{align*}
Q_2 &= m_{text{w}}(T_{text{w}} – 0text{textdegree}text{C}) \
Q_2 &approx 342400text{ J}
end{align*}
$$
We see that $Q_2 > Q_1$ by far, meaning all the ice will melt, and the final temperature $T_{text{f}} > 0text{textdegree}text{C}$
To find it we simply write:
$$
begin{align*}
Q_1 = m_{text{w}} , c_{text{w}} (T_{text{w}} – T_{text{f}})
end{align*}
$$
rearranging for $T_{text{f}}$ we have:
$$
begin{align*}
T_{text{f}} = T_{text{w}} – frac{Q_1}{m_{text{w}} , c_{text{w}}}
end{align*}
$$
Plugging in the values we get:
$$
begin{align*}
T_{text{f}} = 20.5text{textdegree}text{C} – frac{19600text{ J}}{3.99text{ kg} cdot 4186 , frac{text{J}}{text{kg}text{textdegree}text{C}}} approx 19.3text{textdegree}text{C}
end{align*}
$$
T_{text{f}} = 19.3text{textdegree}text{C} text{ , there is no ice}
$$
– The mass of water $m_{text{w}} = 80text{ g}$
– The mass of aluminum $m_{text{A}} = 155text{ g}$
– The initial temperature of water $T_{text{w}} = 15text{textdegree}text{C}$
– The initial temperature of aluminum $T_{text{A}} = -196text{textdegree}text{C}$
– The latent heat of fusion of water $lambda = 335 cdot 10^{3} , frac{text{J}}{text{kg}}$
– The specific heat capacity of water $c_{text{w}} = 4186 , frac{text{J}}{text{kg}text{textdegree}text{C}}$
– The specific heat capacity of aluminum $c_{text{A}} = 653 , frac{text{J}}{text{kg}text{textdegree}text{C}}$
Comparing the total thermal energy needed to cool and freeze all the water:
$$
begin{align*}
Q_1 &= m_{text{w}} , c_{text{w}} (T_{text{w}} – 0text{textdegree}text{C}) + m_{text{w}} , lambda \
Q_1 &approx 31800text{ J}
end{align*}
$$
and the thermal energy the aluminum can absorb heating to $0text{textdegree}text{C}$
$$
begin{align*}
Q_2 &= m_{text{A}} , c_{text{A}} (0text{textdegree}text{C} – T_{text{A}}) \
Q_2 &approx 19840text{ J}
end{align*}
$$
We see that $Q_2 < Q_1$, which means not all the ice will melt, and the final temperature $T_{text{f}} = 0text{textdegree}text{C}$
To find it we simply write:
$$
begin{align*}
Q_2 = m_{text{w}} , c_{text{w}} (T_{text{w}} – 0text{textdegree}text{C}) + Delta m , lambda
end{align*}
$$
rearranging for $Delta m$:
$$
begin{align*}
Delta m = frac{Q_2 – m_{text{w}} , c_{text{w}} , T_{text{w}}}{lambda}
end{align*}
$$
Plugging in the numbers:
$$
begin{align*}
Delta m &= frac{19740text{ J} – 0.08text{ kg} cdot 4186 , frac{text{J}}{text{kg}text{textdegree}text{C}} cdot 15text{textdegree}text{C} }{335 cdot 10^{3} , frac{text{J}}{text{kg}}} \
Delta m &approx 44.23text{ g}
end{align*}
$$
The photons of the Sun emits carry energy towards the Earth, this form of thermal energy flow can exist without a medium to pass through (unlike conduction and convention)
If it is used outdoors when the temperature is $0text{textdegree}text{C}$, it will cool down. As it temperature lowers it will shrink. This is well known from the theory, objects expand when heated.
As the tape shrink the intervals between successive markings will also shrink. This means that more intervals will be needed to cover the same length, which means the tape will give measurements that are too long.
item
To determine the phase of water at a given point on the figure, we simply find the point with the described values of pressure, and see if the point is above (liquid) or below (gas) the vapor-pressure curve. The results are as follows:
begin{align*}
& 20textdegreetext{C}, 1text{ kPa} – text{gas}\
& 20textdegreetext{C}, 2text{ kPa} – text{gas} \
& 20textdegreetext{C}, 3text{ kPa} – text{liquid}
end{align*}
item
To find the boiling point of water at a given pressure, we simply read the temperature of the intersection point a horizontal line on that pressure, and the vapor-pressure curve. The results are as follows:
begin{align*}
& T(1text{ kPa}) = 8textdegreetext{C} \
& T(2text{ kPa}) = 18textdegreetext{C} \
& T(3text{ kPa}) = 24textdegreetext{C}
end{align*}
end{enumerate}
The posed question is will the rod ever touch the ring, considering the fact they weren’t in touching in the beginning. This depends solely on the value of the coefficient of thermal expansion of the metal the rod is made out of. \\
If said values is higher then the one of copper (of which the ring is made), then a sufficient increase in temperature will see the rod touch ring. \
If the values is lower, then heating will never bring about the touching of the rod and the ring. (The student will note that cooling of the system would give the required end, even though that wasn’t part of the question.) \
if the value is the same, then touching is impossible for any temperature. \\
begin{enumerate}[a)]
item
for copper the values are the same, there is no touch.
item
for aluminum we have $alpha_{text{A}} > alpha_{text{C}}$, meaning they will touch after sufficient heating.
item
for steel we have $alpha_{text{S}} < alpha_{text{C}}$, meaning no amount of heating will lead to touching
end{enumerate}
item
no
item
yes
item
no
end{enumerate}
We use the well knows conversion formula:
$$
begin{equation*}
T_{text{F}} = frac{9}{5} , T_{text{C}} + 32
end{equation*}
$$
Plugging in the values
$$
begin{align*}
T_{text{F}} = frac{9}{5} left(121 right) + 32 approx 250text{textdegree}text{F}
end{align*}
$$
T_{text{F}} = 250text{textdegree}text{F}
$$
$$
begin{align*}
Q_{text{A}} &= 33,500text{ J} \
Q_{text{B}} &= 166,000text{ J} \
m_{text{A}} &= 0.100text{ kg} \
m_{text{B}} &= 0.500text{ kg}
end{align*}
$$
For both liquids we find the latent heat of fusion using the formula that governs phase transitions:
$$
begin{equation*}
Q = lambda , m
end{equation*}
$$
rearranging:
$$
begin{equation*}
lambda = frac{Q}{m}
end{equation*}
$$
Plugging in the numbers for both liquids we get:
$$
begin{align*}
lambda_{text{A}} &= 335 cdot 10^{3} , frac{text{J}}{text{kg}} \
lambda_{text{B}} &= 332 cdot 10^{3} , frac{text{J}}{text{kg}}
end{align*}
$$
we conclude:
$$
begin{align*}
lambda_{text{A}} > lambda_{text{B}}
end{align*}
$$
– The amount of thermal energy removed $Q = 5100text{ J}$
– The mass of methanol vaporized $m = 4.6text{ g}$
As heat is removed from the laptop methanol vaporized.
We find the latent heat of fusion using the formula that governs phase transitions:
$$
begin{equation*}
Q = lambda , m
end{equation*}
$$
rearranging:
$$
begin{equation*}
lambda = frac{Q}{m}
end{equation*}
$$
Plugging in the numbers we get:
$$
begin{align*}
lambda &= frac{5100text{ J}}{0.0046text{ kg}} = 1108700 , frac{text{J}}{text{kg}} \
lambda &approx 1.1 cdot 10^{6} , frac{text{J}}{text{kg}}
end{align*}
$$
lambda approx 1.1 cdot 10^{6} , frac{text{J}}{text{kg}}
$$
$$
begin{gather*}
Q_{text{A}} = Q quad , quad Q_{text{B}} = 2Q quad , quad Q_{text{C}} = 3Q quad , quad Q_{text{D}} = 4Q \
m_{text{A}} = m quad , quad m_{text{B}} = 3m quad , quad m_{text{C}} = 3m quad , quad m_{text{D}} = 4m \
Delta T_{text{A}} = Delta T quad , quad Delta T_{text{B}} = 3Delta T quad , quad Delta T_{text{c}} = Delta T quad , quad Delta T_{text{D}} = 2Delta T
end{gather*}
$$
Now we find the specific heat capacity for all four fluids, using the general formula:
$$
begin{align*}
Q = m , c , Delta T Rightarrow c = frac{Q}{m , Delta T}
end{align*}
$$
the results are as follows:
$$
begin{align*}
c_{text{A}} &= 1 , frac{Q}{m , Delta T} \
c_{text{B}} &= frac{1}{3}frac{Q}{m , Delta T} \
c_{text{C}} &= 1 frac{Q}{m , Delta T} \
c_{text{D}} &= frac{1}{2} frac{Q}{m , Delta T}
end{align*}
$$
rearranging the specific heat capacities in increasing order we get:
$$
begin{align*}
c_{text{B}} < c_{text{D}} < c_{text{A}} < c_{text{C}}
end{align*}
$$
c_{text{B}} < c_{text{D}} < c_{text{A}} < c_{text{C}}
$$
– The mass of water $m = 150text{ g}$
– The time interval of heating $t = 2.5text{ min}$
– The rate of thermal energy addition $P = 55 ; frac{text{J}}{text{s}}$
– The specific heat capacity of water $c = 4186 ; frac{text{J}}{text{kg}text{textdegree}text{C}}$
First we express the total thermal energy added using the rate of thermal energy addition (heating rate):
$$
begin{equation*}
Q = P cdot t
end{equation*}
$$
The addition of thermal energy causes an increase in temperature:
$$
begin{equation*}
Q = m , c , Delta T
end{equation*}
$$
combining those formulas we find the change in temperature:
$$
begin{gather*}
P , t = Q = m , c , Delta T \
Delta T = frac{P , t}{m , c}
end{gather*}
$$
Plugging in the numbers:
$$
begin{align*}
Delta T = frac{55 ; frac{text{J}}{text{s}} left(2.5 cdot 60text{ s} right) }{0.150text{ kg} cdot 4186 ; frac{text{J}}{text{kg}text{textdegree}text{C}} } approx 13.14text{textdegree}text{C}
end{align*}
$$
Delta T = 13.14text{textdegree}text{C}
$$
– The mass of water $m = 180text{ g}$
– The time interval of heating $t = 3.5text{ min}$
– The change in temperature $Delta T = 12text{textdegree}text{C}$
– The specific heat capacity of water $c = 4186 ; frac{text{J}}{text{kg}text{textdegree}text{C}}$
First we express the total thermal energy added using the rate of thermal energy addition (heating rate):
$$
begin{equation*}
Q = P cdot t
end{equation*}
$$
The addition of thermal energy causes an increase in temperature:
$$
begin{equation*}
Q = m , c , Delta T
end{equation*}
$$
combining those formulas we find the heating rate:
$$
begin{gather*}
P , t = Q = m , c , Delta T \
P = frac{m , c , Delta T}{t}
end{gather*}
$$
plugging in the numbers:
$$
begin{align*}
P &= frac{0.18text{ kg} cdot 4186 ; frac{text{J}}{text{kg}text{textdegree}text{C}} cdot 12text{textdegree}text{C} }{3.5 cdot 60text{ s}} = 43.056 ; frac{text{J}}{text{s}} \
P &approx 43.1 ; frac{text{J}}{text{s}}
end{align*}
$$
P = 43.1 ; frac{text{J}}{text{s}}
$$
– The mass of steam $m = 1.5text{ kg}$
– The initial temperature $T_{text{i}} = 110text{textdegree}text{C}$
– The final temperature $T_{text{f}} = 0text{textdegree}text{C}$
– The specific heat capacity of water $c_{text{w}} = 4186 ; frac{text{J}}{text{kg}text{textdegree}text{C}}$
– The specific heat capacity of steam $c_{text{s}} = 2010 ; frac{text{J}}{text{kg}text{textdegree}text{C}}$
– The latent heat of fusion of water $L_{text{f}} = 335 cdot 10^{3} ; frac{text{J}}{text{kg}}$
– The latent heat of evaporation of water $L_{text{v}} = 2260 cdot 10^{3} ; frac{text{J}}{text{kg}}$
The total thermal energy that needs to be removed consists of four elements.
The cooling steam, the condensation of steam, the heating of liquid water, and the freezing. These are found routinely using the laws that govern the heating of objects, and phase transitions:
$$
begin{align*}
Q_1 &= m , c_{text{s}} (T_{text{i}} – 100text{textdegree}text{C}) \
Q_2 &= m , L_{text{v}} \
Q_3 &= m , c_{text{w}} (100text{textdegree}text{C} – 0text{textdegree}text{C}) \
Q_4 &= m , L_{text{f}}
end{align*}
$$
The signs in $Q_1$ and $Q_2$ are chosen in such a way that they be positive. Bear in mind that thermal energy is actually leaving the system.
Plugging in the numbers we calculate:
$$
begin{align*}
Q_1 &= 1.5text{ kg} cdot 2010 ; frac{text{J}}{text{kg}text{textdegree}text{C}} cdot (110 – 100)text{textdegree}text{C} = 30.15text{ kJ} \
Q_2 &= 1.5text{ kg} cdot 2260 cdot 10^{3} ; frac{text{J}}{text{kg}} = 3390text{ kJ} \
Q_3 &= 1.5text{ kg} cdot 4186 ; frac{text{J}}{text{kg}text{textdegree}text{C}} cdot 100text{textdegree}text{C} = 627.9text{ kJ} \
Q_4 &= 1.5text{ kg} cdot 335 cdot 10^{3} ; frac{text{J}}{text{kg}} = 5025text{ kJ}
end{align*}
$$
adding all those we get the total thermal energy:
$$
begin{align*}
Q &= Q_1 +Q_2 + Q_3 + Q_4 \
Q &= 30.15text{ kJ} + 3390text{ kJ} + 627.9text{ kJ} + 5025text{ kJ} = 9073text{ kJ}
end{align*}
$$
Q = 9073text{ kJ}
$$
– The mass of steam $m = 45text{ kg}$
– The change in temperature $Delta T = 12text{textdegree}text{C}$
– The specific heat capacity of water $c_{text{w}} = 4186 ; frac{text{J}}{text{kg}text{textdegree}text{C}}$
– The area of the solar collector $A = 5.5text{ m}^2$
– The power of sunlight $T = 520 ; frac{text{w}}{text{m}^2}$
First we use the area of the collector and power per unit of area of sunlight to find the rate of thermal energy collection:
$$
begin{align*}
P = frac{Delta Q}{Delta t} = I cdot A = 520 ; frac{text{w}}{text{m}^2} cdot 5.5text{ m}^2 = 2860text{ W}
end{align*}
$$
We can think about $P$ as the power of an “effective sun heater”
Considering this, we find the time this “heater” needs to work to collect enough energy for the heating of water:
$$
begin{align*}
Q &= P , t \
Q &= m , c_{text{w}} , Delta T
end{align*}
$$
equating the right sides, we rearrange for time, and find:
$$
begin{align*}
t =frac{m , c_{text{w}} , Delta T }{P}
end{align*}
$$
Plugging in the numbers we find:
$$
begin{align*}
t = frac{45text{ kg} cdot 4186 ; frac{text{J}}{text{kg}text{textdegree}text{C}} cdot 12text{textdegree}text{C} }{2860text{ W} } = 790.36text{ s}
end{align*}
$$
converting to minutes:
$$
begin{align*}
t approx 13.2text{ min}
end{align*}
$$
t = 13.2text{ min}
$$
– The mass of water $m_{text{w}} = 1.12text{ kg}$
– The mass of ice $m_{text{i}} = 0.13text{ kg}$
– The initial temperature $T_{text{i}} = 0text{textdegree}text{C}$
– The final temperature $T_{text{f}} = 15text{textdegree}text{C}$
– The specific heat capacity of water $c_{text{w}} = 4186 ; frac{text{J}}{text{kg}text{textdegree}text{C}}$
– The latent heat of fusion of water $L_{text{f}} = 335 cdot 10^{3} ; frac{text{J}}{text{kg}}$
Our process consists first of the melting of ice, and then the raising of temperature of the total mass $m = m_{text{i}} + m_{text{w}}$ from $0text{textdegree}text{C}$ to $T_{text{f}}$
The processes are described seperately:
$$
begin{align*}
Q_1 &= m_{text{i}} , T_{text{f}} \
Q_2 &= (m_{text{w}} + m_{text{i}}) c_{text{w}} (T_{text{f}} – 0text{textdegree}text{C})
end{align*}
$$
Plugging in the numbers we get:
$$
begin{align*}
Q_1 &= 0.13text{ kg} cdot 335 cdot 10^{3} ; frac{text{J}}{text{kg}} = 43550text{ J} \
Q_2 &= (1.12 + 0.13)text{ kg} cdot 4186 ; frac{text{J}}{text{kg}text{textdegree}text{C}} left(15text{textdegree}text{C} – 0text{textdegree}text{C} right) = 78487text{ J}
end{align*}
$$
Now we find the total thermal energy needed:
$$
begin{align*}
Q &= Q_1 + Q_2 \
Q &= 43550text{ J} + 78487text{ J} = 122text{ kJ}
end{align*}
$$
Q = 122text{ kJ}
$$
– The initial diameter $L_{text{iA}} = 10.0text{ cm}$
– The initial rod length $L_{text{iS}} = 9.99text{ cm}$
– Coefficient of thermal expansion of aluminum $alpha_{text{A}} = 24 cdot 10^{-6} text{ k}^{-1}$
– Coefficient of thermal expansion of steel $alpha_{text{S}} = 12 cdot 10^{-6}text{ k}^{-1}$
begin{enumerate}[a)]
item
Fist we will notice that the coefficient of thermal expansion of aluminum is higher than that of steel. \
This means that with further heating of the system the difference in the diameter of the aluminum sheet and steel rod will only increase further. Touching can be accomplished only through removal of thermal energy (cooling).
item
Now we find the change in temperature explicitly. As the system cools down the aluminum sheet must shrink more than the rod for the initial difference in length, this gives the following condition:
begin{align*}
Delta L_{text{A}} – Delta L_{text{s}} = L_{text{iA}} – L_{text{iS}}
end{align*}
Where the individual changes in length due to thermal expansion (shrinking) are found to be:
begin{align*}
Delta L_{text{A}} &= alpha_{text{A}} , L_{text{iA}} , Delta T \
Delta L_{text{S}} &= alpha_{text{S}} , L_{text{iS}} , Delta T
end{align*}
combining all the previous results w solve for $Delta T$:
begin{gather*}
(alpha_{text{A}} , L_{text{iA}} – alpha_{text{s}} , L_{text{iS}} )Delta T = L_{text{iA}} – L_{text{iS}} \
Delta T = frac{L_{text{iA}} – L_{text{is}} }{alpha_{text{A}} , L_{text{iA}} – alpha_{text{S}} , L_{text{iS}} }
end{gather*}
Plugging in the values we get:
begin{align*}
Delta T = frac{10text{ cm} – 9.99text{ cm}}{24 cdot 10^{-6}text{ k}^{-1} cdot 10text{ cm} – 12 cdot 10^{-6}text{ k}^{-1} cdot 9.99text{ cm} } = 83.25textdegreetext{C}
end{align*}
end{enumerate}
item
decreased
item
$Delta T = 83.25textdegreetext{C}$
end{enumerate}
– The mass of piece of steel $m_{text{s}} = 0.33text{ kg}$
– The mass of the chunk of lead $m_{text{l}} = 0.51text{ kg}$
– The temperature of the piece of steel $T_{text{s}} = 42text{textdegree}text{C}$
– The specific heat capacity of steel $c_{text{s}} = 448 ; frac{text{J}}{text{kg}text{textdegree}text{C}}$
– The specific heat capacity of lead $c_{text{l}} = 128 ; frac{text{J}}{text{kg}text{textdegree}text{C}}$
begin{enumerate}[a)]
item
We know that the piece of steel releases thermal energy and cools down from $42textdegreetext{C}$ to $22textdegreetext{C}$. Since the temperature of the water doesn’t change we conclude that internal energy released by steel is absorbed by the chunk of lead. Hence the initial temperature of lead was less than $22textdegreetext{C}$
item
The exact temperature of lead is found using the heat balance.
begin{equation*}
Q_{text{s}} + Q_{text{l}} = 0
end{equation*}
and the law that governs heating of objects:
begin{align*}
Q_{text{s}} &= m_{text{s}} , c_{text{s}} Delta T_{text{s}} = m_{text{s}} , c_{text{s}} (22textdegreetext{C} – T_{text{s}}) \
Q_{text{l}} &= m_{text{l}} , c_{text{l}} , Delta T_{text{l}} = m_{text{l}} , c_{text{l}} (22textdegreetext{C} – T_{text{l}})
end{align*}
combining the results and solving for $T_{text{l}}$ we have:
begin{gather*}
m_{text{s}} , c_{text{s}} (22textdegreetext{C} – T_{text{s}}) + m_{text{l}} , c_{text{l}}(22textdegreetext{C} – T_{text{l}}) = 0 \
T_{text{l}} = frac{m_{text{s}} , c_{text{s}}(22textdegreetext{C} – T_{text{s}}) + m_{text{l}} , c_{text{l}} cdot 22textdegreetext{C} }{m_{text{l}} , c_{text{l}} } \
T_{text{l}} = 22textdegreetext{C} + frac{m_{text{s}} , c_{text{s}} }{m_{text{l}} , c_{text{l}}} left(22textdegreetext{C} – T_{text{s}} right)
end{gather*}
Plugging in the values we get:
begin{align*}
T_{text{l}} &=22textdegreetext{C} frac{0.33text{ kg}}{0.51text{ kg}} cdot frac{448 ; frac{text{J}}{text{kg}textdegreetext{C}} }{128 ; frac{text{J}}{text{kg}textdegreetext{C}} } left(22textdegreetext{C} – 42textdegreetext{C} right) \
T_{text{l}} &approx – 23.3textdegreetext{C}
end{align*}
end{enumerate}
item
less than $22textdegreetext{C}$
item
$T_{text{l}}$ = – 23.3textdegreetext{C}
end{enumerate}
$$
begin{align*}
frac{185}{60text{ s}} = frac{text{N}}{13text{ s}} Rightarrow text{N} = frac{13text{ s}}{60text{ s}} cdot 185 = 40.083 approx 40.08
end{align*}
$$
Now we will use the chirp rate-temperature relationship:
$$
begin{align*}
text{N} = left(5.63 cdot 10^{10} right)e^{-frac{(6290text{ k})}{T}}
end{align*}
$$
To find $T$ we take the neutral logarith of both sides:
$$
begin{align*}
ln left(frac{text{N}}{5.63 cdot 10^{10}} right) = – frac{6290text{ K}}{T}
end{align*}
$$
rearranging for $T$:
$$
begin{align*}
T = – frac{6290text{ K}}{ln left(frac{text{N}}{5.63 cdot 10^{10}} right)}
end{align*}
$$
Plugging in the values:
$$
begin{align*}
T = – frac{6290text{ K}}{ln left(frac{40.08}{5.63 cdot 10^{10}} right)} = 298.63text{ K}
end{align*}
$$
Now we convert the temperature to the Celsius scale as follows:
$$
begin{align*}
T_{text{c}} &= (T – 273.15)text{textdegree}text{C} \
T_{text{c}} &= (298.63 – 273.15)text{textdegree}text{C} \
T_{text{c}} &= 25.48text{textdegree}text{C}
end{align*}
$$
T_{text{c}} = 25.48text{textdegree}text{C}
$$
In the contact point between two sticks there is friction. This friction effectively turns the kinetic energy of the sticks into thermal energy of the particles near the contact point, raising their temperature. If this conversion of energy is done at a sufficient rate to overcome the losses through conduction etc, the temperature will become high enough to ignite the wood.
If steel sticks were used instead of wood sticks the method wouldn’t be effective, because the coefficient of friction between steel is much lower than the one corresponding to wood.
Wooden sticks are more effective than steel rods because they have higher coefficient of friction.
– The initial length of the Blackbird $L_{text{I}} = 32.74text{ m}$
– he change in length of the Blackbird $Delta L = 20text{ cm}$
– The coefficient of thermal expansion of the Blackbird $alpha = 22 cdot 10^{-6}text{ k}^{-1}$
– The temperature at takeoff $T_{text{i}} = 23text{textdegree}text{C}$
During flight the Blackbird gets extremely hot because of air resistance.
This leads to noticeable thermal expansion. Using the well known law:
$$
begin{equation*}
Delta L = alpha , L_{text{i}} , Delta T
end{equation*}
$$
where the change in temperature is expressed as follows:
$$
begin{align*}
Delta T = T_{text{f}} – T_{text{i}}
end{align*}
$$
rearranging for $T_{text{f}}$ we get:
$$
begin{align*}
T_{text{f}} = T_{text{i}} + frac{Delta L}{alpha , L_{text{i}}}
end{align*}
$$
plugging in the numbers we get:
$$
begin{align*}
T_{text{f}} &= 23text{textdegree}text{C} + frac{0.20text{ m}}{22 cdot 10^{-6}text{ k}^{-1} cdot 32.74text{ m}} \
T_{text{f}} &= 23text{textdegree}text{C} + 277.67text{textdegree}text{C} \
T_{text{f}} &= 300.67text{textdegree}text{C} approx 300text{textdegree}text{C}
end{align*}
$$
So the solution is C. $300text{textdegree}text{C}$
– The initial length of the Blackbird $L_{text{i}} = 32.74text{ m}$
– The coefficient of thermal expansion of the Blackbird $22 cdot 10^{-6}text{ k}^{-1}$
– The temperature at takeoff $T_{text{i}} = 23text{textdegree}text{C}$
– The final temperature $T_{text{f}} = 120text{textdegree}text{C}$
During flight the Blackbird gets extremely hot because of air resistance.
This leads to noticeable thermal expansion. Using the well known law:
$$
begin{equation*}
Delta L = alpha , L_{text{i}} , Delta T
end{equation*}
$$
where the change in temperature is expressed as follows:
$$
begin{align*}
Delta T = T_{text{f}} – T_{text{i}}
end{align*}
$$
combining these we get:
$$
begin{align*}
Delta L = alpha , L_{text{i}} (T_{text{f}} – T_{text{i}})
end{align*}
$$
the final length is found as follows:
$$
begin{align*}
L_{text{f}} &= L_{text{i}} + Delta L \
L_{text{f}} &= L_{text{i}} + alpha , L_{text{i}} (T_{text{f}} – T_{text{i}}) = L_{text{i}} (1 + alpha (T_{text{f}} – T_{text{i}}))
end{align*}
$$
Plugging in the numbers we get:
$$
begin{align*}
L_{text{f}} &= 32.74text{ m} left(1 + 22 cdot 10^{-6}text{ k}^{-1}(120text{textdegree}text{C} – 23text{textdegree}text{C}) right) \
L_{text{f}} &= 32.74text{ m} cdot 1.002134 \
L_{text{f}} &approx 32.81text{ m}
end{align*}
$$
Converting this to inches:
$$
begin{align*}
L_{text{f}} = 32.81text{ m} cdot 39.37 ; frac{text{in}}{text{m}} = 1291.73text{ in}
end{align*}
$$
And then to feet:
$$
begin{align*}
L_{text{f}} = 1281.73text{ in} cdot 0.08333 ; frac{text{ft}}{text{in}} = 107.6437text{ ft} = 107text{ ft} , 7.7text{ in}
end{align*}
$$
The closest option we can chose is C. $107text{ ft} , 7.5text{ in}$
