All Solutions
Section 10: Chapter 10
tt{Conversion between a Celsius Temperature and a Kelvin Temperature follows this rule:}
$$
$$
T=T_c+273.15
$$
with $T$ being the temperature in Kelvin and $T_c$ being the temperature in Celsius, we can see that the relationship is linear and therefore there is no temperature value where the two scales agree.
tt{No}
$$
$$
T_F=frac{9}{5}T_C+32
$$
Where $T_F$ is the temperature value in Fahrenheit and $T_C$ is the temperature value in Celsius.
We are looking for a temperature in Fahrenheit that is 3 times greater than its Celsius counterpart:
$$
T_F=3T_C
$$
$$
Rightarrow 3T_C=frac{9}{5}T_C+32
$$
$$
T_C=frac{32}{3-frac{9}{5}}=26.66^otext{ C}
$$
$$
Rightarrow T_F=frac{9}{5}26.66^otext{C}+32=boxed{79.98^otext{F}}
$$
T_F=79.98^otext{F}
$$
tt{Most materials contract when cooled and expand when heated:
\So when the hollow pipe is cooled its diameter decreases because the matter is contracting}
$$
The general volume formula of a cylinder is :
$$
v=Pi r^2h
$$
with $r$ being the radius of the cylinder base and $h$ the height.
We can see clearly that the volume is proportional to the radius, therefore if the radius decreases as in our case because of the material contraction, then the volume decreases as well.
tt{decrease}
$$
This type of thermal transfer is called “conduction” and it takes place when higher energy particles (from the hotter object) collide with and jostle lower energy particles (from the colder object )
Metals are in general better heat conductors than wood therefore heat exchange happens faster, thus giving the person the cooler feeling of the metal.
tt{ (a)The definition of Specific Heat Capacity, $c$ is :
$$c=frac{Q}{mDelta T}$$
$$Rightarrow Delta T=frac{Q}{cm}$$
where $Q$ is the heat exchanged, $Delta T$ is a temperature difference, and $m$ is the mass. \
in our case, object 1 has a mass $m$ and object 2 has mass $2m$. \
Given the two objects are made of the same material, they have the same heat capacity, and since they are in contact the same amount of heat lost/gained by object 1 is equal to the same heat lost/gained by object 2. \
this leaves us with one variable in the equation which is the mass $m$, we can notice that $m$ is inversely proportional to the temperature difference meaning if $muparrow$ then $Delta Tdownarrow$
We know object 1 has a smaller mass than 2 therefore the temperature difference in object one is {color{#4257b2}{greater}} than in object 2. \
(b) Explanation C
}
$$
tt{(a) Greater , (b) C}
$$
tt{Joule’s device is based on the energy conservation principle, in this experiment, the work done by both masses when falling move the paddles and heat the water, in essence, we can say that the heat received by the water is equal to the work done by the two masses, this yields: }
$$
$$
Q=2*mgh=2*0.95*0.48*9.81=8.95J
$$
Knowing that 6200J is needed to increase the temperature by $1^oC$ we can deduce that:
$$
Delta T=frac{8.95}{6200}=0.00144^oC
$$
Delta T =0.00144^oC
$$
tt{The ball is free falling and thus converting its potential energy to Kinetic energy, Assuming that at the moment of impact all kinetic energy accumulated was transferred to heat $Q$, we can deduce that: }
$$
$$
PE=KE=Q
$$
$$
mgh=Q text{ (1)}
$$
on the other hand, based on the definition of Specific Heat Capacity, $c$ we have:
$$
c=frac{Q}{mDelta T}Rightarrow Q=mcDelta T text{ (2)}
$$
with $m$ being the mass and $Delta T$ the change in temperature.
Substituting (2) into (1) yields:
$$
mcDelta T=mgh
$$
$$
Rightarrow Delta T = frac{gh}{c}
$$
Given that the specific heat of the lead is $128 frac{J}{kg.^oC}$, we calculate:
$$
Delta T = frac{9.81*5.43}{128}=0.42^oC
$$
Delta T =0.42^oC
$$
tt{The water under any temperature needs to reach what we call “phase equilibrium”, basically the water contains particles with different levels of energy, the most energetic ones are able to escape the liquid and start creating vapor, in a closed system the vapor continue to condensate until it reaches an equilibrium with the liquid underneath it, this means the number of particles that are exiting the liquid is equal to the one’s entering it. \
In our case the drop of water is placed in an open space thus particles continue to shoot out of the liquid and since equilibrium can not be reached all liquid evaporates.}
$$
tt{evaporation happens due to the phase equilibrium not being reached}
$$
Since this system is isolated, all energy is conserved and we can assume that all heat released by iron block $Q_{iron}$ is received by the water $Q_{water}$, this yields}$
$$
begin{align}
Q_{water}=-Q_{iron}
end{align}
$$
The specific heat capacity Definition is :
$$
begin{align}
C=frac{Q}{mDelta t} \
Q=cmDelta t
end{align}
$$
Substituting (2) into (1) for the water and iron, yields:
$$
begin{align*}
c_{water}m_{water}(T_{equi}-T_{water})&=-c_{iron}m_{iron}(T_{equi}-T_{iron}) \
T_{equi}(c_{water}m_{water}+c_{iron}m_{iron})&=c_{iron}m_{iron}T_{iron}+c_{water}m_{water}T_{water} \
T_{equi}&= frac{c_{iron}m_{iron}T_{iron}+c_{water}m_{water}T_{water}}{c_{water}m_{water}+c_{iron}m_{iron}} \
T_{equi}&= frac{560*0.825*352+4186*0.04*20}{4186*0.04+560*0.825}=263.68^oC
end{align*}
$$
T_{equi}=263.8^oC
$$
When the water starts cooling down the same process happens between the water and the cup.
Since this system is isolated, all energy is conserved and we can assume that all heat released by the water/cup is received by Ice cube water, this yields:
$$
begin{align}
Q_{ice}+Q_{water}+Q_{Alu}=0
end{align}
$$
The specific heat capacity Definition is :
$$
begin{align}
C=frac{Q}{mDelta t} \
Q=cmDelta t
end{align}
$$
Substituting with (2) in (1) , yields:
$$
begin{align*}
c_{ice}m_{ice}(T_{equi}-T_{ice})+c_{water}m_{water}(T_{equi}-T_{water})+c_{Alu}m_{Alu}(T_{equi}-T_{Alu}) &=0 \
T_{equi}=frac{c_{ice}m_{ice}T_{ice}+c_{water}m_{water}T_{water}+c_{Alu}m_{Alu}T_{Alu}}{c_{ice}m_{ice}+c_{water}m_{water}+c_{Alu}m_{Alu}} \
T_{equi}=frac{2090*35*0+4186*110*23+900*62*23}{2090*35+4186*110+900*62}=boxed{20.14^otext{C}}
end{align*}
$$
(b) According to the specific heat capacity definition, the specific heat capacity $c$ and the temperature difference $Delta t$ are inversely proportional. The specific heat capacity of silver is lower than that of Aluminium therefore the temperature difference for the same heat exchange will be greater, meaning that silver will score even lower equilibrium temperature.
Note*: All values of specific heat used in this exercise can be found in $text{color{#4257b2}{Table 10.2}}$.
tt{Conversion between Degrees Celsius and Degrees Fahrenheit is as follows:}
$$
$$
T_F=frac{9}{5}T_C+32
$$
To determine which tempreature scale has the greater value above $-40^oC$, we can determine the general difference function by substracting $T_C$:
$$
T_F-T_C=frac{9}{5}T_C+32-T_C
$$
$$
T_F-T_C=frac{4}{5}T_C+32
$$
We can see clearly that the found function validates the exercise statement at $-40^oC$:
$$
T_F-T_C=0
$$
$$
frac{4}{5}T_C+32=0
$$
$$
T_C=-40^oC
$$
The derivative of the new found function is $frac{4}{5} Rightarrow$ this is an increasing linear function therefore for every value after $-40^oC$ the difference outcome will be positive:
$$
T_F-T_C>0
$$
$$
T_F>T_C
$$
T_F>T_C
$$
Lava lamps are a term used to describe masses of oil that are submerged in water within a container. This container is also equipped with a lamp at the bottom, Naturally, the oil is denser than the water and therefore sink to the bottom where it gets heated by the lamp, thus it expands and becomes less dense than the water allowing it to float where it gets cooler again and sink in a continuous cycle. The exchange of heat between the oil and it’s surrounding occurs due to its physical movement in the system: This is convection.
tt{The movement of oil allows it to exchange heat with its surrondings.}
$$
}
Sublimation: change of physical state from solid-state to gaseous state
Both processes are similar in the sense that both represent a change in phase and also end up in a gaseous state, they are also similar since when both processes happen, adding more heat does not change temperature but rather change the phase of the substance. This is of course the latent heat.
Both processes are different in the initial phase of the process (liquid vs solid) and the value of the latent heat.
tt{They are similar in the end phase and the need of latent heat, they are different in the intial phase and the value of latent heat}
$$
tt{Conversion between Degrees Celsius and Degrees Fahrenheit:}
$$
$$
T_F=frac{9}{5}T_C+32 text { (1)}
$$
Conversion between a Celsius Temperature and a Kelvin Temperature:
$$
T_c=T-273.15 text{ (2)}
$$
Substituting (2) into (1) :
$$
T_F=frac{9}{5}(T-273.15)+32
$$
$$
T_F=frac{9}{5}T-459.67
$$
The goal is to determine the temperature value that is the same on both the Kelvin and Fahrenheit scales
$$
T_F=TRightarrow T_F-T=0
$$
Therefore the answer is:
$$
T_F-T=frac{9}{5}T-459.67-T=0
$$
$$
frac{4}{5}T-459.67=0
$$
$$
T=459.67*frac{5}{4}=boxed{574.58K}
$$
T=574.58K
$$
$$
1C=1kcal
$$
The mechanical equivalent of 1kcal or 1C is 4186J, we can then deduce the amount of energy gained from the 525 Calories cheeseburger:
$$
E=525*4186=2.2*10^{6}J
$$
If we assume that the energy gained from the cheeseburger was totally converted to mechanical energy then:
$$
E=mgh
$$
$$
h=frac{E}{mg}=frac{2.2*10^{6}}{0.145*9.81}=1.54*10^{6}m
$$
(b) We can assume that at the time of release all the energy gained is transformed into kinetic energy, this yields:
$$
E=frac{1}{2}mv^2
$$
$$
v=sqrt{frac{2E}{m}}=sqrt{frac{2*2.2*10^{6}}{0.145}}=boxed{5508.6 frac{m}{s}}
$$
tt{(a) $h=1.54*10^6m$, (b) $v=5508.6frac{m}{s}$}
$$
tt{when a substance is about to have a phase change, in our case from ice to water, the temperature ceases to rise and the heat added contributes to the phase change, this state lasts until the transformation is totally complete. The amount of heat needed to complete this change can be calculated as:}
$$
$$
Q=L_fm
$$
where $m$ is the mass of the substance and the $L_f$ is the fusion latent heat specific to the substance needed to convert it from solid to liquid.
For a mass $m=0.55kg$ of water the heat needed to complete the fusion is :
$$
Q=33.5*10^4*0.55=18.45*10^4J
$$
Each time we will determine the fraction of heat provided compared to the heat needed, this fraction should also represent the mass of ice compared to the total mass.
(a) For : $Q_1=5*10^4$
$$
frac{Q_1}{Q}=frac{m_1}{m}
$$
$$
m_1=frac{Q_1m}{Q}=frac{5*10^4*0.55}{18.45*10^4}=0.149kg
$$
(b) For : $Q_2=1*10^5$
$$
m_2=frac{Q_2m}{Q}=frac{10^5*0.55}{18.45*10^4}=0.298kg
$$
(b) For : $Q_3=1.5*10^5$
$$
m_3=frac{Q_3m}{Q}=frac{1.5*10^5*0.55}{18.45*10^4}=0.447kg
$$
tt{(a) $m_1=0.149kg$, (b) $m_2=0.298kg$, (c) $m_3=0.447kg$}
$$
tt{To make steam, we first have to bring the water temperature form $50^oC$ to $100^oC$, therefore we will have to calculate the heat needed to raise the temperature using the definition of the specific heat equation:}
$$
$$
c=frac{Q}{mDelta t}
$$
$$
Q_1=c_{water}mDelta t=4186*0.22*(100-50)=0.46*10^5J
$$
When the water reaches $100^oC$ it begins the vaporization, in this transitional phase adding more heat is needed to make the change fully, this transformation though happens without raise in temperature, we can determine the amount of heat needed using the Latent heat of vaporization specific to water:
$$
Q_2=mL_v=0.22*22.6*10^5=4.972*10^5J
$$
We supplied the water with $Q=5.6*10^5J$, after raising the temperature of the water and changing it to steam, the remaining heat $Q_3$ will contribute to raising the temperature of the steam:
$$
Q_3=Q-Q_2-Q_1=(5.6-4.972-0.46)*10^5=0.168*10^5J
$$
Finally, we can determine the temperature reached by the steam as follows:
$$
Q_3=c_{steam}m(T_f-100)
$$
$$
T_f=frac{Q_3}{C_{steam}m}+100=frac{0.168*10^5}{2010*0.22}+100=boxed{142.1^oC}
$$
T_f=142.1^oC
$$
tt{First we need to determine the amount of heat needed to raise the temperature of water by $5^oC$, to do so we will use the specific heat equation }
$$
$$
c=frac{Q}{mDelta t}
$$
$$
Q=mcDelta t
$$
$$
Q=0.65*4186*5=1.36*10^4J
$$
Assuming that all work done by turning the crank is converted to heat, we can state that:
$$
Q=W
$$
On the other hand, per the definition power is the time derivative of work:
$$
P=frac{Delta W}{Delta t} Rightarrow Delta W =PDelta t
$$
this yields:
$$
Q=PDelta t
$$
$$
Rightarrow Delta t=frac{Q}{p}=frac{1.36*10^4}{0.18*745.7}=boxed{101.3s}
$$
Note*: $1hplongrightarrow 745.7Watt$
Delta t =101.3s
$$
$$
m=Lrho_{eau}=1000*2.1=2100g=2.1kg
$$
We can proceed by calculating the amount of heat received by the water by using the specific heat definition:
$$
c=frac{Q}{mDelta t}
$$
$$
Q=cmDelta t
$$
We know that the water started at $22^oC$ and the experiment ended when the water started boiling which means the water reached $100^oC$:
$$
Rightarrow Q= 4186*2.1*(100-22)=6.86*10^5J
$$
We can calculate the rate at which the thermal energy was transferred to the pot as:
$$
frac{Q}{Delta t }=frac{6.86*10^5}{8.5*60}=1345frac{J}{s}
$$
1345frac{J}{s}
$$
