MODULE 5
QUESTIONS
5.1 What is the difference between atoms and molecules? How do compounds and elements differ?
ANS:
Atoms are the smallest particle of an element that has the properties of that element while
molecules are a group of two or more atoms chemically bonded together. Compounds are a group of
two or more different elements chemically bonded together while elements are composed of only one
type of atom.
5.2 Based on the submicroscopic structure of solids, liquids, and gases explain why gases are easy to
compress but liquids and solids are very difficult to compress?
ANS:
Gas particles are so far from each other that allows this comparable compressibility. Solids
and liquids have closer distances between each of their particles that do not allow compression.
5.3 If a bottle is filled with liquid with no air space at the top and is tightly capped, it will break when
warmed up. If a few cubic centimetres of air are left above the liquid, the bottle will not break. Explain
both situations.
ANS:
Warming up the liquid will cause its particles to move faster and cause expansion. If there is
not enough space for this event, the bottle will give in. Enough space for the expansion will allow the
expansion to take place without breaking the container.
5.5 What is the difference between density and mass? Give an example of how the density of an
object can be used identify the substance of which it is composed.
ANS:
Density is the amount of mass an object has in a certain amount of volume while mass is the
amount matter an object has. This can be compared to the list of substances with known densities
5.6 What is the relationship between temperature and heat?
ANS:
The heat of an object is the total energy of all molecular motion inside the object since the
hotter an object is, the faster the motion of the molecules inside it. Temperature is the measure of the
average heat in an object. Therefore, the hotter an object gets, the higher its temperature.
5.7 Heat is a form of energy, but it is not stored as heat. Explain what happens to heat energy that
enters an object.
ANS:
Heat is energy in transfer, not a property of the system. Thus it is not contained within the
boundary of the system and cannot be stored. As heat energy enters an object, it is disseminated into
other forms of energy that can be stored or used.
5.8 Does the presence of large amounts of water in the body help keep body temperature stable?
Explain why or why not.
ANS:
Yes. Water has high specific heat capacity which enables it to absorb heat and to give it off
when needed. Body temperature fluctuations are stabilized by water.
5.9 What would be more effective in keeping a bed warm in an unheated house, a hot water bottle or
an equal mass of brick with the same temperature?
ANS:
A hot water bottle would be more effective. It has high specific heat capacity that can give off
more heat that a brick.
5.10 Temperatures in coastal areas are generally more moderate than temperatures inland. For
example, temperatures in San Francisco are neither as hot in the summer nor as cold in the winter as
in Sacramento, 90 miles inland. Explain why this is so.
ANS:
The bodies of water near the coastal areas keep the temperature moderate. They are able to
absorb heat in the surroundings and give it off when it is cold.
5.11 Why is an ice bag at 0°C a more effective coolant that an equal amount of water at 0°C?
ANS:
The ice can absorb latent heat whilst absorbing heat energy. This means that it takes up more
heat in order for it to melt and absorb some more to balance the temperature between it and its
surroundings.
5.12 In cooking, more heat is required to keep an uncovered pot of water boiling than one that is
covered. Explain why.
ANS:
The heat needed to raise the temperature of the water and boil in an uncovered pot escapes
as vapour and thus more heat must be transferred. The covered pot contains the heat produced and
prevents the vapour from escaping.
5.15 Exercising in very cold weather can freeze lung tissue, but exposed skin may not be affected.
Why does the temperature of the lungs drop more than that of the skin?
ANS:
The low body temperature still warms the skin and therefore the skin temperature may not
reflect it. On the other hand, the air that enters the lungs must be humidified first so that it can hold
more water or vapour. Cold air is harder to humidify thus can freeze the tissues.
5.16 A cold room with high humidity feels less chilly than a dry room with the same temperature.
Why?
ANS:
High humidity in cold weather increases the conduction of heat from the body. This takes off
more heat through moist on the skin and the surroundings.
5.17 The relative humidity of air in homes is usually very low in the winter. Why is this?
ANS:
The temperature and the dew point are much closer to each other so the relative humidity or
the capacity of the air to hold water is low.
5.19 How do the heats of fusion and vaporization of water tend to lessen climatic temperature
extremes?
ANS:
These are transitional properties of water prevents the energy from changing quickly from one
state to another, becoming an energy surplus. This energy surplus causes climactic extremes.
5.20 When camping out and sleeping on the ground, can you be warm with little or no insulation
beneath you? Explain.
ANS:
Little to no insulation would mean that more heat would be taken off from one’s body,
especially when it is cold. The humidity would be low since the temperature and dew points at night
usually are close to each other and therefore more water is held in the air. Moisture from the body
contains some body heat that are easily transferred to the surroundings, making one feel colder.
5.23 Why do desert dwellers wear loose-fitting white clothing that covers most of the body, rather than
little or no clothing? (Note that this clothing is an advantage during both hot days and cold desert
evenings.)
ANS:
This clothing protects them from the harmful radiation from the sun in hot days. In cold
evenings, the clothing prevents rapid evaporation of moisture from skin that causes transfer of body
heat onto the surroundings.
5.24 Will a swimming pool with dark-colored plaster be warmer than one with white plaster? Explain.
ANS:
Yes. Dark colors absorb more heat and thus can give off more to the pool water. White ones
reflect the heat and entails loss of heat, causing cooler water.
5.26 People who exercise vigorously outdoors on a cold day risk a reduced body temperature when
they stop. Why is this?
ANS:
Exercising causes sweating even on a cold day. When one stop from the activity, the
moisture on the skin evaporates rapidly and causes transfer of body heat onto the surroundings. This
reduces the body temperature.
5.29 Why is wet fur such a poor insulator?
ANS:
The water content of the wet fur allows heat transfer much faster than a dry one. Water can
be easily evaporated and heat is lost.
PROBLEMS
Section 5.1
5.1 (I) (a) Normal body temperature is 37°C. Convert this to Fahrenheit. (b) What is the temperature in
degrees Celsius of a person with a fever of 104°F?
SOL:
9
°F = °C + 32
5
9
°F = 37°C + 32
5
ANS:
T = 98.6°F
5.2 (I) (a) Room temperatures are kept at 68°F in the winter to conserve energy. What I this in
degrees Celsius? (b) On A summer day the air-conditioner is set to come one when the temperature
rises above 27°C. How hot is this in degrees Fahrenheit?
(a)
SOL:
5
°C = (°F − 32)
9
5
°C = (68°F − 32)
9
(b)
ANS:
T = 20°C
SOL:
°F = °C + 32
9
5
9
°F = 27°C + 32
5
ANS:
T = 80.6°F
5.3 (I) What Fahrenheit temperature corresponds to absolute zero?
SOL:
9
°F = (K − 273.15) + 32
5
9
°F = (0 − 273.15) + 32
5
ANS:
T = −459.67°F
5.4 (I) One of the properties that makes tungsten a good material for light bulb filaments is its high
melting point of 3410°C. What temperature is this on the Fahrenheit scale?
SOL:
9
°F = °C + 32
5
9
°F = 3410°C + 32
5
ANS:
T = 6170°F
5.5 (I) Frozen alcohol makes as good a candle as wax, with one disadvantage: Alcohol melts at 114°C. What Fahrenheit temperature is this?
SOL:
9
°F = °C + 32
5
9
°F = 114°C + 32
5
ANS:
T = 237.2°F
5.6 (I) The temperature of the surface of the sun is about 8000°F. What is this temperature on the
Celsius and Kelvin scales?
SOL:
5
°C = (°F − 32)
9
5
°C = (8000°F − 32)
9
ANS:
T = 4426.667°C
5
K = (°F − 32) + 273.15
9
5
K = (8000°F − 32) + 273.15
9
T = 4699.817K
5.7 (II) At what temperature do the Fahrenheit and Celsius scales have the same numerical value?
SOL:
9
°F = °C + 32
5
9
°F = (−40)°C + 32
5
−40°F = −40°C
ANS:
T = −40°
5.8 (I) A gap must be left between steel railroad rails to allow for thermal expansion. How large a gap
is needed if the maximum temperature reached is 50°C more than the temperature at which the rails
were laid? The length of a rail is 10m.
SOL:
ΔL = α L Δ T = 12 × 10−6 (10𝑚)(50°C)
ANS:
6 × 10−3 m
5.9 (I) The Golden Gate Bridge has an overall length of approximately 2800 m. If he bridge
experiences temperature extremes from -20° to +40°C, what will its change in length be? The bridge
is made primarily of steel.
SOL:
ΔL = α L Δ T = 12 × 10−6 (2800𝑚)(60°C)
ANS:
2.02 m
5.10 (I) The Great Pyramid of Cheops is 145 m high on a cold winter day when its temperature is 5°C.
How high is it in the summer when its temperature is 20°C? (Because of its size the pyramid warms
and cools down slowly and does not reach the temperature extremes of the area.) You may assume
that the coefficient of expansion is the same as that for concrete.
SOL:
ΔL = α L Δ T = 12 × 10−6 (145𝑚)(15°C)
ΔL = 0.261 m; H20°C = ΔL + H5°C = 0.261 + 145 m
ANS:
H20°C = 145.261 m
Sections 5.2 and 5.3
5.13 (I) How many calories of heat would be required to raise the temperature of a 45-kg person by
2.0°C? How many food calories is this?
1000𝑔
Q = mc∆ T = 45kg(
ANS:
Q = 74.7 kcal; 74.7 food Calories
1 𝑘𝑔
)(0.83
𝑐𝑎𝑙
SOL:
𝑔°𝐶
)(2.0°𝐶)
5.14 (I) When a scalpel is sterilized, its temperature may rise to 150°C. Calculate the amount of heat
in calories that must be removed from a 30-g steel scalpel to reduce its temperature from 150° t 20°C.
SOL:
Q = mc∆ T = 30g(0.11
ANS:
Q = 429 cal
𝑐𝑎𝑙
𝑔°𝐶
)(130°𝐶)
5.15 (I) How many calories of heat are needed to thaw out a 0.30-kg package of frozen vegetables
originally at 0°C, assuming the heat of fusion the same as that of water?
𝑐𝑎𝑙
SOL:
Q = H𝑓 m = 80
ANS:
Q = 24,000calories
𝑔
(300g)
5.17 (I) A burn produced by live steam at 100°C is more severe than one produced by the same
amount of water at 100°C. To verify this (a) calculate the heat that must be removed from 5.0 g of
water at 100°C to lower its temperature to 34°C (skin temperature); (b) calculate the heat that must be
removed from 5.0 g of steam at 100°C to condense it and then lower its temperature to 34°C, and
compare this with the answer to part (a).
(a)
SOL:
Q = mc∆ T = 5g(1
𝑐𝑎𝑙
𝑔°𝐶
)(66°𝐶)
(b)
ANS:
Q = 330 cal
SOL:
Condensation of water = Q = H𝑣 m = 540
𝑐𝑎𝑙
𝑔
(5g) = 2700 calories
Warming water from 34°C to 100°C = Q = mc∆ T = 5g (1
𝑐𝑎𝑙
𝑔°𝐶
) (66°𝐶) = 330 cal
Total for condensation and cooling = 3,030 calories
3030 𝑐𝑎𝑙
Compare:
ANS:
330 𝑐𝑎𝑙
= 9.1818
The amount of heat that must be removed from steam is more than 9 times as much
as that from water.
5.20 (II) A 0.50-kg block of material is heated from 20° to 35°C by the addition of 4210 cal of heat.
Calculate the specific heat of the block and identify the substance of which it is composed, assuming
that it is made of pure substance.
SOL:
Q = mc∆ T → c =
ANS:
Q = 0.5613
𝑘𝑐𝑎𝑙
𝑘𝑔•°𝐶
𝑄
𝑚∆ T
=
4.210𝑘𝑐𝑎𝑙
0.50𝑘𝑔(15°𝐶)
= 0.5613
𝑘𝑐𝑎𝑙
𝑘𝑔•°𝐶
→ 𝑂𝑎𝑘 𝑤𝑜𝑜𝑑
5.22 (II) Following strenuous exercise a person has a temperature of 40°C and is giving off heat at the
rate of 50 cal/sec. (a) What is the rate of heat loss in watts? (b) How long will it take for this person’s
temperature to return to 37°C if his mass is 90 kg?
(a)
(b)
𝑐𝑎𝑙
Watts = 50
ANS:
Rate of heat loss = 209.2 watts
SOL:
Q = mcℎ𝑢𝑚𝑎𝑛 ∆ T = 90 kg (3500
𝑠𝑒𝑐
(
4.184 𝑤𝑎𝑡𝑡𝑠
SOL:
1
Q = P𝑐𝑜𝑜𝑙𝑖𝑛𝑔 t → t =
ANS:
𝑐𝑎𝑙
𝑠𝑒𝑐
) = 209.2
𝑄
𝑃𝑐𝑜𝑜𝑙𝑖𝑛𝑔
=
𝐽
𝑘𝑔•°𝐶
) (40°𝐶 − 37°𝐶) = 945000 J
945000 J
209.2 𝑤𝑎𝑡𝑡𝑠
= 4517.2084 𝑠
t = 4517.2084 seconds
5.26 (III) Warming colds hands by rubbing them together is a time-honoured art. If a woman rubs her
hands back and forth for a total of 20 rubs. Calculate the temperature increase of her hands given the
following information: The force exerted in each of the 20 rubs is 40 N, the distance in each rub is
0.075 m, and the total mass of the hands is 1.5 kg.
SOL:
Let N be the number of hand rubs and F be the average frictional force of a hand rub
Q = NFd = mc∆ T → ∆ T =
ANS:
𝑁𝐹𝑑
𝑚𝑐
=
20(40𝑁)(0.075)
1.5𝑘𝑔(3500
𝐽
)
𝑘𝑔•°𝐶
= 0.0114°C
Q = 0.0114°C
5.27 (III) An ice cube having a mass of 50 g and an initial temperature of -20°C is placed in 400g of
30°C water. What is the final temperature of the mixture if the effects of the container can be
neglected?
SOL:
mice cice ∆ T + (∆ Hfusion • mols) + mmelted ice cmelted ice ∆ T = −(mwater cwater ∆ T)
[50g (2.01
J
kJ
50g
1000 J
I
) (25°C)] + [(6.01
)(
)(
)] + [50g (4.184
) (T𝑓 − 0)]
g • °C
mol 18.02
1 kJ
g • °C
I
= −[(400g) (4.184
) (T𝑓 − 30)]
g • °C
ANS:
T𝑓 = 16.4752°C
5.30 (I) What is the relative humidity on a day when the temperature is 25°C and the air contains
20.0g/m3 of water vapour?
vapor density in air
SOL:
Relative Humidity =
ANS:
Relative Humidity = 87%
𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝑒𝑑𝑛𝑠𝑖𝑡𝑦
𝑔
• 100 =
20 3
𝑚
𝑔
23 3
𝑚
• 100
5.34 (III) What is the dew point on a day when the relative humidity is 39% and the temperature is
20°C?
SOL:
𝑅𝐻 = 100% ×
𝐸
𝐸𝑠
𝐿
E0 = 0.611 kPa
𝐿
1
1
𝑅𝑣
𝑇0
𝑇𝑑
E = E0 × [( ) × ( −
𝑅𝑣
= 5423 K
)]
𝐿
1
1
𝑅𝑣
𝑇0
𝑇
E𝑠 = E0 × [( ) × ( − )]
𝑇0 = 273 K
1
1
− )]
273K 𝑇𝑑
1
1
0.611×[(5423K)×(
−
)]
273K 293K
0.611×[(5423K)×(
39% = 100% ×
ANS:
= 280.4633 K
T𝑑 = 7.4633°C
PROBLEMS
Section 5.4
5.36 (I) Calculate the rate of heat conduction in watts out of an animal with 3.0-cm thick fur. The
surface area of the animal is 1.5 m3 , its skin temperature is 35°C, and the temperature of the
surrounding air is 0°C. Heat losses due to convection and conduction can be neglected, and the
thermal conductivity of the fur can be assumed to be the same as that for air. Note that this is not
much different than for a human in a warm room, as seen in Example 5.8.
SOL:
ANS:
Q
t
=
k AΔ T
L
=
0.023 Joules (1.5m2 )(35°C)
sec•m•°C•0.03m
= 40.3
Joules
sec
= 40.3 watts
Heat conduction = 40.3 watts
5.37 (I) Calculate the rate of heat conduction through the walls of a house if they are 8.0 cm thick and
have twice the thermal conductivity as glass wool. The total area of the walls is 120 m2 and the inside
temperature is 20°C, while the outside temperature is 5°C. Note that this is only conduction through
the walls and does not include conduction through windows or ceiling.
SOL:
ANS:
𝑄
𝑡
=
𝑘𝐴(𝑇2 −𝑇1 )
𝑑
𝐽
=
[2(0.042𝑠•𝑚•°𝐶)](120𝑚2 )(20°𝐶−5°𝐶)
0.08𝑚
Rate of heat conduction = 1890 watts
5.38 (I) If a small iceberg with a mass of 20 million kg moves south from the Arctic, how much heat is
required to melt the iceberg? Note that this amount of energy was released earlier at the origin of the
iceberg.
J
SOL:
Q = H𝑓 n = 334 • 2.0 × 1010 𝑔
ANS:
Q = 668 × 1010 𝐽
𝑔
5.39 (I) What is the rate of heat loss by radiation from a man completely clothed in white (head & foot)
if his skin temperature is 34°C and the surrounding temperature is 10°C? His surface area is 1.4
m2 and the emissivity of the clothing is 0.2.
SOL:
Rate of heat loss is given by
Q
t
Q
t
ANS:
= σ eA (Tℎ𝑜𝑡 4 − T𝑐𝑜𝑙𝑑 4 ); the constant σ =
=
5,67×10−8 Joules
sec•m2 •k4
5,67×10−8 Joules
sec•m2 •k4
(0.2)(1.4m2 )(307k 4 − 283k 4 )
Rate of heat loss = 32.9 watts
5.40 (I) What is the rate of heat loss by radiation from a black roof of area 250 m2 if its temperature is
20°C and that of the surrounding is 10°C? The emissivity of the roof is 0.95.
SOL:
Rate of heat loss is given by
Q
t
Q
t
ANS:
= σ eA (Tℎ𝑜𝑡 4 − T𝑐𝑜𝑙𝑑 4 ); the constant σ =
=
5,67×10−8 Joules
sec•m2 •k4
5,67×10−8 Joules
sec•m2 •k4
(0.95)(250m2 )(293k 4 − 283k 4 )
Rate of heat loss = 12,900 watts
MAGDAMIT, MOILIN JOYCE I.
SUPPLEMENTAL PHYSICS
MARCH 2018
PROF. RICO ANTONIO MARIANO
MODULE 6
QUESTIONS
6.1 What is a fluid?
ANS:
Fluid is a substance that continually deforms or flows under an applied shear stress. It is a
subset of the phases of matter and include liquids, gases, plasmas, and to some extent, plastic solids.
6.2 Why are liquids and solids more difficult to compress than gases?
ANS:
The atoms and molecules in gases are much more spread out than in solids or liquids. They
vibrate and move freely at high speeds.
6.3 What factors other than the volume a gas occupies affect its pressure? Explain.
ANS:
Temperature and density of gas affects the pressure. Increasing temperature means that the
gas particles will move faster and collide with each other more frequent and this will in turn increase
pressure. The density of gas is the number of particles a gas has. Denser gases have more particles
and therefore have increased pressures.
6.4 Define gauge pressure, total pressure and pressure due to the weight of a fluid.
ANS:
Gauge pressure is zero-referenced against ambient air pressure, so it is equal to absolute
pressure minus atmospheric pressure. Total pressure is the pressure experienced by an object
submerged in any body of water due to the pressure caused by the water and atmosphere above the
submerged object. The pressure due to weight of a fluid is the pressure due to the pulling of the water
body by the gravity.
6.5 Why is it that you are able to get up from sunbathing if the weight of the atmosphere is resting on
your body?
ANS:
The weight of the atmosphere is pressing on one’s entire body from all directions. The weight
doesn’t just press one towards the Earth.
6.8 Give an example in which the density of an object could be used to identify the substance of
which it is composed. Could the density of an object composed of many substances also be sued to
identify which substances are in it? Explain.
ANS:
Mass and volume of the object may be used to determine density and this can be compared
to a list of substances with known densities. As for objects composed of many substances, density
may be used to characterize and estimate their composition.
6.9 Why can’t an absolute pressure in a gas be less than zero?
ANS:
This is because absolute pressure refers to a vacuum, the absence of any matter. There
would not be any more matter to take off from a system to have it in a negative pressure.
6.10 Explain in terms of equation 6.4 why you can notice a change in atmospheric pressure when
taking an elevator in a tall building. About how large is the total h creating atmospheric pressure at
sea level?
ANS:
The change becomes noticeable since it takes place in a time interval that is fast enough for
our bodies to detect while not being able to adapt to the pressure change.
6.13 Why is mercury rather than water used in barometers and in blood pressure measurements? If a
manometer is used to measure spinal column pressure, why is it preferable to use a saline solution
rather than mercury in the manometer?
ANS:
Mercury needs only a couple of column inches to exert pressure equal to that of the
atmosphere since it is denser than water. Also, mercury is more sensitive to atmospheric pressure
changes and can record even at temperatures below 0°C. Spinal column, unlike the atmosphere, only
exerts minute pressure so it prefers an indicator that it is easily moved.
6.14 How are units of pressure such as millimetres of mercury, centimetres of water, and inches of
mercury converted into units of force per unit area, such as newtons per square meter?
ANS:
This is done by computation using the density of the liquid used a measure of pressure, the
acceleration due to gravity, and the height of column in which the liquid is placed.
6.15 Why can some people float in a swimming pool if they take a deep breath and sink if they empty
their lungs? Why is it difficult to swim under water on the Great Salt Lake?
ANS:
Taking a deep breath and holding the air inside one’s lungs makes the body less dense than
the surrounding body of water, making it float. Emptying one’s lungs causes their body to return to its
normal state, being denser than water, and thus sink. The salinity of the Great Salt Lake is high that it
increases one’s buoyancy by being denser than the person, making it harder to sink or swim under
water.
6.16 Does the fact that more force is needed to pull the plug in a bathtub when it is full than when it is
empty contradict Archimedes’ principle? Explain.
ANS:
No. In fact, Archimedes’ principle doesn’t apply here since the plug is not submerged in water
but rather the water is on top of it. The pressure caused by the weight of the water makes it harder for
the plug to be pulled out.
6.18 What are the differences between laminar and turbulent flow? What can cause turbulence to
occur?
ANS:
Laminar flow is the flow of a fluid when each particle of the fluid follows smooth paths that
never interfere with one another. Turbulent flow is an irregular flow of fluid characterized by tiny
whirlpool regions.
6.19 Write down Poiseuille’s law for laminar flow. What are the factors upon which the flow rate
depends and to which factor is the flow rate most sensitive?
ANS:
Poiseuille’s law for laminar flow states that the change in pressure between two ends of a
pipe is directly proportional to the length of the pipe, the volumetric flow rate, and the dynamic
viscosity multiplied by 8 and inversely proportional to the pipe radius raised to the fourth power
multiplied by 𝜋. The volume flow rate is most sensitive to the radius of the pipe.
6.27 Why is it that a shower curtain is pulled into the spray of the shower?
ANS:
Colder dense air outside and hot less dense air inside causes higher air pressure on the
outside to force the shower curtain inwards to equalize the air pressure.
6.30 What is surface tension and what is its cause? Give an example of a surface tension effect
observed in everyday life.
ANS:
Surface tension is the property of a liquid that allows it to resist an external force. This is
caused by the cohesive forces between liquid molecules that holds them together. For instance,
surface tension can be observed in early morning dew drops as they are observed in their spherical
shape.
6.31 Birds such as ducks, geese and swans have larger densities than water, yet they are able to sit
on the surface of water. Explain this ability by considering the role of surface tension and the inability
of water to wet their feathers (Note that these birds cannot float in soapy water.)
ANS:
The inability of water to wet the feathers of some birds is due to the waxy coating these
feathers have. This provides the notion that the water will have more cohesion with its molecules than
adhesion to the waxy feathers. The water would rather stick with each other than let the bodies of the
birds get through.
6.32 Many insects do not drown when temporarily immersed in water because they are able to expel
it from their breathing passages. If these same insects are immersed in soapy water, they drown.
Explain why.
ANS:
Soapy water decreases surface tension by binding at the end of each water molecule and
being in between them. Surface tension is important to regulate the pressure inside and outside their
breathing passages. Decreased surface tension causes the collapse of these passages.
6.33 Why doesn’t soapy water, unlike plain water, bead up on a waxed surface?
ANS:
This is because the soap makes the water have less cohesion with each of its molecules and
more adhesion with the surface thus form sheets rather beads.
6.35 Why is it important that the surface tension in the alveoli of the lungs be neither too large nor too
small?
ANS:
This may cause difficulty in inhaling and exhaling air or, worse, may proceed to the collapse
of the small alveoli. The surface tension must be enough for a person not to need forceful breathing.
PROBLEMS
Section 6.1
6.1 (I) As a woman walks, her entire weight is momentarily supported on the heel of one shoe.
Calculate the pressure exerted on the floor by a high-heel shoe worn by a 50-kg woman if the heel
has an area of 2.0cm2. Express the pressure in newtons per square meter and atmospheres.
SOL:
𝑃=
𝐹
𝐴
=
𝑚
𝑠
1𝑚2
50𝑘𝑔(9.8 2 )
2.0𝑐𝑚2 (
2)
= 2.45 × 106
𝑁
𝑚2
10,000𝑐𝑚
𝑁
𝑚2
ANS:
→ 𝑎𝑡𝑚 = 2.45 × 106
𝑃 = 2.45 × 106
𝑁
𝑚2
𝑁
𝑚2
•
1 𝑎𝑡𝑚
𝑁
1.01×105 2
𝑚
= 24.3 𝑎𝑡𝑚
𝑜𝑟 24.3 𝑎𝑡𝑚
6.2 (I) What pressure is exerted by the tip of a nail stuck with a force of 20,000N? Assume the tip is a
1.5mm-radius circle.
𝐹
𝑃=
ANS:
𝑃 = 2.83 × 109
𝐴
=
20,000𝑁
SOL:
(1.5×10−3 𝑚2 )2 π
𝑁
𝑚2
6.3 (I) What pressure is created when 2.0 liters of air are forced into a bicycle tire originally at
atmospheric pressure and containing 0.45 liter of air? The inside volume of the tire remains 0.45L.
SOL:
ANS:
𝑃1
𝑉1
=
𝑃2
𝑃2 =
𝑉2
𝑃1 𝑉2
𝑉1
=
1𝑎𝑡𝑚•2.0𝐿
0.45𝐿
𝑃2 = 4.4444 𝑎𝑡𝑚
6.4 (II) A pressure cooker is shaped like a can with a lid 25cm in diameter. If the pressure in the
cooker can reach 3.0 atm, how much force must the latches holding the lid onto the pot be designed
to withstand?
𝑁
𝑚2 )
1 𝑎𝑡𝑚
(12.5×10−2 𝑚)2 π
1.01×105
𝐹
SOL:
𝑃=
ANS:
𝐹 = 4.12 × 106
𝐴
→𝐹=
𝑃
𝐴
=
3.0𝑎𝑡𝑚−1.0𝑎𝑡𝑚(
𝑁
𝑚2
6.5. (III) Assuming car tires are perfectly flexible and support the weight of a car by air pressure alone,
calculate the total area of the tires in contact with the ground for a 1000-kg car having a tire pressure
of 2.5 x 105 N/m2
𝐹
𝑃=
ANS:
𝐴 = 0.392𝑚2
𝐴
=
𝑚𝑎
SOL:
𝐴
→𝐴=
𝑚𝑎
𝑃
𝑚
=
1000𝑘𝑔•9.8 2
𝑠
𝑁
2.5×105 2
𝑚
Section 6.2 & 6.3
6.6 (I) Calculate the force that must be exerted on the master cylinder of a hydraulic system to lift a
10,000 kg truck with the slave cylinder. The master cylinder has a diameter of 1.50cm and the slave
has a diameter of 30.0cm.
SOL:
ANS:
𝐹1
𝐴1
=
𝐹2
𝐴2
𝐹2 = 245 𝑁
𝐹2 =
1.5𝑐𝑚 2
) 𝜋]
2
30.0𝑐𝑚 2
(
) 𝜋
2
10,000𝑘𝑔[(
𝑚
= 25𝑘𝑔(9.8 2 )
𝑠
6.7 (I) The deepest place in any ocean on earth is in the Marianas Trench near the Philippines.
Calculate the pressure at the bottom of this trench due to the ocean water, given that is depth is 11.0
km and assuming that the density of ocean water is 1.025 g/cm3.
SOL:
𝑃 = ℎ𝜌𝑔 = 11000𝑚 • 1025
ANS:
𝑃 = 110495000
𝑘𝑔
𝑚3
• 9.8
𝑚
𝑠2
𝑁
𝑚2
6.8 (a) Calculate the distance to the top of the atmosphere, assuming that the density of air is
constant from sea level up. (b) The atmosphere ends at an altitude of about 130 km. Calculate its
average density and compare this with its density at sea level.
(a)
(b)
SOL:
𝑃 = ℎ𝜌𝑔
ANS:
ℎ = 8440.2332 𝑚
SOL:
𝑃 = ℎ𝜌𝑔
𝜌𝑠𝑒𝑎 𝑙𝑒𝑣𝑒𝑙
𝜌130𝑘𝑚
ANS:
ℎ=
𝜌=
𝑃
𝜌𝑔
𝑃
ℎ𝑔
𝑁
=
101,325 2
𝑚
𝑘𝑔
𝑚
1.225 3 •9.8 2
𝑚
𝑠
𝑁
=
101,325 2
𝑚
𝑚
130000𝑚•9.8 2
𝑠
= 0.0795
𝑘𝑔
𝑚3
𝑘𝑔
=
1.225 3
𝑚
𝑘𝑔
0.0795 3
𝑚
= 15.4089
The air at sea level is more than 15 times denser than the average density of air
throughout the atmosphere.
6.9 (II) (a) Calculate the pressure in newtons per square meter due to whole blood in an IV system,
such as the one shown in Figure 6.9, if h= 1.5m. (b) Noting that there is an open tube, so that
atmospheric pressure is exerted on the blood in the bottle, calculate the total pressure exerted at the
needle by the blood.
(a)
(b)
𝑘𝑔
SOL:
𝑃 = ℎ𝜌𝑔 = 1.5𝑚 • 1060
ANS:
𝑃 = 15,582
SOL:
𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑃𝑎𝑡𝑚 + 𝑃 = 101,325
ANS:
𝑃𝑡𝑜𝑡𝑎𝑙 = 116,907
𝑚3
• 9.8
𝑚
𝑠2
𝑁
𝑚2
𝑁
𝑚2
+ 15,582
𝑁
𝑚2
𝑁
𝑚2
6.10 Water towers are used to store water above the level of homes. If a user observes that the static
water pressure at home is 3.0 x 105 N/m2, how high above the home is the surface of the water in the
tower?
SOL:
𝑃 = ℎ𝜌𝑔
ANS:
ℎ = 30.6122 𝑚
ℎ=
𝑁
𝑃
=
𝜌𝑔
3.0×105 2
𝑚
𝑘𝑔
𝑚
1,000 3 •9.8 2
𝑚
𝑠
Section 6.4
6.18 A certain ideal hygrometer has a uniform density of 0.90 g/cm 3. What fraction of the hydrometer
will be submerged if it is placed in a fluid of density 0.95 g/cm 3?
𝑃𝑜𝑏𝑗𝑒𝑐𝑡
SOL:
% 𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑 =
ANS:
% 𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑 = 94.7%
𝑃𝑓𝑙𝑢𝑖𝑑
• 100 =
𝑔
𝑐𝑚3
𝑔
0.95 3
𝑐𝑚
0.90
• 100
6.19 What is the density of a piece of wood that floats in water with 70% of its volume submerged?
SOL:
% 𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑 =
ANS:
𝑃𝑜𝑏𝑗𝑒𝑐𝑡 = 70
𝑃𝑜𝑏𝑗𝑒𝑐𝑡
𝑃𝑓𝑙𝑢𝑖𝑑
• 100
𝑃𝑜𝑏𝑗𝑒𝑐𝑡 =
% 𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑•𝑃𝑓𝑙𝑢𝑖𝑑
100
=
70•1
𝑔
𝑐𝑚3
100
𝑔
𝑐𝑚3
6.20 Calculate the density of a fluid in which a hydrometer of uniform density 0.80 g/cm 3 floats with
90% of its volume submerged.
SOL:
% 𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑 =
ANS:
𝑃𝑓𝑙𝑢𝑖𝑑 = 8.8889
𝑃𝑜𝑏𝑗𝑒𝑐𝑡
𝑃𝑓𝑙𝑢𝑖𝑑
• 100
𝑃𝑓𝑙𝑢𝑖𝑑 =
𝑃𝑜𝑏𝑗𝑒𝑐𝑡
% 𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑
• 100 =
0.80
𝑔
𝑐𝑚3
90%
• 100
𝑔
𝑐𝑚3
6.21 Calculate the density of a person who floats in fresh water with 5.0% of her volume above the
surface.
SOL:
% 𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑 =
ANS:
𝑃𝑜𝑏𝑗𝑒𝑐𝑡 = 95
𝑃𝑜𝑏𝑗𝑒𝑐𝑡
𝑃𝑓𝑙𝑢𝑖𝑑
• 100
𝑃𝑜𝑏𝑗𝑒𝑐𝑡 =
% 𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑•𝑃𝑓𝑙𝑢𝑖𝑑
100
=
95•1
𝑔
𝑐𝑚3
100
𝑔
𝑐𝑚3
Section 6.5
6.29 If the flow rate in an IV setup is two cubic centimeters/min, how long does it takes to empty a 1.0
L bottle of IV solution?
𝑉𝑜𝑙𝑢𝑚𝑒
SOL:
𝐹𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 =
ANS:
𝑡 = 500 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
𝑡𝑖𝑚𝑒
𝑡𝑖𝑚𝑒 =
𝑉𝑜𝑙𝑢𝑚𝑒
𝐹𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
=
1.0 𝐿
𝑐𝑚3
2.0
𝑚𝑖𝑛
•
1000𝑐𝑚3
1𝐿
6.34 If the radius of a blood vessel is reduced by cholesterol deposits to 90% of its original value, flow
will decrease. By what factor would the pressure difference along the vessel have to be increased to
get the flow rate up to its original value?
SOL:
Flow Rate =
∆Pπr4
8ηL
∆P =
Flow rate•8ηL
πr4
New radii are 0.90 of their original size. The ∆P is inversely proportional to r 4 .
∆P
1∆P
∆P′ = 0.94 = 0.94 = 1.52∆P
ANS:
The pressure difference have to be increased by a factor of 1.52.
6.35 By what factor the radii of the arteries have to decrease to reduce blood flow rate by a factor of
2?
SOL:
Flow Rate =
∆Pπr4
8ηL
new flow rate
original flow rate
= 0.50
𝑟 ′ = 𝑟 • 0.50.25 = 0.841𝑟
ANS:
The radii have to decrease by a factor of 0.841.
Flow rate•8ηL 0.25
)
∆Pπ
r=(
Section 6.6
6.38 The aorta is 1.0cm in radius and the average speed of blood in the aorta is 30cm/sec in a normal
resting adult. Calculate the blood flow rate in the aorta in cubic centimeters per second and liters per
minute.
SOL:
𝐹𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 =
𝑐𝑚3
𝑠
ANS:
→
𝐿
𝑚𝑖𝑛
𝑉𝑜𝑙𝑢𝑚𝑒
𝑡𝑖𝑚𝑒
= 94.2
= 𝐴𝑟𝑒𝑎 •
𝑐𝑚3
𝑠
𝑡𝑖𝑚𝑒
1𝐿
1000𝑐𝑚3
•
𝐵𝑙𝑜𝑜𝑑 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 = 94.2
𝐿𝑒𝑛𝑔𝑡ℎ
𝑐𝑚3
𝑠
•
̅ = (1.0𝑐𝑚)2 π • 30
= 𝐴𝑟𝑒𝑎 • 𝑣
60 𝑠
1 𝑚𝑖𝑛
= 5.65
𝑐𝑚
𝑠
= 94.2
𝑐𝑚3
𝑠
𝐿
𝑚𝑖𝑛
𝐿
𝑜𝑟 5.65 𝑚𝑖𝑛
6.39 If turbulent flow begins in the aorta when the average blood speed reaches 80cm/sec, what is
the maximum flow rate that can be achieved without turbulence in an aorta 0.90cm in radius? Give
the flow rate in cubic centimeters per second and liters per minute.
SOL:
𝐹𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 =
𝑐𝑚3
𝑠
ANS:
𝑉𝑜𝑙𝑢𝑚𝑒
𝑡𝑖𝑚𝑒
= 𝐴𝑟𝑒𝑎 •
𝐿
→ 𝑚𝑖𝑛 = 203.5752
𝑐𝑚3
𝑠
𝐿𝑒𝑛𝑔𝑡ℎ
𝑡𝑖𝑚𝑒
̅ = (0.90𝑐𝑚)2 π • 80
= 𝐴𝑟𝑒𝑎 • 𝑣
1𝐿
60 𝑠
𝑐𝑚
𝑠
= 203.5752
𝑐𝑚3
𝑠
𝐿
• 1000𝑐𝑚3 • 1 𝑚𝑖𝑛 = 12.2145 𝑚𝑖𝑛
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 = 203.5752
𝑐𝑚3
𝑠
𝐿
𝑜𝑟 12.2145 𝑚𝑖𝑛
6.40 What is the average speed of water in a water main 0.25 m in diameter that is carrying
1000liters/min?
𝑉𝑜𝑙𝑢𝑚𝑒
SOL:
𝐹𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 =
ANS:
𝑣̅ = 33.9531
𝑡𝑖𝑚𝑒
= 𝐴𝑟𝑒𝑎 •
𝐿𝑒𝑛𝑔𝑡ℎ
𝑡𝑖𝑚𝑒
̅→𝑣
̅=
= 𝐴𝑟𝑒𝑎 • 𝑣
𝐹𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
𝐴𝑟𝑒𝑎
𝐿
=
1000𝑚𝑖𝑛•
1000𝑐𝑚3 1𝑚𝑖𝑛
• 60𝑠
1𝐿
2
(12.5𝑐𝑚) 𝜋
𝑐𝑚
𝑠
6.41 Calculate the flow rate of blood through a capillary in cubic centimeters per second, given that
the capillary has a diameter of 4.0 x 10 -4 cm and the average speed of the blood in the capillary is
3.0x10-2 cm/sec.
𝑉𝑜𝑙𝑢𝑚𝑒
SOL:
𝐹𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 =
ANS:
𝐹𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 = 3.78 × 10−9
𝑡𝑖𝑚𝑒
= 𝐴𝑟𝑒𝑎 •
𝐿𝑒𝑛𝑔𝑡ℎ
𝑡𝑖𝑚𝑒
2
̅ = (2.0 × 10−4 𝑐𝑚) π • 3.0 × 10−2
= 𝐴𝑟𝑒𝑎 • 𝑣
𝑐𝑚
𝑠
𝑐𝑚3
𝑠
6.42 Fire hoses have nozzles that increase the speed of the water to reach higher in a fire. (a) if the
flow rate in the fire hose is 5000L/min and the opening of the fire hose nozzle is 2.0cm in diameter,
what is the speed of the water as it leaves the nozzle? (b) if the diameter of the hose itself is 10cm,
what is the speed of the water in the hose?
(a)
SOL:
𝐹𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 =
𝑣̅ =
(b)
𝐹𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
𝐴𝑟𝑒𝑎
𝑉𝑜𝑙𝑢𝑚𝑒
=
𝑡𝑖𝑚𝑒
5000𝐿/𝑚𝑖𝑛
(1.0𝑐𝑚)2 π
ANS:
𝑣̅ = 26525.8238
SOL:
𝑣̅ =
ANS:
𝑣̅ = 265.2582
𝐹𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
𝐴𝑟𝑒𝑎
=
= 𝐴𝑟𝑒𝑎 •
•
𝐿𝑒𝑛𝑔𝑡ℎ
𝑡𝑖𝑚𝑒
1000𝑐𝑚3
1𝐿
•
̅
= 𝐴𝑟𝑒𝑎 • 𝑣
1𝑚𝑖𝑛
60𝑠
𝑐𝑚3
𝑠
5000𝐿/𝑚𝑖𝑛
(10𝑐𝑚)2 π
𝑐𝑚3
𝑠
•
1000𝑐𝑚3
1𝐿
•
1𝑚𝑖𝑛
60𝑠
MAGDAMIT, MOILIN JOYCE I.
SUPPLEMENTAL PHYSICS
MARCH 2018
PROF. RICO ANTONIO MARIANO
MODULE 7
7.1 When a blockage of flow occurs in the body, such as in the eye, brain, or gastrointestinal system,
why does the pressure upstream build to higher than normal levels?
ANS:
It is because the pressure downstream is reduced. This increases the velocity of which the
blood flows through these systems.
7.3 New parents spend a great deal of time burping their babies. How does belching help relieve the
babies’ cramps and allow them to eat more at one feeding?
ANS:
The air that a baby may have swallowed upon feeding occupies space in his tummy and
therefore lesser amount of food can be accommodated. Also, belching helps to ease the pressure this
gas exerts to the baby.
7.7 Define diffusion, osmosis, dialysis, filtration, and active transport.
ANS:
Diffusion is the movement of substances due to random thermal molecular motion. Osmosis
is the transport of water through a semipermeable membrane from a region of high concentration to a
region of low concentration. Dialysis is the transport of any molecule though a semipermeable
membrane due to its concentration difference. Filtration occurs when back pressure is sufficient to
reverse the normal direction of substances through membranes. Active transport is a process in which
a living membrane expends energy to move substances across it.
7.8 There are no blood vessels in the cornea. Suggest a method for oxygen to reach cornea from the
atmosphere through the tear layer. Is there a potential problem for people who wear nonporous
contact lenses? If so, explain why it is possible to wear contact during the day but not while sleeping.
ANS:
The oxygen dissolves in the tears and diffuses throughout the cornea. There is potential
problem for people wearing nonporous contact lenses. During the day, our eyelids are open and
enough oxygen can be dissolved in the tear layer, but when we sleep, our eyelids are closed and it is
much harder for the cornea to receive oxygen, even more if nonporous contact lenses are used.
7.9 Water molecules are small compared to most other molecules. Is this consistent with the fact that
most membranes are permeable to water?
ANS:
Yes. The smaller molecules can easily pass or relatively easy to pass through membranes
unlike the larger ones since the pores dictates how large of a molecule it allows to.
7.11 The salinity of ocean water is greater than that of blood, and the sanity of fresh water is less than
that of blood. If the red blood cells in the lungs of a drowning victim have burst because of osmotic
transfer of water into them, was the victim drowned in the ocean or in fresh water?
ANS:
The victim drowned in fresh water. Since the blood have burst, this may be only due to the
fact that the blood has higher concentration of solutes than that of the surrounding fluid.
7.14 Why is it inadvisable to drink ocean water? What is the net effect on the water content of the
body if a large amount of salt water is consumed?
ANS:
Ocean water contains large amounts of salt that when upon intake, water from the body will
try to dilute it. The overall water content of the body will decrease due to this.
7.16 Why don’t people breathe by absorbing oxygen through the skin and diffusing carbon dioxide out
through the skin? (Actually a small percentage of normal oxygen intake is accomplished by this
method.)
ANS:
The oxygen cannot penetrate through the skin although a small percentage of respiration is
accomplished through this method. The skin has cellular barriers and basement membranes that
prevent this from happening.
7.17 Explain how the periodic variations in blood pressure due to heartbeats are smoothed out by the
time blood reaches the venous system (the veins).
ANS:
Blood vessels are equipped with layers of muscle that can control the blood pressure
fluctuations. When the veins detects a drop of blood pressure, the muscles help it constrict to
decrease vessel radius and increase pressure. Increase in blood pressure causes the muscles to
relax and the vessels dilate to increase radius and decrease the pressure.
7.18 If a person’s diastolic pressure rises during exercise, it is considered a sign that the arterioles are
not dilating. How would this cause the diastolic pressure to rise during exercise?
ANS:
The variations in blood pressures can be regulated by the blood vessels through dilating or
constricting. If the arterioles do not dilate and the blood pressure shoots up, as in this case, when
exercising, this increased blood pressure is not regulated back to the normal level, making the
diastolic pressure rise.
7.19 What causes blood velocity to increase when blood goes from the capillaries to the veins?
ANS:
The pressure from exiting a smaller radius to a larger one causes the increase of blood flow
since the restriction in the flow of the fluid is lifted.
7.20 The veins on the back of a person’s hands bulge noticeably when the hand is lower than the
heart but not when the hand is held over the head. Explain why.
ANS:
There is much less pressure when the blood is going back to the heart than when the blood
leaves the heart since the heart puts pressure on the latter but pumping through the arteries. Bulging
indicates that the decrease in pressure causes a slower blood flow.
7.21 It seems natural that newborn infants should have lower blood pressure than full-grown adults.
Why don’t infants need blood pressure as high as an adult’s?
ANS:
There is much smaller body area that needs the blood compare with that of adults. To
achieve a normal blood flow rate, the pressure must take into consideration the area where it is
applied.
7.22 Some people suffer from a disease in which they have an abnormally high blood cell count,
meaning that the number of blood cells per unit volume of liquid is very high. What is the likely effect
of this on the viscosity of the blood and its flow rate?
ANS:
The viscosity of blood gets higher ad thus slows its flow rate. More solutes in the blood makes
it heavier and more interaction between the vessel walls and blood components happen.
7.23 What is the effect of turbulence on flow rate? What are some circumstances in which turbulence
is easily detected by the sound it makes?
ANS:
Turbulence makes flow rate slower since it disrupts the organize flow. For instance, the sound
a river makes when its waters are disrupted by the rocks in its path makes it easy to detect that the
water is experiencing turbulence.
7.25 If a fluid is flowing through a tube at a given flow rate, will the tube heat up more when the flow is
turbulent or when it is laminar? Explain.
ANS:
When there is laminar flow, the rate of which the fluid is moving is not disrupted and so it can
move fast. Moving faster creates friction and this causes transfer energy to the form of heat.
7.27 If you wished to pump air into a patient’s lungs by squeezing on a balloon (connected to their
ventilator); would you choose a balloon of large or small diameter to get the maximum pressure? To
get the maximum volume?
ANS:
To get maximum pressure, a balloon with a smaller diameter would be needed since the air
will have smaller space to move. To get maximum volume, it is pertinent that a balloon with larger
diameter is used.
7.28 Large balloons, such as weather balloons, do not need large pressures to be inflated. Why is
this?
ANS:
There is so much space for the air to move around so the air molecules collide with the walls
of the balloon and with each other less frequent.
7.29 Major veins passing through leg muscles are not as large in radius as some other major vessels
in the body. Does this hinder or aid the action of the “muscle pump” in returning blood to the heart?
(Figure 7.16)
ANS:
This aids the action of returning blood to the heart in the sense that blood exiting from the
capillaries whose radii are small would get to increase the blood flow rate.
7.32 During the inhalation do the lungs pull air or does the atmosphere push air in?
ANS:
The lungs pull air in during inhalation. When the thoracic walls expand when inhaling, the
lungs also expand and there becomes a space that the air can fill.