1–50E

1–50E A pressure gage connected to a tank reads 50 psi at
a location where the barometric reading is 29.1 mm Hg.
Determine the absolute pressure in the tank. Take rHg
848.4 lbm/ft
3
. Answer: 64.3 psia

1–49

1–49 A vacuum gage connected to a tank reads 15 kPa at a
location where the barometric reading is 750 mm Hg. Determine the absolute pressure in the tank. Take rHg 13,590
kg/m
3
. Answer: 85.0 kPa
ANS

1–48

1–48 Consider a 70-kg woman who has a total foot imprint
area of 400 cm2. She wishes to walk on the snow, but the
snow cannot withstand pressures greater than 0.5 kPa. Determine the minimum size of the snowshoes needed (imprint
area per shoe) to enable her to walk on the snow without
sinking.
ANS

1–47E

1–47E A 200-pound man has a total foot imprint area of 72
in2. Determine the pressure this man exerts on the ground if
(a) he stands on both feet and (b) he stands on one foot.
ANS

1–46E

1–46E Show that 1 kgf/cm2 14.223 psi.
ANS

1–45

1–45 The absolute pressure in water at a depth of 5 m is
read to be 145 kPa. Determine (a) the local atmospheric pressure, and (b) the absolute pressure at a depth of 5 m in a liquid whose specific gravity is 0.85 at the same location.
ANS
1-45The absolute pressure inwater ata specified depth isgiven. The localatmospheric pressure and the
absolutepressure atthe samedepth ina differentliquid are tobe determined.
Assumptions  The liquid and water are incompressible.
Properties The specific gravityof the fluid isgiven tobe SG = 0.85. Wetake the densityof water tobe
1000 kg/m
3
. Thendensity ofthe liquidis obtainedbymultiplying its specific gravity by the density of
water,

1–44

1–44 The gage pressure in a liquid at a depth of 3 m is read
to be 28 kPa. Determine the gage pressure in the same liquid
at a depth of 9 m.
ANS
1-44The gage pressure ina liquid ata certaindepth isgiven. The gage pressure in the same liquid at a
differentdepth istobe determined.
Assumptions  The variation of the densityof the liquid withdepth isnegligible.
Analysis The gage pressure attwo differentdepths of a liquid can be expressed as

1–43 Determine the atmospheric pressure at a location
where the barometric reading is 750 mm Hg. Take the density
of mercury to be 13,600 kg/m3.
ANS
1-43The barometricreading ata location isgiven inheightof mercury column. The atmospheric pressure
is tobedetermined.

1–42

1–42 The water in a tank is pressurized by air, and the
pressure is measured by a multifluid manometer as shown in
Fig. P1–42. Determine the gage pressure of air in the tank if
h1 0.2 m,h2 0.3 m, and h3
0.46 m. Take the densities
of water, oil, and mercury to be 1000 kg/m
3, 850 kg/m3, and13,600 kg/m3, respectively.

ANS

1-42 The pressure ina pressurized water tank ismeasured bya multi-fluid manometer. The gage pressure 

ofair inthe tankis tobedetermined. 

Assumptions The air pressure inthe tankis uniform(i.e., its variationwithelevationis negligible due to its 

low density), and thus we can determine the pressure at the air-water interface. 

Properties  The densities of mercury, water, and oil are given to be 13,600, 1000, and 850 kg/m

3

, 

respectively. 

AnalysisStarting with the pressure at point 1 at the air-water interface, and moving along the tube by 

adding (as we go down) or subtracting (as we go up) the  gh ρ terms until we reach point 2, and setting the 

result equal to Patm

since the tube isopen tothe atmosphere gives 

1–41E

1–41E A manometer is used to measure the air pressure in
a tank. The fluid used has a specific gravity of 1.25, and the
differential height between the two arms of the manometer is
28 in. If the local atmospheric pressure is 12.7 psia, determine the absolute pressure in the tank for the cases of the
manometer arm with the (a) higher and (b) lower fluid level
being attached to the tank.
ANS
1-41E The pressure ina tank ismeasured witha manometer bymeasuring the differentialheightof the
manometer fluid. The absolutepressure inthe tank isto be determined for the cases of the manometer arm
withthe higher and lower fluid levelbeing attached tothe tank .

1–40

1–40 A vacuum gage connected to a chamber reads 35 kPa
at a location where the atmospheric pressure is 92 kPa.
Determine the absolute pressure in the chamber.
ANS

1–39C

1–39C Consider two identical fans, one at sea level and the
other on top of a high mountain, running at identical speeds.
How would you compare (a) the volume flow rates and
(b) the mass flow rates of these two fans?
ANS
1-39CThe densityof air atsea levelishigher than the densityof air on top of a high mountain. Therefore,
the volume flow rates of the two fans running at identical speeds will be the same, but the mass flow rate of
the fanat sea level will behigher.

1–38C

1–38C Express Pascal’s law, and give a real-world example
of it.
ANS
1-38C  Pascal’s principlestates that the pressure appliedtoaconfinedfluidincreases the pressure
throughout by the same amount. Thisisa consequence of the pressure ina fluid remaining constantinthe
horizontaldirection. An example of Pascal’s principle isthe operation of the hydrauliccar jack.

1–37C

1–37C A tiny steel cube is suspended in water by a string.
If the lengths of the sides of the cube are very small, how
would you compare the magnitudes of the pressures on the
top, bottom, and side surfaces of the cube?
ANS
1-37CIf the lengths of the sides of the tinycube suspended inwater by a string are very small, the
magnitudes ofthe pressures onall sides ofthe cubewill bethe same.

1–37C

1–37C A tiny steel cube is suspended in water by a string.
If the lengths of the sides of the cube are very small, how
would you compare the magnitudes of the pressures on the
top, bottom, and side surfaces of the cube?
ANS
1-37CIf the lengths of the sides of the tinycube suspended inwater by a string are very small, the
magnitudes ofthe pressures onall sides ofthe cubewill bethe same.

1–36C

1–36C Someone claims that the absolute pressure in a liquid of constant density doubles when the depth is doubled.
Do you agree? Explain.
ANS
1-36CNo, the absolutepressure ina liquid of constantdensitydoes notdouble when the depth isdoubled.
It is the gage pressure thatdoubles when the depth isdoubled.

1–35C

1–35C Explain why some people experience nose bleeding
and some others experience shortness of breath at high elevations.
ANS
1-35CThe atmospheric pressure, which is the externalpressure exerted on the skin, decreases with
increasing elevation. Therefore, the pressure is lower at higher elevations. As a result, the difference
between the blood pressure inthe veins and the air pressure outside increases. Thispressure imbalance may
cause some thin-walled veins such as the ones inthe nose toburst, causing bleeding. The shortness of
breath iscaused bythe lower air densityathigher elevations, and thus lower amountof oxygen per unit
volume.

1–34C

Pressure, Manometer, and Barometer
1–34C What is the difference between gage pressure and
absolute pressure?
ANS
Pressure, Manometer, and Barometer
1-34CThe pressure relative tothe atmospheric pressure is calledthe gage pressure, and the pressure
relative toanabsolute vacuumis called absolute pressure.

1–33

1–33 Consider two closed systems A and B. System A contains 3000 kJ of thermal energy at 20°C, whereas system B
contains 200 kJ of thermal energy at 50°C. Now the systems
are brought into contact with each other. Determine the direction of any heat transfer between the two systems.
ANS
1-33Two systems having differenttemperatures and energy contents are brought in contact. The direction
ofheat transfer is tobedetermined.
Analysis Heat transfer occurs fromwarmer tocooler objects.  Therefore, heat will betransferredfrom
systemB to systemA until both systems reach the sametemperature.

1–32E

1–32E The temperature of a system drops by 45°F during a
cooling process. Express this drop in temperature in K, R,
and °C.
ANS
1-32EA temperature change isgiven in °F. Itistobe expressed in °C, K, andR.
Analysis Thisproblemdeals withtemperature changes, which are identicalin Rankine and Fahrenheit
scales. Thus,
∆T(R) = ∆T(°F) = 45R
The temperature changes in Celsius and Kelvin scales are also identical, and are related to the changes in
Fahrenheitand Rankine scales by
∆T(K) = ∆T(R)/1.8 = 45/1.8 = 25 K
and  ∆T(°C) = ∆T(K) = 25°C

1–31

1–31 The temperature of a system rises by 15°C during a
heating process. Express this rise in temperature in kelvins
ANS
1-31A temperature change isgiven in °C.Itistobe expressed inK.
Analysis This problemdeals with temperature changes, which are identical inKelvinandCelsius scales.
Thus,  ∆T(K]= ∆T(°C) = 15 K

1–30E

1–30E Consider a system whose temperature is 18°C.
Express this temperature in R, K, and °F.
ANS
1-30EA temperature isgiven in °C.Itistobe expressed in °F, K, and R.
Analysis Using the conversion relations between the various temperature scales,
T(K]= T(°C)+ 273 = 18°C+ 273 = 291 K
T(°F] = 1.8T(°C)+ 32 = (1.8)(18) + 32 = 64.4°F
T(R]= T(°F) + 460 = 64.4 + 460 = 524.4 R

1–29

1–29 The deep body temperature of a healthy person is
37°C. What is it in kelvins?
ANS
1-29A temperature isgiven in °C.Itistobe expressed inK.
Analysis The Kelvinscale is relatedtoCelsius scale by
T(K]= T(°C)+ 273
Thus,  T(K]= 37°C+ 273 = 310 K

1–28C

1–28C Consider an alcohol and a mercury thermometer
that read exactly 0°C at the ice point and 100°C at the steam
point. The distance between the two points is divided into
100 equal parts in both thermometers. Do you think these
thermometers will give exactly the same reading at a temperature of, say, 60°C? Explain.
ANS
1-28C  Probably, butnotnecessarily. The operation of these two thermometers isbased on the thermal
expansion of a fluid. If the thermalexpansion coefficientsof both fluids varylinearlywithtemperature,
then both fluids will expandat the samerate with temperature, andboththermometers will always give
identical readings. Otherwise, the two readings maydeviate.

1–27C

1–27C What are the ordinary and absolute temperature
scales in the SI and the English system?
ANS
1-27C They are celsius(°C)and kelvin(K) inthe SI, and fahrenheit(°F) and rankine (R) inthe English
system.

1–26C

Temperature
1–26C What is the zeroth law of thermodynamics?
ANS
1-26CThe zeroth law of thermodynamics states that two bodies are in thermal equilibriumif both have the
sametemperature reading, evenif they are not in contact.

1–25

1–25 The density of atmospheric air varies with elevation, decreasing with increasing altitude. (a) Using
the data given in the table, obtain a relation for the variation of
density with elevation, and calculate the density at an elevation
of 7000 m. (b) Calculate the mass of the atmosphere using the
correlation you obtained. Assume the earth to be a perfect
sphere with a radius of 6377 km, and take the thickness of the
atmosphere to be 25 km.
z, km r, kg/m
3
6377 1.225
6378 1.112
6379 1.007
6380 0.9093
6381 0.8194
6382 0.7364
6383 0.6601
6385 0.5258
6387 0.4135
6392 0.1948
6397 0.08891
6402 0.04008
ANS
1-25 EES The variation of densityof atmospheric air withelevation isgiven intabular form. A relation for
the variation of densitywithelevation istobe obtained, the densityat7 kmelevation istobe calculated,
and the mass of the atmosphere using the correlation istobe estimated.
Assumptions1Atmospheric air behaves as an ideal gas. 2 The earthisperfectlysphere with a radius of
6377 km, and the thickness of the atmosphere is25 km.
PropertiesThe densitydataare given intabular formas
r, km  z, km  ρ, kg/m3
6377  0  1.225
6378  1  1.112
6379  2  1.007
6380  3  0.9093
6381  4  0.8194
6382  5  0.7364
6383  6  0.6601
6385  8  0.5258
6387  10  0.4135
6392  15  0.1948
6397  20  0.08891
6402  25  0.04008

EES Solutionfor final result:
a=1.2025166
b=-0.10167
c=0.0022375
r=6377
h=25
m=4*pi*(a*r^2*h+r*(2*a+b*r)*h^2/2+(a+2*b*r+c*r^2)*h^3/3+(b+2*c*r)*h^4/4+c*h^5/5)*1E+9

1–24C

1–24C What is specific gravity? How is it related to density?
ANS
1-24CThe specific gravity, or relative density, and isdefined as the ratioof the densityof a substance to
the densityof somestandard substance ata specified temperature (usually water at 4°C, for which ρH2O=
1000 kg/m
3
). That is,  SG H2O /ρ ρ = .  When specific gravityisknown, densityisdetermined from
H2O SG ρ ρ × = .

1–23C

1–23C What is a steady-flow process?
ANS
1-23C  A process issaidtobe steady-flow ifitinvolves no changes withtime anywhere withinthe system
or atthe systemboundaries.

1–22C

1–22C Is the state of the air in an isolated room completely
specified by the temperature and the pressure? Explain.
ANS
1-22C Yes, because temperature and pressure are two independent properties and the air in an isolated
roomis a simple compressible system.

1–21C

1–21C What is the state postulate?
ANS
1-21CThe state of a simple compressible systemis completely specified bytwo independent, intensive
properties.

1–20C

1–20C Define the isothermal, isobaric, and isochoric
processes.
ANS
1-20CA process during which the temperature remains constantiscalled isothermal;a process during
whichthe pressure remains constantis calledisobaric; and a process during which the volume remains
constantiscalled isochoric.

1–19C

1–19C What is a quasi-equilibrium process? What is its
importance in engineering?
ANS
1-19CA process duringwhicha systemremains almost inequilibriumat all times is called a quasiequilibrium process. Manyengineeringprocesses canbeapproximatedas beingquasi-equilibrium. The
work output of a device is maximumand the work input to a device is minimumwhen quasi-equilibrium
processes are used instead ofnonquasi-equilibriumprocesses.

1–18C

1–18C For a system to be in thermodynamic equilibrium,
do the temperature and the pressure have to be the same
everywhere?
ANS
1-18CFor a systemto be in thermodynamic equilibrium, the temperature has to be the same throughout
butthe pressure does not. However, there should be no unbalanced pressure forces present. The increasing
pressure withdepth ina fluid, for example, should be balanced byincreasing weight.

1–17C

1–17C What is the difference between intensive and extensive properties?
ANS
1-17CIntensive properties do notdepend on the size (extent)of the systembutextensive properties do.

1–16C

1–16C A can of soft drink at room temperature is put into
the refrigerator so that it will cool. Would you model the can
of soft drink as a closed system or as an open system?
Explain.
ANS
1-16CA can of softdrink should be analyzed as a closed systemsince no mass is crossing the boundaries
of the system.

1–15C

Systems, Properties, State, and Processes
1–15C A large fraction of the thermal energy generated in
the engine of a car is rejected to the air by the radiator
through the circulating water. Should the radiator be analyzed
as a closed system or as an open system? Explain.

ANS
Systems, Properties, State, and Processes
1-15CThe radiator should be analyzed as an open systemsince mass iscrossing the boundaries of the
system.

Application Areas of Thermodynamics

All activities in nature involve some interaction between energy and matter;
thus, it is hard to imagine an area that does not relate to thermodynamics in
some manner. Therefore, developing a good understanding of basic principles
of thermodynamics has long been an essential part of engineering education.
Thermodynamics is commonly encountered in many engineering systems
and other aspects of life, and one does not need to go very far to see some
application areas of it. In fact, one does not need to go anywhere. The heart
is constantly pumping blood to all parts of the human body, various energy
conversions occur in trillions of body cells, and the body heat generated is
constantly rejected to the environment. The human comfort is closely tied to
the rate of this metabolic heat rejection. We try to control this heat transfer
rate by adjusting our clothing to the environmental conditions.
Other applications of thermodynamics are right where one lives. An ordinary house is, in some respects, an exhibition hall filled with wonders of
thermodynamics (Fig. 1–4). Many ordinary household utensils and appliances are designed, in whole or in part, by using the principles of thermodynamics. Some examples include the electric or gas range, the heating and
air-conditioning systems, the refrigerator, the humidifier, the pressure
cooker, the water heater, the shower, the iron, and even the computer and
the TV. On a larger scale, thermodynamics plays a major part in the design
and analysis of automotive engines, rockets, jet engines, and conventional or
nuclear power plants, solar collectors, and the design of vehicles from ordinary cars to airplanes (Fig. 1–5). The energy-efficient home that you may be
living in, for example, is designed on the basis of minimizing heat loss in
winter and heat gain in summer. The size, location, and the power input of
the fan of your computer is also selected after an analysis that involves
thermodynamics.

1–14

1–14 The value of the gravitational acceleration gdecreases
with elevation from 9.807 m/s2
at sea level to 9.767 m/s2at an altitude of 13,000 m, where large passenger planes cruise.
Determine the percent reduction in the weight of an airplane
cruising at 13,000 m relative to its weight at sea level.
ANS
1-14Gravitational acceleration g and thus the weightof bodies decreases withincreasing elevation. The
percentreduction inthe weightof an airplane cruising at13,000 mistobe determined.
PropertiesThe gravitational acceleration g isgiven tobe 9.807 m/s2atsea leveland 9.767 m/s2at an
altitude of 13,000 m.
Analysis Weight is proportional to the gravitational acceleration g, and thus the
percent  reduction in  weight  isequivalenttothe percentreduction inthe
gravitational acceleration, which is determined from

1–13

1–13 Solve Prob. 1–12 using EES (or other) software.
Print out the entire solution, including the
numerical results with proper units.
ANS
1-13 EES Problem1-12 isreconsidered. The entire EES solution istobe printed out, including the
numericalresultswithproper units.
Analysis The problemissolved using EES, and the solution isgiven below.
W=m*g"[N]"
m=5"[kg]"
g=9.79"[m/s^2]"
"The force balance on the rock yields the net force acting on the rock as"
F_net = F_up - F_down"[N]"
F_up=150"[N]"
F_down=W"[N]"
"The acceleration of the rock is determined from Newton's second law."
F_net=a*m
"To Run the program, press F2 or click on the calculator icon from the Calculate menu"
SOLUTION
a=20.21 [m/s^2]
F_down=48.95 [N]
F_net=101.1 [N]
F_up=150 [N]
g=9.79 [m/s^2]
m=5 [kg]
W=48.95 [N]

1–12 A

1–12 A 5-kg rock is thrown upward with a force of 150 N at a location where the local gravitationalacceleration is 9.79 m/s2. Determine the acceleration of therock, in m/s2.
ANS
1-12  [Also solved by EES on enclosed CD]A rock isthrown upward with a specified force. The
acceleration of the rock is to be determined.
Analysis The weight of the rock is

1–11

1–11 The acceleration of high-speed aircraft is sometimes
expressed in g’s (in multiples of the standard acceleration of
gravity). Determine the upward force, in N, that a 90-kg man
would experience in an aircraft whose acceleration is 6 g’s.
ANS
1-11The acceleration of an aircraft is given in g’s. The netupward force acting on a man inthe aircraftis
tobedetermined.
Analysis Fromthe Newton's second law, the force applied is

1-10E

 1-10EAn astronaut took his scales with himto space. Itis to be determined how much he will weigh on
the spring and beamscales in space.
Analysis (a) A spring scale measures weight, which isthe localgravitationalforce applied on a body:

(b) A beamscale compares masses and thus is not affected by the variations in gravitational acceleration.
The beamscale will readwhat it reads onearth,
W=150lbf

1-9

1-9The variation of gravitational acceleration above the sea level is given as a function of altitude. The 

height at which the weightof a body will decrease by 1% is to be determined.

Analysis The weightof a bodyatthe elevation z can be expressed as 

1–8

1–8 Determine the mass and the weight of the air contained in
a room whose dimensions are 6 m 6 m 8 m. Assume the
density of the air is 1.16 kg/m3. Answers: 334.1 kg, 3277 N
ANS
1-8The interior dimensions of a roomare given. The mass and weightof the air inthe room are to be
determined.
Assumptions The densityof air isconstantthroughoutthe room.
Properties The densityof air isgiven tobe ρ= 1.16 kg/m3.Analysis The mass of the air in the roomis

1–7 A

1–7 A 3-kg plastic tank that has a volume of 0.2 m3
is filled with liquid water. Assuming the density of water is 1000
kg/m3, determine the weight of the combined system
ANS
1-7A plastic tankis filledwithwater. The weightofthe combinedsystemis tobedetermined.
Assumptions The densityof water isconstantthroughout.
Properties The densityof water isgiven tobe ρ= 1000 kg/m3.
Analysis The mass ofthe water inthe tankandthe total mass are

1-6C

1–6C What is the net force acting on a car cruising at a
constant velocity of 70 km/h (a) on a level road and (b) on an
uphill road?
ANS
1-6CThere is no acceleration, thus the net force is zero in both cases.

1–5C

1–5C What is the difference between kg-mass and kg
force?
ANS
1-5CKg-mass is the mass unit inthe SIsystemwhereas kg-force is a force unit. 1-kg-force is the force
required to accelerate a 1-kg mass by 9.807 m/s2. In other words, the weightof 1-kg mass atsea levelis1
kg-force.

1–4C

1–4C What is the difference between pound-mass and
pound-force?
ANS
1-4CPound-mass lbmisthe mass unitinEnglish systemwhereas pound-force lbf is the force unit.  One
pound-force is the force required to accelerate a mass of 32.174 lbmby 1 ft/s 2. In other words, the weight ofa 1-lbmmass at sea level is 1lbf.

1-3C

1–3C An office worker claims that a cup of cold coffee on
his table warmed up to 80°C by picking up energy from the
surrounding air, which is at 25°C. Is there any truth to his
claim? Does this process violate any thermodynamic laws?
ANS
1-3CThere isno truth tohis claim.Itviolates the second law of thermodynamics.

1–2C

1–2C Why does a bicyclist pick up speed on a downhill
road even when he is not pedaling? Does this violate the con
servation of energy principle?
ANS
1-2COn a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and
thus the bicyclistpicks up speed. There isno creation of energy, and thus no violation of the conservation
of energyprinciple.

1-1c

1–1C What is the difference between the classical and the
statistical approaches to thermodynamics?
ANS
1-4CPound-mass lbmisthe mass unitinEnglish systemwhereas pound-force lbf is the force unit.  One
pound-force is the force required to accelerate a mass of 32.174 lbmby 1 ft/s 2. In other words, the weight ofa 1-lbmmass at sea level is 1lbf.

1.3

3.J. A. Schooley. Thermometry. Boca Raton, FL: CRC
Press, 1986.
ANS
1-3CThere isno truth tohis claim.Itviolates the second law of thermodynamics.

1.2

2.A. Bejan. Advanced Engineering Thermodynamics.2nd
ed. New York: Wiley, 1997
ANS
1-2COn a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and
thus the bicyclistpicks up speed. There isno creation of energy, and thus no violation of the conservation
of energyprinciple.

1.1

1.American Society for Testing and Materials. Standards
for Metric Practice.ASTM E 380-79, January 1980.
ANS
1-1CClassicalthermodynamics isbased on experimental observations whereas statistical thermodynamics
isbased on the average behavior of large groups of particles.