Thermodynamics 3

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PROBLEMS: PROBLEMS:

1.If air is at pressure, p, of 3200 lbf/ft^2, and at a temperature, T, of 800 ˚R, What is the 1.If air is at pressure, p, of 3200 lbf/ft^2, and at a temperature, T, of 800 ˚R, What is the specific volume, v? (R=5303 ft-lbf/lbm-˚R, and air can be modeled as an ideal gas.) specific volume, v? (R=5303 ft-lbf/lbm-˚R, and air can be modeled as an ideal gas.) A.9.8 ft^3/lbm A.9.8 ft^3/lbm B.11.2 ft^3/lbm B.11.2 ft^3/lbm C.13.33 ft^3/lbm C.13.33 ft^3/lbm D.14.2 ft^3/lbm D.14.2 ft^3/lbm Formula: Formula: pv pv = = RT RT v v = = RT RT / / pp ans ans. C. C

2. Steam at 1000 lbf/ft^2 pressure and 300˚R has specific volume of 6.5 ft^3/lbm and a 2. Steam at 1000 lbf/ft^2 pressure and 300˚R has specific volume of 6.5 ft^3/lbm and a specific enthalpy of 9800 lbf-ft/lbm. Find the internal energy per pound mass of steam. specific enthalpy of 9800 lbf-ft/lbm. Find the internal energy per pound mass of steam. A.2500 lbf-ft/lbm A.2500 lbf-ft/lbm B.3300 lbf-ft/lbm B.3300 lbf-ft/lbm C.5400 lbf-ft/lbm C.5400 lbf-ft/lbm D.6900 lbf-ft/lbm D.6900 lbf-ft/lbm Formula: Formula: h h = = u u + + pV pV u u = = h h – – pVpV ans ans. B. B

3. 3.0 lbm of air are contained at 25 psia and 100 ˚F. given that Rair = 53.35 3. 3.0 lbm of air are contained at 25 psia and 100 ˚F. given that Rair = 53.35 ft-lbf/lbm-˚F, what is the volume of the container?

˚F, what is the volume of the container? A.10.7 ft^3 A.10.7 ft^3 B.14.7 ft^3 B.14.7 ft^3 C.15 ft^3 C.15 ft^3 D.24.9 ft^3 D.24.9 ft^3

Formula: use the ideal gas law Formula: use the ideal gas law

pV = mRT pV = mRT T = (100 + 460) ˚R T = (100 + 460) ˚R V = mRT / p V = mRT / p ans. ans. DD

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4. The compressibility factor, x, is used for predicting the behavior of non-ideal gases. 4. The compressibility factor, x, is used for predicting the behavior of non-ideal gases. How is the compressibility factor defined relative to an ideal gas? (subscript “c” refers to How is the compressibility factor defined relative to an ideal gas? (subscript “c” refers to critical value) critical value) A. z = P / Pc A. z = P / Pc B. z = pV / RT B. z = pV / RT C. z = T / Tc C. z = T / Tc D. z = RT / pV D. z = RT / pV

Hint: for an real gases the compressibility factor, x, is an dimensionless constant given Hint: for an real gases the compressibility factor, x, is an dimensionless constant given by pV= zRT. Therefore z = pV / RT

by pV= zRT. Therefore z = pV / RT ans.

ans. BB

5. From the steam table , Determine the average constant pressure specific heat (c) of 5. From the steam table , Determine the average constant pressure specific heat (c) of steam at 10 kPa and 45.8 ˚C

steam at 10 kPa and 45.8 ˚C A.1.79 kJ/ kg-˚C A.1.79 kJ/ kg-˚C B.10.28 kJ/ kg-˚C B.10.28 kJ/ kg-˚C C.30.57 kJ/ kg-˚C C.30.57 kJ/ kg-˚C D. 100.1 kJ/ kg-˚C D. 100.1 kJ/ kg-˚C Formula: Formula:

�

h = ch = c

�

TT From the steam table From the steam table

At At 47.7 47.7 ˚C ˚C h= h= 2588.1 2588.1 kJ/ kJ/ kgkg At At 43.8 43.8 ˚C ˚C h= h= 2581.1 2581.1 kJ/ kJ/ kgkg ans. ans. AA

6. A 10m^3 vessel initially contains 5 m^3 of liquid water and 5 m^3 of saturated water 6. A 10m^3 vessel initially contains 5 m^3 of liquid water and 5 m^3 of saturated water vapor at 100 kPa. Calculate the internal energy of the system using the steam table. vapor at 100 kPa. Calculate the internal energy of the system using the steam table. A. A. 5 5 x10^5 x10^5 kJkJ B. 8 x10^5 kJ B. 8 x10^5 kJ C. 1 x10^6 kJ C. 1 x10^6 kJ D. 2 x10^6 kJ D. 2 x10^6 kJ

Formula: from the steam table Formula: from the steam table vƒ = 0.001043 m^3 / kg vƒ = 0.001043 m^3 / kg vg = 1.6940 m^3 / kg vg = 1.6940 m^3 / kg u u ƒ ƒ = = 417.3 417.3 kJ/kg kJ/kg ug ug = = 2506 2506 kJ/kgkJ/kg formula: M vap = V vap/vg

formula: M vap = V vap/vg M liq

M liq = = V liq/ V liq/ vƒvƒ u =

u = uƒ uƒ M liq M liq + ug + ug M vapM vap ans.

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7. A vessel with a volume of cubic meter contains liquid water and water vapor ion 7. A vessel with a volume of cubic meter contains liquid water and water vapor ion equilibrium at 600 kPa. The liquid water has mass of 1kg. using the steam table, equilibrium at 600 kPa. The liquid water has mass of 1kg. using the steam table, calculate the

calculate the mass of mass of the water the water vapor.vapor. A. 0.99kg A. 0.99kg B. 1.57 kg B. 1.57 kg C. 2.54 kg C. 2.54 kg D.3.16 kg D.3.16 kg

Formula: from the steam table at 600 kPa Formula: from the steam table at 600 kPa vƒ = 0.001101 m^3 / kg vƒ = 0.001101 m^3 / kg vg vg = = 0.3157 0.3157 m^3 m^3 / / kg kg ans.ans.DD Vtot = mƒ vƒ + mg vg Vtot = mƒ vƒ + mg vg mg = (tot- mƒ vƒ) / vg mg = (tot- mƒ vƒ) / vg ans. ans.DD

8.Calculate the entropy of steam at 60 psiawith a quality of 0.8 8.Calculate the entropy of steam at 60 psiawith a quality of 0.8 A. A. 0.4274 0.4274 BTU/lbm-˚RBTU/lbm-˚R B. B. 0.7303 0.7303 BTU/lbm-˚RBTU/lbm-˚R C. C. 1.1577 1.1577 BTU/lbm-˚RBTU/lbm-˚R D. D. 1.2172 1.2172 BTU/lbm-˚RBTU/lbm-˚R

Formula: from the steam table at 60 psia: Formula: from the steam table at 60 psia:

sƒ = 0.4274 BTU/lbm-˚R sƒ = 0.4274 BTU/lbm-˚R sƒg

sƒg = = 1.2172 1.2172 BTU/lbm-˚R)BTU/lbm-˚R) s =

s = sƒ sƒ + + x sƒg x sƒg where x where x = is = is the qualitythe quality ans.

ans. CC

9. Find the change in internal energy of 5 lbm of oxygen gas when the temperature 9. Find the change in internal energy of 5 lbm of oxygen gas when the temperature changes from

changes from 100 ˚F to 100 ˚F to 120 ˚F. cv= 120 ˚F. cv= 0.157 BTU/lbm-˚R0.157 BTU/lbm-˚R A.14.7 BTU A.14.7 BTU B.15.7 BTU B.15.7 BTU C. 16.8 BTU C. 16.8 BTU D. 15.9 BTU D. 15.9 BTU Formula: Formula:

�

U= mcvU= mcv

�

TT Ans. Ans. BB

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10. Water (specific heat cv= 4.2 kJ/ kg

··

K ) is being heated by a 1500 W heater. What is K ) is being heated by a 1500 W heater. What is the rate of change im temperature of 1kg of the water?

the rate of change im temperature of 1kg of the water? A. 0.043 K/s A. 0.043 K/s B. 0.179 K/s B. 0.179 K/s C. 0.357 K/s C. 0.357 K/s D. 1.50 K/s D. 1.50 K/s Formula: Q = mcv ( Formula: Q = mcv (

�

T)T) Ans. Ans. BB

11. A system weighing 2 kN. Determine the force that accelerate if to 12 m/s^2. 11. A system weighing 2 kN. Determine the force that accelerate if to 12 m/s^2. a. vertically upward when g= 9.7 m/s^2

a. vertically upward when g= 9.7 m/s^2 A. A. 4474.23 4474.23 NN B.5484.23 N B.5484.23 N C.4495.23 N C.4495.23 N D.5488.23 N D.5488.23 N Formula: F = m/k (a +g) Formula: F = m/k (a +g) ans. ans. AA

12. refer to problem # 11. determne the force that accelerates if to 12 m/s^2.horizontally 12. refer to problem # 11. determne the force that accelerates if to 12 m/s^2.horizontally along a frictionless plane.

along a frictionless plane. A.2474.23 N A.2474.23 N B.2574.23 N B.2574.23 N C.3474.23 N C.3474.23 N D.2374.23 N D.2374.23 N Formula : Formula : M = wk / g M = wk / g F = ma / k F = ma / k ans ans. A. A

13. A problem Drum ( 3 ft dia ; 6 ft height ) is field with a fluid whose density is 50 lb/ft^3 13. A problem Drum ( 3 ft dia ; 6 ft height ) is field with a fluid whose density is 50 lb/ft^3 Determine the total volume of the fluid.

Determine the total volume of the fluid.

A. 42.41 ft^3 A. 42.41 ft^3 B.44.35 ft^3 B.44.35 ft^3 C.45.63 ft^3 C.45.63 ft^3 D.41.23 ft^3 D.41.23 ft^3 Formula: Vf = (pi d^2 h) / 4 Formula: Vf = (pi d^2 h) / 4 ans ans. A. A

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14. What is the resulting pressure when one pound of air at 15 psia and 200 ˚F is 14. What is the resulting pressure when one pound of air at 15 psia and 200 ˚F is heated at constant volume to 800 ˚F.

heated at constant volume to 800 ˚F. A.15 psia A.15 psia B.28.6 psia B.28.6 psia C.36.4 psia. C.36.4 psia. D.52.1 psia D.52.1 psia Formula : Formula : T1/p1 = T2/p2 T1/p1 = T2/p2 p2 = p1T2 / T1 p2 = p1T2 / T1 ans. ans.BB

15. What horsepower is required to isothermally compress 800 ft^3 of Air per minute 15. What horsepower is required to isothermally compress 800 ft^3 of Air per minute from 14.7 psia to 120 psia?

from 14.7 psia to 120 psia? A. 28 hp A. 28 hp B.108 hp B.108 hp C.256 hp C.256 hp D.13900 hp] D.13900 hp] Formula: Formula: W= p1V1 W= p1V1 ln ln (p1/p2)(p1/p2) Power = dW / dt Power = dW / dt ans ans. B. B

16. What is the equation for the work done by a constant temperature system? 16. What is the equation for the work done by a constant temperature system? A. A. W W = = mRT mRT ln(V2-V1)ln(V2-V1) B. B. W W = = mR mR ( ( T2-T1 T2-T1 ) ) ln( ln( V2/V1)V2/V1) C. C. W W = = mRT mRT ln ln (V2/V1)(V2/V1) D. D. W W = = RT RT ln ln (V2/V1)](V2/V1)] Formula : W=

Formula : W=

∫∫

pdV lim pdV lim 1,21,2

рр

= mRT / V = mRT / V ans.

ans. CC

17. Twenty grams of oxygen gas are compressed At a constant temperature of 30 ˚C to 17. Twenty grams of oxygen gas are compressed At a constant temperature of 30 ˚C to 5% of their original volume. What work is done on the system.

5% of their original volume. What work is done on the system. A.824 cal A.824 cal B.924 cal B.924 cal C.944 cal C.944 cal D.1124 cal D.1124 cal

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Formula: Formula:

W = -mRT ln (V2/V1) W = -mRT ln (V2/V1)

Where R = (1.98 cal/gmole

··

K) (32 g/gmole)K) (32 g/gmole) ans.

ans. DD

18. Helium ( R= 0.4698 BTU/lbm-˚R ) is compressed isothermally from 14.7 psia and 68 18. Helium ( R= 0.4698 BTU/lbm-˚R ) is compressed isothermally from 14.7 psia and 68 ˚F. The compression ratio is 1:4. Calculate the work done by the gas.

˚F. The compression ratio is 1:4. Calculate the work done by the gas. A. A. –1454 –1454 BTU/lbmBTU/lbm B.-364 BTU/lbm B.-364 BTU/lbm C.-187 BTU/lbm C.-187 BTU/lbm D.46.7 BTU/lbm D.46.7 BTU/lbm Formula: W = RT ln (V2/V1) Formula: W = RT ln (V2/V1) ans ans.B.B

19.Gas is enclosed in a cylinder with a weighted piston as the stop boundary. The gas 19.Gas is enclosed in a cylinder with a weighted piston as the stop boundary. The gas is heated and expands from a volume of 0.04 m^3 to 0.10 m^3 at a constant pressure of is heated and expands from a volume of 0.04 m^3 to 0.10 m^3 at a constant pressure of 200kPa.Calculate the work done by the system.

200kPa.Calculate the work done by the system. A. 8 kJ A. 8 kJ B. 10 kJ B. 10 kJ C.12 kJ C.12 kJ D.14 kJ D.14 kJ Formula: W = p (V2-V1) Formula: W = p (V2-V1) ans. ans.CC

20. refer to problem no.13. Determine the specific volume… 20. refer to problem no.13. Determine the specific volume… A. 0.02 ft^3/lbm A. 0.02 ft^3/lbm B. 0.05 ft^3/lbm B. 0.05 ft^3/lbm C. 1.0 ft^3/lbm C. 1.0 ft^3/lbm D. 1.2 ft^3/lbm D. 1.2 ft^3/lbm Formula : Formula : Vf = ( pi d^2 h) / 4 Vf = ( pi d^2 h) / 4 Pf = mf / vf Pf = mf / vf Specific

Specific volume= volume= Vf Vf / / mfmf ans.

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21.What is the weight of a 66-kg

21.What is the weight of a 66-kgmm man at standard condition? man at standard condition?

(Formula: F (Formula: Fgg = mg / k) = mg / k) a. 66 kg a. 66 kgff b. 66 kg b. 66 kgmm c. 66 lb c. 66 lbmm d. 66 g d. 66 gff a a

22.What is the specific weight of ater at standard condition? 22.What is the specific weight of ater at standard condition?

(Formula: (Formula:

γγ

= =

ρ

g / k)g / k) a. 1000 kg a. 1000 kgmm /m /m33 b. 9.8066 m/s b. 9.8066 m/s22 c. 1000 kg c. 1000 kg /m /mff 33

d. None of the above d. None of the above c c 23 23.7.746 46 °°R = R = __________ __ °°FF a. 254 a. 254 b. 345 b. 345 c. 286 c. 286

d. None of the above d. None of the above c

c

24.A 30-m vertical column of fluid (density 1878 kg/m

24.A 30-m vertical column of fluid (density 1878 kg/m33) is located where g= 9.65 mps) is located where g= 9.65 mps22.. Find the pressure at t6he base of the column.

Find the pressure at t6he base of the column. (Formula: p (Formula: pgg = g = g

ρ

hhgg /k ) /k ) a. 543680 N/m a. 543680 N/m22 b. 543.68 kPa (gage) b. 543.68 kPa (gage) c. Both a & b c. Both a & b

c 25.Ten

25.Ten cu ft of air cu ft of air at 300 pat 300 psia 400sia 400°°F is cooF is cooled to led to 140°140°F at consF at constant volutant volume. Wme. What is thehat is the final pressure? final pressure? (formula: p (formula: p22 = p= p11TT22 / T / T11 ) ) a. 0 a. 0 b. 209 psia b. 209 psia c. - 420 psia c. - 420 psia

d. None of the above d. None of the above b

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26 26.8.876 76 °°R = R = _________ _ °°FF a. 335 a. 335 b. 416 b. 416 c. 400 c. 400

b

27.There are 1.36 kg of gas, for which R = 377 J/kg.k and k=1.25, that undergo a 27.There are 1.36 kg of gas, for which R = 377 J/kg.k and k=1.25, that undergo a nonflow constant volume process from p

nonflow constant volume process from p11 = 551.6 kPa and t = 551.6 kPa and t11 = 60 = 60°°C tC to po p 22 = 1655 kPa. = 1655 kPa.

During the process the gas is internally stirred and there are also added 105.5 kJ of During the process the gas is internally stirred and there are also added 105.5 kJ of heat. Determine t heat. Determine t22.. (Formula: T (Formula: T22 = T = T11p2/ pp2/ p11 ) ) a. 999 K a. 999 K b. 888 K b. 888 K c. 456K c. 456K

d. One of the above d. One of the above a a 28.5 atm = ____ mmHg 28.5 atm = ____ mmHg a. 8300 a. 8300 b. 3800 b. 3800 c. 3080 c. 3080

b

29.A certain gas, with c

29.A certain gas, with cpp = 0.529 Bt = 0.529 Btu/lb.°u/lb.°R and R = 96.2 ftR and R = 96.2 ft.lb/lb.°.lb/lb.°R, expanR, expandd s from 5 cu fts from 5 cu ft

and 80 °

and 80 °F to 15 cu fF to 15 cu ft while the pressure t while the pressure remains coremains co nstant at 15.5 psia. nstant at 15.5 psia. Compute for TCompute for T22..

(Formula: T (Formula: T22 = T = T11VV22 / V / V11 ) ) a a. 4. 4660 °0 °RR b b. 2. 2770 °0 °RR c. c. 161620 20 °°RR

c

30.In the above problem, compute for the mass. 30.In the above problem, compute for the mass.

(Formula: m = p (Formula: m = p11VV11 / RT / RT11 ) ) a. 0.2148 lb a. 0.2148 lb b. 0.2134 lb b. 0.2134 lb c. 0.1248 lb c. 0.1248 lb

d. None of the above d. None of the above a

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31 31.7.71010°°R= R= __________ __ °°CC a. 214 a. 214 b. 121 b. 121 c. 213 c. 213

d. None of the above d. None of the above b b 32 32.2.212 12 °°F = F = _________ _ °°CC a. 200 a. 200 b. 150 b. 150 c. 100 c. 100

c 33.Let a

33.Let a closed system closed system execute execute a state a state change for change for which which the the heat heat is Q is Q = = 100 J 100 J andand work is W = -25 J. Find work is W = -25 J. Find

�

E.E. (Formula: (Formula:

�

E = Q- W )E = Q- W ) a. 125 J a. 125 J b. 123 J b. 123 J c. 126 J c. 126 J

a

34.A pressure gage registers 50 psig in a region where the barometer is 14.25 psia. 34.A pressure gage registers 50 psig in a region where the barometer is 14.25 psia. Find absolute pressure in psia, Pa.

Find absolute pressure in psia, Pa. (Formula;

(Formula; pp ==p p atmatm + + p p gg ) )

a. 433 kPa a. 433 kPa b. 443 kPa b. 443 kPa c. 343 kPa c. 343 kPa

b

35. A mass of 5 kg is 100 m above a given datum where local g = 9.75 m/s

35. A mass of 5 kg is 100 m above a given datum where local g = 9.75 m/s22. Find the. Find the gravitational force in newtons.

gravitational force in newtons. (Formula: F (Formula: Fgg = mg/k ) = mg/k ) a. 48.75 N a. 48.75 N b. 50 N b. 50 N c. 45 N c. 45 N

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36.

36. In the above In the above problem, find the potential problem, find the potential energy of the energy of the mass with respect mass with respect to datum.to datum. (Formula: (Formula: PP = mgz/k )= mgz/k ) a. 4875 j a. 4875 j b. 0.51 j b. 0.51 j c. 0.46 j c. 0.46 j

a 37.

37. The combined mThe combined mass of car aass of car and passengers trand passengers travelling at 72 kmvelling at 72 km/hr is 1500 kg. /hr is 1500 kg. FindFind the kinetic energy of this combined mass.

the kinetic energy of this combined mass. (Formula: K = m (Formula: K = mv v 22 / 2k ) / 2k ) a. 300 kJ a. 300 kJ b. 200 kJ b. 200 kJ c. 500 kJ c. 500 kJ

d. None of the above d. None of the above a a 38. 38. 14.696 psia 14.696 psia = _____ m= _____ mmHgmHg a. 760 a. 760 b. 1 b. 1 c. 350 c. 350

d. None of the above d. None of the above b b 39. 212 39. 212 °°C C = _= ________ K_ K a. 485 a. 485 b. 435 b. 435 c. 498 c. 498

d. None of the above d. None of the above a a 40. 40.212 212 °°F = _F = ________ R_ R a. 567 a. 567 b. 672 b. 672 c. 700 c. 700

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41.

41. An An automautomobile obile tire tire has has a a gauge gauge presspressure ure of of 200kp200kpa a at at 0°0°C C assumassuming ing no no airair leaks and no change of volume of the tire, what is the gauge pressure at leaks and no change of volume of the tire, what is the gauge pressure at 35ºC. 35ºC. a. 298.645 a. 298.645 b. 398.109 b. 398.109 c. 291.167 c. 291.167 d. 281.333 d. 281.333 a a Pg = Pabs - Patm Pg = Pabs - Patm 42.

42. An An ideal ideal gas gas at at 45psig 45psig and and 80ºF 80ºF is is heated heated in in the the close close container container to to 130ºF.130ºF. What is the final pressure?

What is the final pressure? a. 65.10psi a. 65.10psi b. 65.11psi b. 65.11psi c. 65.23psi c. 65.23psi d. 61.16psi d. 61.16psi c c P P11VV11 /T /T11 = P = P22VV22 /T /T2;2; V = Constant V = Constant 43.

43. A A wall wall of of the the firebrick firebrick has has an an inside inside temperature temperature of of 313ºF 313ºF and and an an outsideoutside temperature of 73ºF. What is the difference in the surface temperature in temperature of 73ºF. What is the difference in the surface temperature in Rankin? Rankin? a. 70 a. 70 b. 68 b. 68 c. 72 c. 72 d. 94 d. 94 a a ºR = ºF + 460 ºR = ºF + 460 44.

44. What What is is the the force force required required to to accelerate accelerate a a mass mass of of 30kg 30kg at at a a rate rate of of 15m/s².15m/s². a. 460N a. 460N b. 380N b. 380N c. 560N c. 560N d. 450N d. 450N d d F = ma F = ma 45.

45. How How much much does does an an object object having having the the mass mass of of 100kg 100kg weight weight in in newton.newton. a. 981N a. 981N b. 991N b. 991N c. 981.6N c. 981.6N d. 980.1N d. 980.1N a a F = ma F = ma

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46.

46. The volume The volume of of the gas the gas held at held at constaconstant nt presspressure increases 4cm² ure increases 4cm² at at 0°0°C C toto 5cm². What is the final pressure?

5cm². What is the final pressure? a. 68.65ºC a. 68.65ºC b. 68.25ºC b. 68.25ºC c. c. 7070.0.01°1°CC d. 79.1ºC d. 79.1ºC b b tt22 = T = T22 – T – T11 47.

47. A A certain certain gas gas with with cp = cp = 0.529Btu/lb°0.529Btu/lb°R R and and R R = = 99 6.2ft/lbºR 6.2ft/lbºR expands expands from 5ftfrom 5ft and 80ºF to 15ft while the pressure remains constant at 15.5 psia.

and 80ºF to 15ft while the pressure remains constant at 15.5 psia. a. a. TT22 =1.620ºR, =1.620ºR,

▲

H = 122.83 BtuH = 122.83 Btu b. b. TT22= = 22°°RR,,

▲

H = 122.83 BtuH = 122.83 Btu c. c. TT22= 2.620ºR,= 2.620ºR,

▲

H = 122.83 BtuH = 122.83 Btu d. d. TT22= = 11°°RR,,

▲

H = 122.83 BtuH = 122.83 Btu a a T T22 = V = V22 (t (t22)/V)/V11andand

▲

H = mcp (TH = mcp (T22-T-T11)) 48.

48. A A vacuum is vacuum is connected connected to to a a tank tank reads reads 3kpa 3kpa at at a a location location w/ w/ the the barametricbarametric pressure reading is 75mmhg. Determined the Pabsolute in the tank

pressure reading is 75mmhg. Determined the Pabsolute in the tank a. 70.658kpa a. 70.658kpa b. 68kpa b. 68kpa c. 58.78kap c. 58.78kap d.

d. None None of of the the aboveabove a

a

Pabs = Patm - Pvacuum Pabs = Patm - Pvacuum 49. Calculate:

49. Calculate: a.

a. Mass flMass flow ow rate irate in ln lb/hr.b/hr. b.

b. The velocity The velocity at section at section 2 in 2 in fpsfps a. 800,000lb/hr;625ft/s a. 800,000lb/hr;625ft/s b. 900,000lb/hr;625ft/s b. 900,000lb/hr;625ft/s c. 888,000lb/hr;269ft/s c. 888,000lb/hr;269ft/s d. 700,000lb/hr;269ft/s d. 700,000lb/hr;269ft/s b b m = A1V!/V1 m = A1V!/V1 50.

50. A A 600kg hammer 600kg hammer of of a pila pile e driver driver is is lilted lilted 2m the 2m the pilling pilling head. What head. What is is thethe change of potential energy? If the hammer is realest. What will be its velocity change of potential energy? If the hammer is realest. What will be its velocity and the instant if it sticks the pilling?

and the instant if it sticks the pilling? a. a. 10,772N-m 10,772N-m and and 5.26m/s5.26m/s b. b. 13,200N-m 13,200N-m and and 5.26m/s5.26m/s c. c. 11,772N-m 11,772N-m and and 6.26m/s6.26m/s d. d. 11,772N-m 11,772N-m and and 5.26m/s5.26m/s

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c c

▲

PE = mgo(PE = mgo(

▲

Z)/gcZ)/gc

51. A bayabas falls from a branch 5m above the ground with what speed in meter 51. A bayabas falls from a branch 5m above the ground with what speed in meter

per second does it strike the ground assume g = 10m/s². per second does it strike the ground assume g = 10m/s². a. 11m/s a. 11m/s b. 12m/s b. 12m/s c. 13m/s c. 13m/s d. 10m/s d. 10m/s d d

▲

KE = mV2/2gcKE = mV2/2gc 52.

52. While swimming While swimming a depth of a depth of 13m in a 13m in a fresh water lakfresh water lake a fish e a fish emits an emits an airair bubble of volume 2mm² atmospheric pressure is 100kpa what is the original bubble of volume 2mm² atmospheric pressure is 100kpa what is the original pressure of the bubble.

pressure of the bubble. a. 217.17kpa a. 217.17kpa b. 119kpa b. 119kpa c. 326.15kpa c. 326.15kpa d. 210kap d. 210kap a a Pabs = Pg + Patm Pabs = Pg + Patm 53.

53. Oxygen at 15ºOxygen at 15ºC and 1C and 10.3Mpa gauge 0.3Mpa gauge pressure occupressure occupies 600L. Whapies 600L. What is thet is the occupied by the oxygen at 8.28Mpa gauge pressure and 35ºC?

occupied by the oxygen at 8.28Mpa gauge pressure and 35ºC? a. 789.32L a. 789.32L b. 796.32L b. 796.32L c. 699L c. 699L d. 588L d. 588L b b V V22 = P = P11VV11 /T /T11PP22

54. Water is flowing through a 1 foot diameter pipe at the rate of 10ft/sec. What is 54. Water is flowing through a 1 foot diameter pipe at the rate of 10ft/sec. What is

the volume flow rate of water in ft³/sec? the volume flow rate of water in ft³/sec? a. 7.85 a. 7.85 b. 6.85 b. 6.85 c. 8.85 c. 8.85 d. 5.85 d. 5.85 a a V = A V = A

νν

55. A certain fluid is flowing an a 0.5mx0.3 channel at the rate of 3m/s and has a 55. A certain fluid is flowing an a 0.5mx0.3 channel at the rate of 3m/s and has a

specific volume of 0.012m³/kg. Determined the mass

specific volume of 0.012m³/kg. Determined the mass of water flowing in kg/s.of water flowing in kg/s. a. 267kg/s a. 267kg/s b. 378kg/s b. 378kg/s c. 375kg/s c. 375kg/s d. 456.5kg/s d. 456.5kg/s

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c c m = A m = A

νν

/V /V 56.

56. A gas having A gas having a volume a volume of 100ft³ at 27ºC of 100ft³ at 27ºC is expanded is expanded to 120ft³ by heated atto 120ft³ by heated at constant pressure to what temperature has it been heated to have this new constant pressure to what temperature has it been heated to have this new volume? volume? a. 87C a. 87C b. 85C b. 85C c. 76C c. 76C d. 97C d. 97C a a tt22 = T = T22 – T – T11 57.

57. Water fWater flow to a low to a terminal 3mm terminal 3mm diameter and diameter and has an has an average spaverage speed ofeed of 2m/s.What is the rate of flow in cubic meter/mm?

2m/s.What is the rate of flow in cubic meter/mm? a. 0.0001m³/min a. 0.0001m³/min b. b. 0.076 0.076 m³/minm³/min c. c. 0.085 0.085 m³/minm³/min d. d. 0.097 0.097 m³/minm³/min c c 58.

58. Water flowing at Water flowing at a 6m/s a 6m/s through a 60mthrough a 60mm pipe m pipe is suddenly chais suddenly channeled into anneled into a 30mm pipe. What is the velocity in the small pipe?

30mm pipe. What is the velocity in the small pipe? a. 34m/s a. 34m/s b. 24m/s b. 24m/s c. 15m/s c. 15m/s d. 27m/s d. 27m/s b b 59.

59. A vertical A vertical column of column of water will bwater will be suppe supported to orted to what height what height by standardby standard atmospheric pressure. atmospheric pressure. a. 33.9ft a. 33.9ft b. 45ft b. 45ft c. 67ft c. 67ft d. 25.46ft d. 25.46ft a a ho = Po/Yo ho = Po/Yo

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60.

60. A A fluid fluid flows flows in in a a steady steady manner manner between between two two section section in in a a flow flow line line at at sectionsection 1: A

1: A1 =1 = 1ft², V1ft², V11 = 100fpm, volume1 of 4ft³/lb at sec2: A= 100fpm, volume1 of 4ft³/lb at sec2: A22= 2ft², p = 0.20lb/ft³ calculate the= 2ft², p = 0.20lb/ft³ calculate the

velocity at section 2. velocity at section 2. a. 625fpm a. 625fpm b. 567fpm b. 567fpm c. 356fpm c. 356fpm d.

d. None None of of the the aboveabove a

a

61. The weight of an object is 50lb. What is its mass at standard condition? 61. The weight of an object is 50lb. What is its mass at standard condition? a. 50 lb

a. 50 lbmm b. b. 60 60 lblbmm c. 70 lbc. 70 lbmm d. d. 80 80 lblbmm

a

a formula: formula: m m = = FFggk / gk / g

62. A vertical column of water will be supported to what height by standard atmospheric 62. A vertical column of water will be supported to what height by standard atmospheric pressure. If the Y

pressure. If the Yw w = 62.4lb/ft= 62.4lb/ft33 ppoo = 14.7 psi. = 14.7 psi.

a.

a. 44.9 44.9 ft ft b. b. 33.9 33.9 ft ft c. c. 22.9 22.9 ft ft d. d. 55.9 55.9 ftft b

b formula: formula: hhoo = p = poo / Y / Yww

63. For a certain gas R= 320J/kg.K and c

63. For a certain gas R= 320J/kg.K and cvv= 0.84kJ/kg.K= 0.84kJ/kg.Kcc . Find k? . Find k?

a.

a. 1.36 1.36 b. b. 1.37 1.37 c. c. 1.38 1.38 d. d. 1.391.39 c

c formula: formula: k= k= R R / / ccvv +1 +1

64. Ten

64. Ten cu.ft of acu.ft of air at 300 ir at 300 psia anpsia and 400°d 400°F is coolF is coolee d to 140d to 140°°F at consF at constant votant volume. Wlume. What ishat is the transferred heat?

the transferred heat? a.-120Btu

a.-120Btu b. b. -220Btu -220Btu c.-320Btu c.-320Btu d. d. -420Btu-420Btu d

d formula: formula: Q= Q= mcmcvv(T(T22-T-T11))

65. Utilizing the answer to the previous problem, estimate the overall or average 65. Utilizing the answer to the previous problem, estimate the overall or average increase in temperature (

increase in temperature ( ∆ ∆T T ) of the concrete roof from the energy absorbed from the) of the concrete roof from the energy absorbed from the sun during a 12 hour day. Assume that all of the radiation absorbed goes into heating sun during a 12 hour day. Assume that all of the radiation absorbed goes into heating the roof. The specific heat of concrete is about 900 J/kg, and the density is about 2,300 the roof. The specific heat of concrete is about 900 J/kg, and the density is about 2,300 kg/m kg/m33.. a a. . 77.9 .9 °°C C bb. . 8.8.9°9°C C c. c. 9.9.9°9°C C dd. . 1010.9.9°°CC a formula: a formula:

∆

Q = m cQ = m c

∆

TT

66. The concrete roof of a house is 10 m by 8 m and 10 cm thick (4"). Estimate the total 66. The concrete roof of a house is 10 m by 8 m and 10 cm thick (4"). Estimate the total heat the roof would absorb over the 12 day?

heat the roof would absorb over the 12 day? a. 1.3 x 10

a. 1.3 x 1088 J J b b 2.3 2.3 x x 101088 J J c. c. 3.3 3.3 x x 101088 J J d. d. 4.3 4.3 x x 101088 J J

a formula:

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67. The value for the

∆

U of a system is -120 J. If the system is known to have absorbedU of a system is -120 J. If the system is known to have absorbed 420 J of heat, how much work was done?

420 J of heat, how much work was done? a.

a. -540 -540 J J b. b. -640 -640 J J c. c. -740 -740 J J d. d. -840 -840 JJ

a formula:

∆

U = q + wU = q + w

68. When the pressure on a 1 kg liquid is increased isothermally from 1 bar to 3000 bar 68. When the pressure on a 1 kg liquid is increased isothermally from 1 bar to 3000 bar the Gibbs free energy increases by 360 kJ. Estimate the density of the liquid.

the Gibbs free energy increases by 360 kJ. Estimate the density of the liquid. a. 0.66 kg liter

a. 0.66 kg liter -1-1 b. b. 0.77 0.77 kg kg literliter -1-1 c. c. 0.88 0.88 kg kg literliter -1-1 d. d. 0.99 0.99 kg kg literliter -1-1

b solution: b solution: 69.

69. A car wA car whose mass is hose mass is 2 metric 2 metric tons is tons is accelerated uniformly from accelerated uniformly from stand hill stand hill toto 100kmph in 5 sec. Find the driving force in Newton’s.

100kmph in 5 sec. Find the driving force in Newton’s. a.

a. 11,120N 11,120N b. b. 11,320N 11,320N c. c. 11,420N 11,420N d. d. 11520N11520N a

a formula: formula: F= F= ma ma / / kk

70. An ideal gas of volume 1 liter and pressure 10 bar undergoes a quasistatic adiabatic 70. An ideal gas of volume 1 liter and pressure 10 bar undergoes a quasistatic adiabatic expansion until the pressure drops to 1 bar. Assume

expansion until the pressure drops to 1 bar. Assume γ γ to be 1.4 what is the final to be 1.4 what is the final volume? volume? a. a. 3.18 3.18 l l b. b. 4.18 4.18 l l c. c. 5.18 5.18 l l d. d. 6.18 6.18 ll c solution: c solution:

71. Two masses, one of the 10kg and the other unknown, are placed on a scale in a 71. Two masses, one of the 10kg and the other unknown, are placed on a scale in a region where g = 9.67 m/sec

region where g = 9.67 m/sec22. The combined weight of these two masses is 174.06 N.. The combined weight of these two masses is 174.06 N. Find the unknown mass in kg.

Find the unknown mass in kg. a.

a. 20kg 20kg b. b. 19kg 19kg c. c. 18kg 18kg d. d. 17kg17kg c

c formula: formula: m= m= FFggk / gk / g

72. The flow energy of 5ft

72. The flow energy of 5ft33 of a fluid passing a boundary to a system is 80,000 ft-lb. of a fluid passing a boundary to a system is 80,000 ft-lb. Determine the pressure at this point.

Determine the pressure at this point. a.

a. 222psi 222psi b. b. 333psi 333psi c. c. 444psi 444psi d. d. 111psi111psi d

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73. Find

ии

for steam at 100psia and 600°for steam at 100psia and 600°F.If h = F.If h = 1329.6 and 1329.6 and v = v = 6.2166.216 a.1214Btu

a.1214Btu / / lb lb b. b. 1234 1234 Btu Btu / / lb lb c. 1342 c. 1342 Btu Btu / / lb lb d. d. 1324 1324 Btu Btu / / lblb

a formula:

ии

= h – pv/ J = h – pv/ J

74. What mass of nitrogen is contained in a 10ft

74. What mass of nitrogen is contained in a 10ft33 vessel at a pressure of 840 atm and vessel at a pressure of 840 atm and 820°

820°R? Make a R? Make a computation by ucomputation by using ideal sing ideal gas equatigas equati on.on. a.

a. 194 194 lb lb b. b. 214 214 lb lb c.394 c.394 lb lb d. d. 413 413 lblb c

c formula: formula: m= m= pV pV / / RTRT

75. A rotary compressor receives 6m

75. A rotary compressor receives 6m33 / min of a gas (R=410J / kgK, c / min of a gas (R=410J / kgK, cpp= 1.03kJ / kgK,= 1.03kJ / kgK,

k= 1.67) at 105 k/P

k= 1.67) at 105 k/Paa, 27°aa, 27°C and deliverC and delivers it at 630 s it at 630 kPaa:kPaa:

∆

P = 0,P = 0,

∆

K= 0. Find the work ifK= 0. Find the work if the process is isentropic?

the process is isentropic? a.

a. -1664kJ/min -1664kJ/min b. b. -1774 -1774 kJ/min kJ/min c. c. -1884 -1884 kJ/min kJ/min d. d. -1994 -1994 kJ/minkJ/min a

a formula: Wformula: WSFSF= Q -= Q -

∆

H H m=pm=p11VV11 / RT / RT11 TT22= T= T11(p(p22 /p /p11))(k-1)/k(k-1)/k

76. A c

76. A carnot arnot power power cycle cycle operoperates oates on 2 lb on 2 lb of air bef air between ttween the limhe limits of its of 71°71°F and F and 500°500°F.F. The pressure at the beginning of isothermal expansion is 400 psia and at the end of The pressure at the beginning of isothermal expansion is 400 psia and at the end of isothermal expansion is 185psig. Determine the volume at the end of isothermal isothermal expansion is 185psig. Determine the volume at the end of isothermal compression. compression. a.7.849 ft a.7.849 ft33 b. b. 7.850 7.850 ftft33 c. 7.851 c. 7.851 ftft33 d. d. 7852 7852 ftft33 a a formula: formula: V= V= mRT/ mRT/ P P PP33= P= P22[T[T33 / T / T22]]

77. During a polytropic process, 10 lb of an ideal gas, whose R= 40ft.lb/lb.R and c 77. During a polytropic process, 10 lb of an ideal gas, whose R= 40ft.lb/lb.R and cpp ==

0.25B

0.25Btu/lb.R, ctu/lb.R, changehanges state fro 2s state fro 20psia an0psia and 40°d 40°F to F to 120 psia 120 psia and 340°and 340°F. DeterF. Determine n?mine n? a.

a. 1.234 1.234 b. b. 1.345 1.345 c. c. 1.456 1.456 d. d. 1.3561.356 d

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78. A perfect gas has a value of R= 319.2 J/ kf.K and k= 1.26. If 120 kJ are added to 78. A perfect gas has a value of R= 319.2 J/ kf.K and k= 1.26. If 120 kJ are added to 2.27 kf\g of

2.27 kf\g of this gas at this gas at constant pressure constant pressure when the initial when the initial temp is 32.2°temp is 32.2°C? Find TC? Find T 22..

a.

a. 339.4 339.4 K K b. b. 449.4 449.4 K K c. c. 559.4K 559.4K d. d. 669.4K669.4K a

a formula: formula: ccpp = = kR/ kR/ k-1 k-1 Q= Q= mcmcpp(T(T22-T-T11))

79. A certain gas, with c

79. A certain gas, with cpp= 0.52= 0.529Btu/ lb9Btu/ lb. °. °R and R = 96R and R = 96.2ft.lb.2ft.lb/lb. °/lb. °R, expanR, expandd s from 5 s from 5 cu ftcu ft

and 80°

and 80°F to 15 cu fF to 15 cu ft while the pressure t while the pressure remains conremains con stant at 15.5psia. Compute for stant at 15.5psia. Compute for TT2.2.

a.

a.15152020°°R R b. b. 16162020°°R R c. c. 17172020°°R R d. d. 18182020°°RR b

b formula: formula: TT22= T= T11 V V22 / V / V11

80.

80. A System haA System has a temperats a temperature of 250°ure of 250°F. Convert tF. Convert t his Value to °his Value to °R?R? a

a. . 747400°°R R bb.7.73030°°R R c. c. 72720°0°R R dd. . 71710°0°RR d

d foformrmulula: a: °°R= R= °°F F + + 464600

81) Steam with a specific volume of 0.09596 m³/kg undergoes a constant pressure 81) Steam with a specific volume of 0.09596 m³/kg undergoes a constant pressure process at 1.70 MPa until the specific volume becomes 0.13796 m³/kg. What are (a) the process at 1.70 MPa until the specific volume becomes 0.13796 m³/kg. What are (a) the final temperature, (b)

final temperature, (b)

�

u, (c) W, (d)u, (c) W, (d)

�

s, and (e) Q?s, and (e) Q? a) 265.4° a) 265.4°C, 430.7kJ/kg, 71.4kJ/kg, C, 430.7kJ/kg, 71.4kJ/kg, 1.0327kJ/(kg)(K)1.0327kJ/(kg)(K) , 502.1 , 502.1 kJ/kgkJ/kg b) 204.2° b) 204.2°C, C, -703.2 kJ/kg, -84.15 kJ/kg, -1.7505 kJ-703.2 kJ/kg, -84.15 kJ/kg, -1.7505 kJ /(kg)(K), -787.4 kJ/kg/(kg)(K), -787.4 kJ/kg c) 304.2° c) 304.2°C, C, -803.2 kJ/kg, -89.15 kJ/kg, -2.7505 kJ-803.2 kJ/kg, -89.15 kJ/kg, -2.7505 kJ /(kg)(K), -987.4 kJ/kg/(kg)(K), -987.4 kJ/kg d) 279.4° d) 279.4°C, 439.7kJ/kg, 79.4kJ/kg, C, 439.7kJ/kg, 79.4kJ/kg, 3.0327kJ/(kg)(K)3.0327kJ/(kg)(K) , 602.1 , 602.1 kJ/kgkJ/kg Ans. (a) Ans. (a) 82) Steam with

82) Steam with an enthalpy of an enthalpy of 2843.5 kJ/kg undergoes a 2843.5 kJ/kg undergoes a constant pressure process atconstant pressure process at 0.9 MPa until the enthalpy becomes 2056.1 kJ/kg. What are (a) the initial temperature 0.9 MPa until the enthalpy becomes 2056.1 kJ/kg. What are (a) the initial temperature or quality, (b)

or quality, (b)

�

u, (c) W, (d)u, (c) W, (d)

�

s, and (e) Q?s, and (e) Q? a) 265.4° a) 265.4°C, 430.7kJ/kg, 71.4kJ/kg, C, 430.7kJ/kg, 71.4kJ/kg, 1.0327kJ/(kg)(K)1.0327kJ/(kg)(K) , 502.1 , 502.1 kJ/kgkJ/kg b) 204.2° b) 204.2°C, C, -703.2 kJ/kg, -84.15 kJ/kg, -1.7505 kJ-703.2 kJ/kg, -84.15 kJ/kg, -1.7505 kJ /(kg)(K), -787.4 kJ/kg/(kg)(K), -787.4 kJ/kg c) 304.2° c) 304.2°C, C, -803.2 kJ/kg, -89.15 kJ/kg, -2.7505 kJ-803.2 kJ/kg, -89.15 kJ/kg, -2.7505 kJ /(kg)(K), -987.4 kJ/kg/(kg)(K), -987.4 kJ/kg d) 279.4° d) 279.4°C, 439.7kJ/kg, 79.4kJ/kg, C, 439.7kJ/kg, 79.4kJ/kg, 3.0327kJ/(kg)(K)3.0327kJ/(kg)(K) , 602.1 , 602.1 kJ/kgkJ/kg Ans. (b) Ans. (b) Formula of #1 and #2:

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83) At throttling calorimeter receives steam from a boiler drum at 0.11 MPa and is 83) At throttling calorimeter receives steam from a boiler drum at 0.11 MPa and is superheated by 10 degrees. If the boiler drum pressure is 1.55 MPa, what is the quality superheated by 10 degrees. If the boiler drum pressure is 1.55 MPa, what is the quality of the steam generated by the boiler?

of the steam generated by the boiler? a) 95.20% a) 95.20% b) 70.10% b) 70.10% c) 65.60% c) 65.60% d) 95.56% d) 95.56% Ans. (a) Ans. (a) Formula: h1 = hf1 + x1hfg1 Formula: h1 = hf1 + x1hfg1 84) A

84) A steam calorimeter steam calorimeter receives steam receives steam from a pfrom a pipe at 0.1 ipe at 0.1 MPa and 20MPa and 20°°SH. For a SH. For a pipepipe steam pressure of 2 MPa, what is the quality of the steam?

steam pressure of 2 MPa, what is the quality of the steam? a) 95.56% a) 95.56% b) 70.10% b) 70.10% c) 95.20% c) 95.20% d) 85.10% d) 85.10% Ans. (a) Ans. (a) Formula: h1 = hf1 + x1hfg1 Formula: h1 = hf1 + x1hfg1

85) A 1-kg steam-water mixture at 1.0 MPa is contained in an inflexible tank. Heat is 85) A 1-kg steam-water mixture at 1.0 MPa is contained in an inflexible tank. Heat is added until the

added until the pressure rises pressure rises to 3.5 MPto 3.5 MPa and the a and the temperature to temperature to 400°400°. Determine the. Determine the heat added. heat added. a) a) 1378.7 1378.7 kJkJ b) b) 1348.5 1348.5 kJkJ c) c) 1278,7 1278,7 kJkJ d) d) 1246,5 1246,5 kJkJ Ans. (a) Ans. (a) Formula: Q = (h2 – p2v2) – (h1 – p1v1) Formula: Q = (h2 – p2v2) – (h1 – p1v1) 86) Water

86) Water vapor at 100 vapor at 100 KPa and 150°KPa and 150°C is compressed C is compressed isothermally until half the isothermally until half the vaporvapor has condensed. How much work must be performed on the steam in this compression has condensed. How much work must be performed on the steam in this compression process per kilogram?

process per kilogram? a) a) -1384.7 -1384.7 kJkJ b) b) 1384.7 1384.7 kJkJ c) c) -2384.7 -2384.7 kJkJ

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d)

d) 2384.7 2384.7 kJkJ Ans. (a)

Ans. (a)

87) Wet steam at 1 MPa flowing through a pipe is throttled to a pressure of 0.1 MPa. If 87) Wet steam at 1 MPa flowing through a pipe is throttled to a pressure of 0.1 MPa. If the throttling temperature

the throttling temperature is 110°is 110°C, What is the C, What is the ququ ality of the steam ality of the steam in the pipe?in the pipe? a) 96% a) 96% b) 86% b) 86% c) 76% c) 76% d) 66% d) 66% Ans. (a) Ans. (a)

88) Steam is throttled to 0.1 MPa with 20 degrees of superheat. (a) What is the quality 88) Steam is throttled to 0.1 MPa with 20 degrees of superheat. (a) What is the quality of throttled steam if its pressure is 0.75 MPa (b) What is the enthalpy of the process? of throttled steam if its pressure is 0.75 MPa (b) What is the enthalpy of the process?

a) a) 97.6%, 97.6%, 2713 2713 kJ/kgkJ/kg b) b) -97.6%, -97.6%, 2713 2713 kJ/kgkJ/kg c) c) 87.6%, 87.6%, 3713 3713 kJ/kgkJ/kg d) d) -87.6%, -87.6%, 3713 3713 kJ/kgkJ/kg Ans. (a) Ans. (a)

89) The pressure gauge on a 2000 m³ tank of oxygen gas reads 600kPa. How much 89) The pressure gauge on a 2000 m³ tank of oxygen gas reads 600kPa. How much volumes will the oxygen occupied at pressure of the outside air 100kPa?

volumes will the oxygen occupied at pressure of the outside air 100kPa? a) a) 14026.5 14026.5 m³m³ b) b) 15026.5 15026.5 m³m³ c) c) 13026.5 13026.5 m³m³ d) d) 16026.5 16026.5 m³m³ Ans. (a)Formula: Ans. (a)Formula: P1V1/T1 P1V1/T1 = = P2V2/T2P2V2/T2

90) Assuming compression is according to the Law PV = C, Calculate the initial volume 90) Assuming compression is according to the Law PV = C, Calculate the initial volume of the gas at a pressure of 2 bars w/c will occupy a volume of 6m³ when it is of the gas at a pressure of 2 bars w/c will occupy a volume of 6m³ when it is compressed to a pressure of 42 Bars.

compressed to a pressure of 42 Bars. a) 130m³ a) 130m³ b) 136m³ b) 136m³ c) 120m³ c) 120m³ d) 126m³ d) 126m³ Ans. (d) Ans. (d) Formula: P1V1/T1 = P2V2/T2 Formula: P1V1/T1 = P2V2/T2

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91) A Gas tank registers 1000kPa. After some gas has been used, the gauge registers 91) A Gas tank registers 1000kPa. After some gas has been used, the gauge registers 500kPa. What percent of the gas remains in the tank?

500kPa. What percent of the gas remains in the tank? a) 64.40% a) 64.40% b) 74.60% b) 74.60% c) 58.40% c) 58.40% d) 54.60% d) 54.60% Ans. (d) Ans. (d)

Formula: Pabs = Patm + Pgage & % = P2/P1 * 100% Formula: Pabs = Patm + Pgage & % = P2/P1 * 100%

92) The volume of a gas under standard atmospheric pressure & 76cmHg is 200m³. 92) The volume of a gas under standard atmospheric pressure & 76cmHg is 200m³. What is the volume when pressure is 80cmHg if the temperature is unchanged?

What is the volume when pressure is 80cmHg if the temperature is unchanged? a) a) 180 180 in³in³ b) b) 170 170 in³in³ c) c) 160 160 in³in³ d) d) 190 190 in³in³ Ans. (d) Ans. (d) Formula: P2V2 = P1V1 Formula: P2V2 = P1V1

93) While swimming at depth of 120m in a fresh water lake, A fish emits an air bubbles 93) While swimming at depth of 120m in a fresh water lake, A fish emits an air bubbles of volume 2.0mm³ atmospheric pressure is 100kPa. What is the pressure of the bubble? of volume 2.0mm³ atmospheric pressure is 100kPa. What is the pressure of the bubble?

a) 217.7kPa a) 217.7kPa b) 317.7kPa b) 317.7kPa c) 417.7kPa c) 417.7kPa d) 517.7kPa d) 517.7kPa Ans. (a) Ans. (a) Formula: P = Formula: P =

δδ

hh

94) How many joules of work is the equivalent of 15000 cal of heat? 94) How many joules of work is the equivalent of 15000 cal of heat?

a) a) 62850 62850 joulesjoules b) b) 3579.95 3579.95 joulesjoules c) c) 14995.81 14995.81 joulesjoules d) d) 15004.19 15004.19 joulesjoules Ans. (a) Ans. (a) Formula: J = Work/Heat Formula: J = Work/Heat

J = mechanical equivalent of heat whose value is 4.19 joules/calorie J = mechanical equivalent of heat whose value is 4.19 joules/calorie

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95) Two thick slices of bread, when completely oxidized by the body, can supply 95) Two thick slices of bread, when completely oxidized by the body, can supply 200,000 cal of heat. How much work is this equivalent to?

200,000 cal of heat. How much work is this equivalent to? a) a) 4,190,000 4,190,000 joulesjoules b) b) 8,390,000 8,390,000 joulesjoules c) c) 839,000 839,000 joulesjoules d) d) 419 419 000 000 joulesjoules Ans. (d) Ans. (d) Formula: J = Work/Heat Formula: J = Work/Heat

J = mechanical equivalent of heat whose value is 4.19 joules/calorie J = mechanical equivalent of heat whose value is 4.19 joules/calorie 96) 3 horsepower (hp) = _____________watts? 96) 3 horsepower (hp) = _____________watts? a) a) 1492 1492 wattswatts b) b) 2238 2238 wattswatts c) c) 746 746 wattswatts d) d) 2238 2238 kilowattskilowatts Ans. (b) Ans. (b) Formula: 1hp = 746 watts Formula: 1hp = 746 watts

97) How many Newton’s (N) in 900,000 dynes? 97) How many Newton’s (N) in 900,000 dynes?

a) a) 8 8 Newton’sNewton’s b) b) 9 9 Newton’sNewton’s c) c) 7 7 Newton’sNewton’s d) d) 6 6 Newton’sNewton’s Ans. (b) Ans. (b)

Formula: 1 Newton (N) = 100,000dynes Formula: 1 Newton (N) = 100,000dynes

98) Calculate the power output in horsepower of an 80-kg man that climbs a flight of 98) Calculate the power output in horsepower of an 80-kg man that climbs a flight of stairs 3.8 m high in 4.0 s. stairs 3.8 m high in 4.0 s. a) a) 744.8 744.8 hphp b) b) 0.998 0.998 hphp c) c) 746 746 hphp d) d) 1.998 1.998 hphp Ans. (b) Ans. (b) Formula: Power = Fd/t = mgh/t Formula: Power = Fd/t = mgh/t F = W = mg F = W = mg d = h d = h

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99) How many calories of heat will be needed to raise the temperature of 200 g of iron 99) How many calories of heat will be needed to raise the temperature of 200 g of iron fr

from om 2727°°C tC to o 8080°°C? C? (c (c = 0= 0.1.11 c1 calal/g. /g. °°C)C) a) 1.16kcal a) 1.16kcal b) 2166cal b) 2166cal c) 3.16kcal c) 3.16kcal d) 4166cal d) 4166cal Ans. (a) Ans. (a) Formula: H = mc Formula: H = mc

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TT 100) 1

100) 100g of ir00g of iron was hon was heated teated to 100°o 100°C and mixeC and mixed witd wit h 22g of h 22g of water at 4water at 40°0°C. The finC. The finalal temperature of

temperature of the mixture was 6the mixture was 60°0°C. Show that the C. Show that the heat given off heat given off by the iron by the iron equalsequals the heat absorbed by the water.

the heat absorbed by the water. a) a) 440 440 calcal b) b) 540 540 calcal c) c) 340 340 calcal d) d) 640 640 calcal Ans. (a) Ans. (a)

Formula: H (given off by iron) = H (absorbed by water), Formula: H (given off by iron) = H (absorbed by water),

mc