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 p1 = 551.6 kPa and t1 = 60°C to p2 = 1655 kPa. During the process the gas is internally stirred and there are also added 105.5 kJ of heat. Determine t2. (Formula: T2= T1p2/ p1)  a. 999 K  b. 888 K  c. 456 K  d. One of the above

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 p1 = 551.6 kPa and t1 =
60°C to p2 = 1655 kPa. During the
process the gas is internally stirred and
there are also added 105.5 kJ of heat.
Determine t2.
(Formula: T2= T1p2/ p1)
 a. 999 K
 b. 888 K
 c. 456 K
 d. One of the above
by

1 Answer

Answer :

999 K
by

Related questions

What is the resulting pressure when one pound of air at 15 psia and 200 ˚F is heated at constant volume to 800 ˚F?  A.15 psia  B. 28.6 psia  C. 36.4 psia.  D. 52.1 psia Formula : T1/p1 = T2/p2 p2= p1T2 / T1

Description : What is the resulting pressure when one pound of air at 15 psia and 200 ˚F is heated at constant volume to 800 ˚F?  A.15 psia  B. 28.6 psia  C. 36.4 psia.  D. 52.1 psia Formula : T1/p1 = T2/p2 p2= p1T2 / T1

Last Answer : 28.6 psia

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 this gas at constant pressure when the initial temp is 32.2°C? Find T2.  a. 339.4 K  b. 449.4 K  c. 559.4K  d. 669.4K formula: cp = kR/ k-1 Q= mcp(T2-T1)

Description : 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 this gas at constant pressure when the initial temp is 32.2°C? Find T2.  a. 339.4 K  b. 449.4 K  c. 559.4K  d. 669.4K formula: cp = kR/ k-1 Q= mcp(T2-T1)

Last Answer : 339.4 K

Which of the following is the Ideal gas law (equation)?  A. V/T = K  B. V= k*(1/P)  C. P1/T1 = P2/T2  D. PV = nRT

Description : Which of the following is the Ideal gas law (equation)?  A. V/T = K  B. V= k*(1/P)  C. P1/T1 = P2/T2  D. PV = nRT

Last Answer : PV = nRT

Which of the following is the mathematical representation of the Charles’s law?  A. V1/V2= P2/P1  B. V1/T1=V2/T2  C. V1/T2=V2/T1  D. V1/V2=√P2/√P1

Description : Which of the following is the mathematical representation of the Charles’s law?  A. V1/V2= P2/P1  B. V1/T1=V2/T2  C. V1/T2=V2/T1  D. V1/V2=√P2/√P1

Last Answer : V1/T1=V2/T2

From the steam table, determine the average constant pressure specific heat (c) of steam at 10 kPa and45.8 ˚C  A.1.79 kJ/ kg-˚C  B.10.28 kJ/ kg-˚C  C.30.57 kJ/ kg-˚C  D. 100.1 kJ/ kg-˚C Formula: h = c T ∆ ∆ From the steam table At 47.7 ˚C h= 2588.1 kJ/ kg At 43.8 ˚C h= 2581.1 kJ/ kg

Description : From the steam table, determine the average constant pressure specific heat (c) of steam at 10 kPa and45.8 ˚C  A.1.79 kJ/ kg-˚C  B.10.28 kJ/ kg-˚C  C.30.57 kJ/ kg-˚C  D. 100.1 kJ/ kg-˚C Formula: h = c T ∆ ∆ From the steam table At 47.7 ˚C h= 2588.1 kJ/ kg At 43.8 ˚C h= 2581.1 kJ/ kg

Last Answer : 1.79 kJ/ kg-˚C

The pressure gauge on a 2000 m³ tank of oxygen gas reads 600 kPa. How much volumes will the oxygen occupied at pressure of the outside air 100 kPa?  a) 14026.5 m³  b) 15026.5 m³  c) 13026.5 m³  d) 16026.5 m³ Formula: P1V1/T1 =P2V2/T2

Description : The pressure gauge on a 2000 m³ tank of oxygen gas reads 600 kPa. How much volumes will the oxygen occupied at pressure of the outside air 100 kPa?  a) 14026.5 m³  b) 15026.5 m³  c) 13026.5 m³  d) 16026.5 m³ Formula: P1V1/T1 =P2V2/T2

Last Answer : 14026.5 m³

The heat supplied to the gaS at constant volume is (where m = Mass of gas, cv = Specific heat at constant volume, cp = Specific heat at constant pressure, T2 – T1 = Rise in temperature, and R = Gas constant)  A. mR(T2 – T1)  B. mcv(T2 – T1)  C. mcp(T2 – T1)  D. mcp(T2 + T1)

Description : The heat supplied to the gaS at constant volume is (where m = Mass of gas, cv = Specific heat at constant volume, cp = Specific heat at constant pressure, T2 – T1 = Rise in temperature, and R = Gas constant)  A. mR(T2 – T1)  B. mcv(T2 – T1)  C. mcp(T2 – T1)  D. mcp(T2 + T1)

Last Answer : Answer: B

To obtain integrated form of Clausius-Clapeyron equation, ln (P2/P1) = (∆HV/R) (1/T1- 1/T2) from the exact Clapeyron equation, it is assumed that the (A) Volume of the liquid phase is negligible compared to that of vapour phase (B) Vapour phase behaves as an ideal gas (C) Heat of vaporisation is independent of temperature (D) All (A), (B) & (C)

Description : To obtain integrated form of Clausius-Clapeyron equation, ln (P2/P1) = (∆HV/R) (1/T1- 1/T2) from the exact Clapeyron equation, it is assumed that the (A) Volume of the liquid phase is negligible compared to ... gas (C) Heat of vaporisation is independent of temperature (D) All (A), (B) & (C)

Last Answer : (D) All (A), (B) & (C)

Ten cu. ft of air at 300psia and 400°F is cooled to 140°F at constant volume. What is the transferred heat?  a.-120Btu  b. -220Btu  c.-320Btu  d. -420Btu formula: Q= mcv(T2-T1)

Description : Ten cu. ft of air at 300psia and 400°F is cooled to 140°F at constant volume. What is the transferred heat?  a.-120Btu  b. -220Btu  c.-320Btu  d. -420Btu formula: Q= mcv(T2-T1)

Last Answer : -420Btu

Ten cu ft. of air at 300 psia 400°F is cooled to 140°F at constant volume. What is the final pressure? (formula: p2 = p1T2/T1)  a. 0  b. 209 psia  c. - 420 psia  d. None of the above

Description : Ten cu ft. of air at 300 psia 400°F is cooled to 140°F at constant volume. What is the final pressure? (formula: p2 = p1T2/T1)  a. 0  b. 209 psia  c. - 420 psia  d. None of the above

Last Answer : 209 psia

A 1-kg steam-water mixture at 1.0 MPa is contained in an inflexible tank. Heat is added until the pressure rises to 3.5 MPa and the temperature to 400°. Determine the heat added.  a) 1378.7 kJ  b) 1348.5 kJ  c) 1278,7 kJ  d) 1246,5 kJ Formula: Q = (h2 – p2v2) –(h1 –p1v1)

Description : A 1-kg steam-water mixture at 1.0 MPa is contained in an inflexible tank. Heat is added until the pressure rises to 3.5 MPa and the temperature to 400°. Determine the heat added.  a) 1378.7 kJ  b) 1348.5 kJ  c) 1278,7 kJ  d) 1246,5 kJ Formula: Q = (h2 – p2v2) –(h1 –p1v1)

Last Answer : 1378.7 kJ

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.  A. 5 x10^5 kJ  B. 8x10^5 kJ  C. 1 x10^6 kJ  D. 2 x10^6 kJ Formula: fromthe steamtable vƒ = 0.001043 m^3 / kg vg = 1.6940 m^3 / kg u ƒ= 417.3 kJ/kg ug= 2506kJ/kg formula: Mvap = V vap/vg M liq = Vliq/ vƒ u =uƒM liq + ug M vap

Description : 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.  A. 5 x10^5 kJ  B. 8x10^5 kJ  C. 1 ... 3 kJ/kg ug= 2506kJ/kg formula: Mvap = V vap/vg M liq = Vliq/ vƒ u =uƒM liq + ug M vap

Last Answer : 2 x10^6 kJ

For a certain gas R = 320 J/kg.K and cv= 0.84kJ/kg.K. Find k?  a. 1.36  b. 1.37  c. 1.38  d. 1.39 formula: k= R / cv+1

Description : For a certain gas R = 320 J/kg.K and cv= 0.84kJ/kg.K. Find k?  a. 1.36  b. 1.37  c. 1.38  d. 1.39 formula: k= R / cv+1

Last Answer : 1.38

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

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

Last Answer : 126m³

A gas is enclosed in a cylinder with a weighted piston as the top boundary. The gas is heated and expands from a volume of 0.04 m3 to 0.10 m3 at a constant pressure of 200 kPa. Find the work done on the system.  a. 5 kJ  b. 15 kJ  c. 10 kJ  d. 12 kJ

Description : A gas is enclosed in a cylinder with a weighted piston as the top boundary. The gas is heated and expands from a volume of 0.04 m3 to 0.10 m3 at a constant pressure of 200 kPa. Find the work done on the system.  a. 5 kJ  b. 15 kJ  c. 10 kJ  d. 12 kJ

Last Answer : 12 kJ

A certain gas with cp = 0.529Btu/lb°R and R = 96.2ft/lbºR expands from 5 ft and 80ºF to 15 ft while the pressure remains constant at 15.5 psia.  a. T2=1.620ºR, ∫H = 122.83 Btu  b. T2 = 2°R, ∫H = 122.83 Btu  c. T2 = 2.620ºR, ∫H = 122.83 Btu  d. T2 = 1°R, ∫H = 122.83 Btu T2= V2(t2)/V1 and ∫H = mcp (T2-T1)

Description : A certain gas with cp = 0.529Btu/lb°R and R = 96.2ft/lbºR expands from 5 ft and 80ºF to 15 ft while the pressure remains constant at 15.5 psia.  a. T2=1.620ºR, ∫H = 122.83 Btu  b. T2 = 2°R, ∫H = 122.83 Btu  c. ... , ∫H = 122.83 Btu  d. T2 = 1°R, ∫H = 122.83 Btu T2= V2(t2)/V1 and ∫H = mcp (T2-T1)

Last Answer : T2=1.620ºR, ∫H = 122.83 Btu

A gas having a volume of100 ft³ at 27ºC is expanded to 120 ft³by heated at constant pressure to what temperature has it been heated to have this new volume?  a. 87°C  b. 85°C  c. 76°C  d. 97°C t2= T2–T1

Description : A gas having a volume of100 ft³ at 27ºC is expanded to 120 ft³by heated at constant pressure to what temperature has it been heated to have this new volume?  a. 87°C  b. 85°C  c. 76°C  d. 97°C t2= T2–T1

Last Answer : 87°C

The volume of the gas held at constant pressure increases 4 cm² at 0°C to 5cm². What is the final pressure?  a. 68.65ºC  b. 68.25ºC  c. 70.01°C  d. 79.1ºC t2= T2–T1

Description : The volume of the gas held at constant pressure increases 4 cm² at 0°C to 5cm². What is the final pressure?  a. 68.65ºC  b. 68.25ºC  c. 70.01°C  d. 79.1ºC t2= T2–T1

Last Answer : 981 N

What is the equation for the work done by a constant temperature system?  A. W = mRTln(V2-V1)  B. W = mR( T2-T1 ) ln( V2/V1)  C. W = mRTln (V2/V1)  D. W = RT ln (V2/V1) Formula : W=∫ pdV lim1,2 ∫ = mRT / V

Description : What is the equation for the work done by a constant temperature system?  A. W = mRTln(V2-V1)  B. W = mR( T2-T1 ) ln( V2/V1)  C. W = mRTln (V2/V1)  D. W = RT ln (V2/V1) Formula : W=∫ pdV lim1,2 ∫ = mRT / V

Last Answer : W = mRTln (V2/V1)

What horse power is required to isothermally compress 800 ft^3 of Air per minute from 14.7 psia to 120 psia?  A. 28 hp  B.108 hp  C.256 hp  D.13900 hp Formula: W= p1V1 ln (p1/p2) Power = dW / dt

Description : What horse power is required to isothermally compress 800 ft^3 of Air per minute from 14.7 psia to 120 psia?  A. 28 hp  B.108 hp  C.256 hp  D.13900 hp Formula: W= p1V1 ln (p1/p2) Power = dW / dt

Last Answer : 108 hp

Water (specific heat cv= 4.2 kJ/ kg ∙ K ) is being heated by a 1500 W h eater. What is the rate of change in temperature of 1kg of the water?  A. 0.043 K/s  B. 0.179 K/s  C. 0.357 K/s  D. 1.50 K/s Formula: Q = mcv ( T)

Description : Water (specific heat cv= 4.2 kJ/ kg ∙ K ) is being heated by a 1500 W h eater. What is the rate of change in temperature of 1kg of the water?  A. 0.043 K/s  B. 0.179 K/s  C. 0.357 K/s  D. 1.50 K/s Formula: Q = mcv ( T)

Last Answer : 0.179 K/s

An ideal gas at 45psig and 80ºF is heated in the close container to 130ºF. What is the final pressure?  a. 65.10 psi  b. 65.11 psi  c. 65.23 psi  d. 61.16 psi P1V1/T1= P2V2/T2;V = Constant

Description : An ideal gas at 45psig and 80ºF is heated in the close container to 130ºF. What is the final pressure?  a. 65.10 psi  b. 65.11 psi  c. 65.23 psi  d. 61.16 psi P1V1/T1= P2V2/T2;V = Constant

Last Answer : 65.23 psi

The flow energy of 150 L of a fluid passing a boundary to a system is 110 kJ. Determine the pressure at this point  a. 733.33 kPa  b. 833.33 kPa  c. 933.33 kPa  d. 633.33 kPa

Description : The flow energy of 150 L of a fluid passing a boundary to a system is 110 kJ. Determine the pressure at this point  a. 733.33 kPa  b. 833.33 kPa  c. 933.33 kPa  d. 633.33 kPa

Last Answer : 733.33 kPa

A 30-m vertical column of fluid (density 1878 kg/m3 ) is located where g= 9.65 mps2 . Find the pressure at the base of the column. (Formula: pg= gρhg/k )  a. 543680 N/m2  b. 543.68 kPa (gauge)  c. Both a & b  d. None of the above

Description : A 30-m vertical column of fluid (density 1878 kg/m3 ) is located where g= 9.65 mps2 . Find the pressure at the base of the column. (Formula: pg= gρhg/k )  a. 543680 N/m2  b. 543.68 kPa (gauge)  c. Both a & b  d. None of the above

Last Answer : Both a & b

A certain gas, with cp = 0.529Btu/ lb. °Rand R = 96.2ft.lb/lb. °R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5psia. Compute for T2.  a.1520°R  b. 1620°R  c. 1720°R  d. 1820°R formula: T2= T1V2/V1

Description : A certain gas, with cp = 0.529Btu/ lb. °Rand R = 96.2ft.lb/lb. °R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5psia. Compute for T2.  a.1520°R  b. 1620°R  c. 1720°R  d. 1820°R formula: T2= T1V2/V1

Last Answer : 1620°R

A certain gas, with cp = 0.529Btu/lb.°R and R = 96.2 ft.lb/lb.°R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5 psia. Compute for T2. (Formula: T2= T1V2/V1)  a. 460°R  b. 270°R  c. 1620 °R  d. None of the above

Description : A certain gas, with cp = 0.529Btu/lb.°R and R = 96.2 ft.lb/lb.°R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5 psia. Compute for T2. (Formula: T2= T1V2/V1)  a. 460°R  b. 270°R  c. 1620 °R  d. None of the above

Last Answer : 1620 °R

A vessel with a volume of cubic meter contains liquid water and water vapor ion equilibrium at 600 kPa. The liquid water has mass of1kg. Using the steam table, calculate the mass of the water vapor.  A. 0.99kg  B. 1.57 kg  C. 2.54 kg  D. 3.16 kg Formula: from the steam table at 600kPa vƒ = 0.001101 m^3 / kg vg = 0.3157 m^3 / kg Vtot = mƒ vƒ + mg vg mg = (tot-mƒ vƒ) / vg

Description : A vessel with a volume of cubic meter contains liquid water and water vapor ion equilibrium at 600 kPa. The liquid water has mass of1kg. Using the steam table, calculate the mass of the water vapor.  A. 0.99kg  B. 1.57 kg  C ... / kg vg = 0.3157 m^3 / kg Vtot = mƒ vƒ + mg vg mg = (tot-mƒ vƒ) / vg

Last Answer : 3.16 kg

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

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

Last Answer : 12 kJ

The combined mass of car and passengers travelling at 72 km/hr is 1500 kg. Find the kinetic energy of this combined mass. (Formula: K =mv2 / 2k )  a. 300 kJ  b. 200 kJ  c. 500 kJ  d. None of the above

Description : The combined mass of car and passengers travelling at 72 km/hr is 1500 kg. Find the kinetic energy of this combined mass. (Formula: K =mv2 / 2k )  a. 300 kJ  b. 200 kJ  c. 500 kJ  d. None of the above

Last Answer : 300 kJ

For heat engine operating between two temperatures (T1>T2), what is the maximum efficiency attainable?  A. Eff = 1 – (T2/T1)  B. Eff = 1 - (T1/T2)  C. Eff = T1 - T2  D. Eff = 1 - (T2/T1)^2

Description : For heat engine operating between two temperatures (T1>T2), what is the maximum efficiency attainable?  A. Eff = 1 – (T2/T1)  B. Eff = 1 - (T1/T2)  C. Eff = T1 - T2  D. Eff = 1 - (T2/T1)^2

Last Answer : Eff = 1 – (T2/T1)

Water vapor at 100 KPa and 150°C is compressed isothermally until half the vapor has condensed. How much work must be performed on the steam in this compression process per kilogram?  a) -1384.7 kJ  b) 1384.7 kJ  c) -2384.7 kJ  d) 2384.7 kJ

Description : Water vapor at 100 KPa and 150°C is compressed isothermally until half the vapor has condensed. How much work must be performed on the steam in this compression process per kilogram?  a) -1384.7 kJ  b) 1384.7 kJ  c) -2384.7 kJ  d) 2384.7 kJ

Last Answer : -1384.7 kJ

A gas is compressed in a cylinder by a movable piston to a volume onehalf its original volume. During the process 300 kJ heat left the gas and internal energy remained same. The work done on gas in Nm will be  (a) 300 Nm  (b) 300,000 Nm  (c) 30 Nm  (d) 3000 Nm  (e) 30,000 Nm.

Description : A gas is compressed in a cylinder by a movable piston to a volume onehalf its original volume. During the process 300 kJ heat left the gas and internal energy remained same. The work done on gas in Nm will be  (a) 300 Nm  (b) 300,000 Nm  (c) 30 Nm  (d) 3000 Nm  (e) 30,000 Nm.

Last Answer : Answer : b

Specific heat capacity in SI unit.  a. kJ / kg.k  b. kJ / kg  c. kN / kg  d. None of the above

Description : Specific heat capacity in SI unit.  a. kJ / kg.k  b. kJ / kg  c. kN / kg  d. None of the above

Last Answer : kJ / kg.k

A 0.064 kg of octane vapor (MW = 114) is mixed with0.91 kg of air (MW = 29.0) in the manifold of an Engine. The total pressure in the manifold is 86.1 kPa, and a temperature is 290 K. assume octane behaves ideally. What is the partial pressure of the air in the mixture in KPa?  a. 46.8  b. 48.6  c. 84.6  d. 64.8

Description : A 0.064 kg of octane vapor (MW = 114) is mixed with0.91 kg of air (MW = 29.0) in the manifold of an Engine. The total pressure in the manifold is 86.1 kPa, and a temperature is 290 K. assume octane behaves ideally. What is ... of the air in the mixture in KPa?  a. 46.8  b. 48.6  c. 84.6  d. 64.8

Last Answer : 84.6

What is the value of the universal gas constant in kJ/kmol K? ∙  A. 10.73  B. 1.986  C. 8.314  D. 1545

Description : What is the value of the universal gas constant in kJ/kmol K? ∙  A. 10.73  B. 1.986  C. 8.314  D. 1545

Last Answer : 8.314

Utilizing the answer to the previous problem, estimate the overall or average increase in temperature ( ΔT) of the concrete roof from the energy absorbed from the sun during a12hour 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 kg/m3 .  a. 7.9 °C  b. 8.9°C  c. 9.9°C  d. 10.9°C formula: ΔQ = m c ΔT

Description : Utilizing the answer to the previous problem, estimate the overall or average increase in temperature ( ΔT) of the concrete roof from the energy absorbed from the sun during a12hour day. Assume that all of the radiation absorbed goes into ... °C  b. 8.9°C  c. 9.9°C  d. 10.9°C formula: ΔQ = m c ΔT

Last Answer : 7.9 °C

What is the latent heat of fusion of water at 1 atm?  A. 331.1 kJ/kg  B. 332.6 kJ/kg  C. 333.7 kJ/kg  D. 330.7 kJ/kg

Description : What is the latent heat of fusion of water at 1 atm?  A. 331.1 kJ/kg  B. 332.6 kJ/kg  C. 333.7 kJ/kg  D. 330.7 kJ/kg

Last Answer : 333.7 kJ/kg

While swimming at depth of120m 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?  a) 217.7 kPa  b) 317.7 kPa  c) 417.7 kPa  d) 517.7 kPa Formula: P= δh

Description : While swimming at depth of120m 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?  a) 217.7 kPa  b) 317.7 kPa  c) 417.7 kPa  d) 517.7 kPa Formula: P= δh

Last Answer : 217.7 kPa

A certain fluid is flowing in a 0.5m x 0.3 channel at the rate of 3 m/s and has a specific volume of 0.012 m³/kg. Determined the mass of water flowing in kg/s.  a. 267 kg/s  b. 378 kg/s  c. 375 kg/s  d. 456.5 kg/s m = Aν/V

Description : A certain fluid is flowing in a 0.5m x 0.3 channel at the rate of 3 m/s and has a specific volume of 0.012 m³/kg. Determined the mass of water flowing in kg/s.  a. 267 kg/s  b. 378 kg/s  c. 375 kg/s  d. 456.5 kg/s m = Aν/V

Last Answer : 375 kg/s

A pressure gage registers 50 psig in a region where the barometer is 14.25 psia. Find absolute pressure in psia, Pa. (Formula; p = patm+ pg)  a. 433 kPa  b. 443 kPa  c. 343 kPa  d. None of the above

Description : A pressure gage registers 50 psig in a region where the barometer is 14.25 psia. Find absolute pressure in psia, Pa. (Formula; p = patm+ pg)  a. 433 kPa  b. 443 kPa  c. 343 kPa  d. None of the above

Last Answer : 443 kPa

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.  A.824 cal  B.924 cal  C.944 cal  D.1124 cal Formula: W = -mRTln (V2/V1) Where R = (1.98 cal/gmole·K) (32 g/gmole)

Description : 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.  A.824 cal  B.924 cal  C.944 cal  D.1124 cal Formula: W = -mRTln (V2/V1) Where R = (1.98 cal/gmole·K) (32 g/gmole)

Last Answer : 1124 cal

What is the latent heat of vaporization of water at 1 atm?  A. 2314.8 kJ/kg  B. 2257.1 kJ/kg  C. 2511.7 kJ/kg  D. 2429.8 kJ/kg

Description : What is the latent heat of vaporization of water at 1 atm?  A. 2314.8 kJ/kg  B. 2257.1 kJ/kg  C. 2511.7 kJ/kg  D. 2429.8 kJ/kg

Last Answer : 2257.1 kJ/kg

The specific heat at constant volume is  A. the amount of heat required to raise the temperature of unit mass of gas through one degree, at constant pressure  B. the amount of heat required to raise the temperature of unit mass of gas through one degree, at constant volume  C. the amount of heat required to raise the temperature of 1 kg of water through one degree  D. any one of the above

Description : The specific heat at constant volume is  A. the amount of heat required to raise the temperature of unit mass of gas through one degree, at constant pressure  B. the amount of heat required to raise ... to raise the temperature of 1 kg of water through one degree  D. any one of the above

Last Answer : Answer: B

Oxygen at 15ºC and 10.3 Mpa gauge pressure occupies 600L. What is the occupied by the oxygen at 8.28 Mpa gauge pressure and 35ºC?  a. 789.32 L  b. 796.32 L  c. 699 L  d. 588 L V2= P1V1/T1P2

Description : Oxygen at 15ºC and 10.3 Mpa gauge pressure occupies 600L. What is the occupied by the oxygen at 8.28 Mpa gauge pressure and 35ºC?  a. 789.32 L  b. 796.32 L  c. 699 L  d. 588 L V2= P1V1/T1P2

Last Answer : 796.32 L

Let a closed system execute a state change for which the heat is Q = 100 J and work is W = -25 J. Find E. ∆ (Formula: E = Q- W) ∆  a. 125 J  b. 123 J  c. 126 J  d. None of the above

Description : Let a closed system execute a state change for which the heat is Q = 100 J and work is W = -25 J. Find E. ∆ (Formula: E = Q- W) ∆  a. 125 J  b. 123 J  c. 126 J  d. None of the above

Last Answer : 125 J

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?  a) 97.6%,2713 kJ/kg  b) -97.6%, 2713 kJ/kg  c) 87.6%,3713 kJ/kg  d) -87.6%, 3713 kJ/kg

Description : 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?  a) 97.6%,2713 kJ/kg  b) -97.6%, 2713 kJ/kg  c) 87.6%,3713 kJ/kg  d) -87.6%, 3713 kJ/kg

Last Answer : 97.6%,2713 kJ/kg

The Product of the displacement of the body and the component if the force in the direction of the displacement.  a. Nonflow Work  b. Flow Work  c. Work  d. None of the above

Description : The Product of the displacement of the body and the component if the force in the direction of the displacement.  a. Nonflow Work  b. Flow Work  c. Work  d. None of the above

Last Answer : Work

A classification of system in which mass does not cross its boundaries.  a. Nonflow System  b. Open System  c. Closed System  d. None of the above

Description : A classification of system in which mass does not cross its boundaries.  a. Nonflow System  b. Open System  c. Closed System  d. None of the above

Last Answer : Closed System

First law of thermodynamics  (a) enables to determine change in internal energy of the system  (b) does not help to predict whether the system will or not undergo a change  (c) does not enable to determine change in entropy  (d) provides relationship between heat, work and internal energy  (e) all of the above.

Description : First law of thermodynamics  (a) enables to determine change in internal energy of the system  (b) does not help to predict whether the system will or not undergo a change  (c) does not enable ... entropy  (d) provides relationship between heat, work and internal energy  (e) all of the above.

Last Answer : Answer : e

In a constant pressure process steam is generated from 10 bar and 0.8 dry condition till it become dry and saturated. Determine amount of heat added per kg of steam. From steam table at 10 bar Tsat = 179.9°C hf = 762.6 kl/kg, hg = 2776.2 kJ/kg

Description : In a constant pressure process steam is generated from 10 bar and 0.8 dry condition till it become dry and saturated. Determine amount of heat added per kg of steam. From steam table at 10 bar Tsat = 179.9°C hf = 762.6 kl/kg, hg = 2776.2 kJ/kg 

Last Answer : Constant pressure process Final condition of steam is dry saturated Enthalpy of dry steam = hg= hf + hfg =2776.2 KJ/Kg .Given Latent heat of vaporization = hfg = hg - hf =2776.2-762.6 =2013.6 KJ/Kg Initial condition ... hfg = 762.6+0.8 X 2013.6 =2373.48 KJ/Kg Heat added = hg-hw = 402.72 KJ/kg