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)

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)
by

1 Answer

Answer :

-420Btu
by

Related questions

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

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

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

Description : 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 ... (Formula: T2= T1p2/ p1)  a. 999 K  b. 888 K  c. 456 K  d. One of the above

Last Answer : 999 K

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 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 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

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

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

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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)

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

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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

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

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Last Answer : 14026.5 m³

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

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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)

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

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

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Find the change in internal energy of 5 lb. of oxygen gas when the temperature changes from 100 ˚F to 120 ˚F. CV = 0.157 BTU/lbm-˚R  A.14.7 BTU  B.15.7 BTU  C. 16.8 BTU  D. 15.9 BTU Formula: U= mcv T

Description : Find the change in internal energy of 5 lb. of oxygen gas when the temperature changes from 100 ˚F to 120 ˚F. CV = 0.157 BTU/lbm-˚R  A.14.7 BTU  B.15.7 BTU  C. 16.8 BTU  D. 15.9 BTU Formula: U= mcv T

Last Answer : 15.7 BTU

In a Carnot cycle, heat is transferred at  (a) constant pressure  (b) constant volume  (c) constant temperature  (d) constant enthaply  (e) any one of the above.

Description : In a Carnot cycle, heat is transferred at  (a) constant pressure  (b) constant volume  (c) constant temperature  (d) constant enthaply  (e) any one of the above.

Last Answer : Answer : c

Change in internal energy in a closed system is equal to heat transferred if the reversible process takes place at constant  (a) pressure  (b) temperature  (c) volume  (d) internal energy  (e) entropy.

Description : Change in internal energy in a closed system is equal to heat transferred if the reversible process takes place at constant  (a) pressure  (b) temperature  (c) volume  (d) internal energy  (e) entropy.

Last Answer : Answer : c

Change in enthalpy in a closed system is equal to heat transferred if the reversible process takes place at constant  (a) pressure  (b) temperature  (c) volume  (d) internal energy  (e) entropy.

Description : Change in enthalpy in a closed system is equal to heat transferred if the reversible process takes place at constant  (a) pressure  (b) temperature  (c) volume  (d) internal energy  (e) entropy.

Last Answer : Answer : a

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Last Answer : 24.9 ft^3

If air is at pressure, p, of 3200 lbf/ft2 , 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.)  A.9.8 ft^3/lbm  B.11.2 ft^3/lbm  C.13.33 ft^3/lbm  D.14.2 ft^3/lbm Formula: pv = RT v = RT / p

Description : If air is at pressure, p, of 3200 lbf/ft2 , 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.)  A.9.8 ft^3/lbm  B.11.2 ft^3/lbm  C.13.33 ft^3/lbm  D.14.2 ft^3/lbm Formula: pv = RT v = RT / p

Last Answer : 13.33 ft^3/lbm

The gas in a constant gas thermometer cooled to absolute zero would have _________.  a. no volume  b. no pressure  c. zero temperature at all scales  d. none of the above

Description : The gas in a constant gas thermometer cooled to absolute zero would have _________.  a. no volume  b. no pressure  c. zero temperature at all scales  d. none of the above

Last Answer : no pressure

A problem Drum ( 3 ft. diameter ; 6 ft. height ) is field with a fluid whose density is 50 lb/ft^3. Determine the total volume of the fluid.  A. 42.41 ft^3  B.44.35 ft^3  C.45.63 ft^3  D.41.23 ft^3 Formula: Vf = (pi d^2 h) / 4

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Last Answer : 42.41 ft^3

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.  A.2500 lbf-ft/lbm  B.3300 lbf-ft/lbm  C.5400 lbf-ft/lbm  D.6900 lbf-ft/lbm Formula: h= u+ pV u= h– pV

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Last Answer : 3300 lbf-ft/lbm

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According to first law of thermodynamics  (a) work done by a system is equal to heat transferred by the system  (b) total internal energy of a system during a process remains constant  (c) internal energy, enthalpy and entropy during a process remain constant  (d) total energy of a system remains constant

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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

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The value for the ΔU of a system is -120 J. If the system is known to have absorbed 420 J of heat, how much work was done?  a. -540 J  b. -640 J  c. -740 J  d. -840 J formula: ΔU = q +w

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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

The flow energy of 5 ft3 of a fluid passing a boundary to a system is 80,000 ft-lb. Determine the pressure at this point.  a. 222 psi  b. 333 psi  c. 444 psi  d. 111 psi formula: Ef= pV

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Last Answer : 111 psi

A vertical column of water will be supported to what height by standard atmospheric pressure. If the Y w = 62.4lb/ft3 po = 14.7 psi.  a. 44.9 ft  b. 33.9 ft  c. 22.9 ft  d. 55.9 ft formula: ho= po/Yw

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Last Answer : 33.9 ft

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)

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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 refers to the amount of heat removed from the cooled space in BTS’s for 1 watt-hour of electricity consumed?  A. Cost efficiency rating  B. Energy efficiency rating  C. Coefficient of performance  D. Cost of performance

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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 door of a running refrigerator inside a room was left open. Which of the following statements is correct?  (a) The room will be cooled to the temperature inside the refrigerator.  (b) The room will be cooled very slightly.  (c) The room will be gradually warmed up.  (d) The temperature of the air in room will remain unaffected.  (e) any one of above is possible depending on the capacity.

Description : The door of a running refrigerator inside a room was left open. Which of the following statements is correct?  (a) The room will be cooled to the temperature inside the refrigerator.  (b) The room ... air in room will remain unaffected.  (e) any one of above is possible depending on the capacity.

Last Answer : Answer : c

The theory that heat consisted of a fluid, which could be transferred from one body to another, but not “created” or “destroyed”.  a. Clausius Theorem  b. Caloric Theory  c. Joules Law  d. Newton’s Law of cooling

Description : The theory that heat consisted of a fluid, which could be transferred from one body to another, but not “created” or “destroyed”.  a. Clausius Theorem  b. Caloric Theory  c. Joules Law  d. Newton’s Law of cooling

Last Answer : Caloric Theory

A process by which heat is transferred through a material without a bulk movement of the material.  a. Convection  b. Conduction  c. Radiation  d. Emission

Description : A process by which heat is transferred through a material without a bulk movement of the material.  a. Convection  b. Conduction  c. Radiation  d. Emission

Last Answer : Conduction

The flow of a fluid when heat is transferred by convection.  a. placidity  b. mass flow  c. convection current  d. heat transfer

Description : The flow of a fluid when heat is transferred by convection.  a. placidity  b. mass flow  c. convection current  d. heat transfer

Last Answer : convection current

A process in which heat energy is transferred by the flow of fluid.  a. Convection  b. Conduction  c. Radiation  d. Sublimation

Description : A process in which heat energy is transferred by the flow of fluid.  a. Convection  b. Conduction  c. Radiation  d. Sublimation

Last Answer : Convection

Entropy is transferred by ______.  A. Work  B. Heat  C. Energy  D. Work and heat

Description : Entropy is transferred by ______.  A. Work  B. Heat  C. Energy  D. Work and heat

Last Answer : Heat

How is heat transferred?  A. By conduction  B. By convection  C. By radiation  D. All of the above

Description : How is heat transferred?  A. By conduction  B. By convection  C. By radiation  D. All of the above

Last Answer : All of the above

If there is no heat transferred during the process, it is called a ______ process.  A. Static  B. Isobaric  C. Polytropic  D. Adiabatic

Description : If there is no heat transferred during the process, it is called a ______ process.  A. Static  B. Isobaric  C. Polytropic  D. Adiabatic

Last Answer : Adiabatic

What is an energy that can be transferred from one object to another causing a change in temperature of each object?  A. Power  B. Heat transfer  C. Heat  D. Work

Description : What is an energy that can be transferred from one object to another causing a change in temperature of each object?  A. Power  B. Heat transfer  C. Heat  D. Work

Last Answer : Heat

Solids and liquids have  (a) one value of specific heat (ft) two values of specific heat  (c) three values of specific heat  (d) no value of specific heat  (e) one value under some conditions and two values under other conditions.

Description : Solids and liquids have  (a) one value of specific heat (ft) two values of specific heat  (c) three values of specific heat  (d) no value of specific heat  (e) one value under some conditions and two values under other conditions.

Last Answer : Answer : a