How many kilocalories of heat are required to heat 750 g of water from 35˚C to 55˚C.  a. 15  b. 1500  c. 1.5 x 10^4  d. 6.3 x 10^4

How many kilocalories of heat are required to heat 750 g of water from 35˚C to 55˚C.
 a. 15
 b. 1500
 c. 1.5 x 10^4
 d. 6.3 x 10^4
by

1 Answer

Answer :

15
by

Related questions

Two 12 ft. sections of aluminum siding are placed end to end on the outside wall of house. How large a gap should be left between the pieces to prevent buckling if the temperature can change by 55˚C?  a. 0.21 m  b. 0.18 m  c. 0.31 in  d. 0.18 in

Description : Two 12 ft. sections of aluminum siding are placed end to end on the outside wall of house. How large a gap should be left between the pieces to prevent buckling if the temperature can change by 55˚C?  a. 0.21 m  b. 0.18 m  c. 0.31 in  d. 0.18 in

Last Answer : 0.18 in

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

If a system absorbs 500 cal of heat at the same time does 400J of work, find the change in internal energy of the system.  a. 1400 J  b. 1700 J  c. 1900 J  d. 1500 J

Description : If a system absorbs 500 cal of heat at the same time does 400J of work, find the change in internal energy of the system.  a. 1400 J  b. 1700 J  c. 1900 J  d. 1500 J

Last Answer : 1700 J

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

Roughly what is the total weight of air in the entire earth?  a. 1 x 10^5 tons  b. 2 x 10^6 tons  c. 6 x 10^15 tons  d. 8 x 10^10 tons

Description : Roughly what is the total weight of air in the entire earth?  a. 1 x 10^5 tons  b. 2 x 10^6 tons  c. 6 x 10^15 tons  d. 8 x 10^10 tons

Last Answer : 6 x 10^15 tons

_________ is the quantity of heat required to raise the temperature of one pound of water 1 F˚ from 63 ˚F to 64 ˚F.  a. one Joule  b. one calorie  c. one watt  d. one BTU

Description : _________ is the quantity of heat required to raise the temperature of one pound of water 1 F˚ from 63 ˚F to 64 ˚F.  a. one Joule  b. one calorie  c. one watt  d. one BTU

Last Answer : one BTU

The amount of heat required to raise the temperature of 1 pound of water by 1°F is called ______.  A. Calorie  B. Joule  C. BTU  D. Kilocalorie

Description : The amount of heat required to raise the temperature of 1 pound of water by 1°F is called ______.  A. Calorie  B. Joule  C. BTU  D. Kilocalorie

Last Answer : BTU

The amount of heat required to raise the temperature of 1kg of water through 1 °C is called ______.  A. Calorie  B. Joule  C. BTU  D. Kilocalorie

Description : The amount of heat required to raise the temperature of 1kg of water through 1 °C is called ______.  A. Calorie  B. Joule  C. BTU  D. Kilocalorie

Last Answer : Kilocalorie

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

The coefficient of performance (COP) is the ratio between the:  A. Power consumption in watts and heat absorbed per hour  B. Heat absorbed per hour and the power consumption in watts  C. Work required and the absorbed heat  D. Absorbed heat and work required

Description : The coefficient of performance (COP) is the ratio between the:  A. Power consumption in watts and heat absorbed per hour  B. Heat absorbed per hour and the power consumption in watts  C. Work required and the absorbed heat  D. Absorbed heat and work required

Last Answer : Absorbed heat and work required

What is defined as the energy required to raise the temperature of a unit mass of a substance by one degree?  A. Latent heat of fusion  B. Molar heat  C. Specific heat capacity  D. Specific heat

Description : What is defined as the energy required to raise the temperature of a unit mass of a substance by one degree?  A. Latent heat of fusion  B. Molar heat  C. Specific heat capacity  D. Specific heat

Last Answer : Specific heat

The amount of heat required to raise the temperature of the unit mass of gas through one degree at constant volume, is called  A.specific heat at constant volume  B.specific heat at constant pressure  C.kilo Joule  D.none of these

Description : The amount of heat required to raise the temperature of the unit mass of gas through one degree at constant volume, is called  A.specific heat at constant volume  B.specific heat at constant pressure  C.kilo Joule  D.none of these

Last Answer : Answer: A

In a non-flow reversible process for which p = (- 3V+ 15) x 105N/m2,V changes from 1 m to 2 m3. The work done will be about  (a) 100 xlOO5 joules  (b) lxlO5 joules  (c) 10 xlO5 joules  (d) 10 xlO5 kilo joules  (e) 10xl04ki\ojoules.

Description : In a non-flow reversible process for which p = (- 3V+ 15) x 105N/m2,V changes from 1 m to 2 m3. The work done will be about  (a) 100 xlOO5 joules  (b) lxlO5 joules  (c) 10 xlO5 joules  (d) 10 xlO5 kilo joules  (e) 10xl04ki\ojoules.

Last Answer : Answer : c

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?  a. 1.3 x 108 J  b 2.3 x 108 J  c. 3.3 x 108 J  d. 4.3 x 108 J formula: ΔQ = ΔQ/Δtx Δt

Description : 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?  a. 1.3 x 108 J  b 2.3 x 108 J  c. 3.3 x 108 J  d. 4.3 x 108 J formula: ΔQ = ΔQ/Δtx Δt

Last Answer : 1.3 x 108 J

A cylinder contains oxygen at a pressure of 10 atm and a temperature of 300 K. The volume of the cylinder is 10 liters. What is the mass of the oxygen in grams? Molecular weight (MW) of oxygen is 32 g/mole?  a. 125.02  b. 130.08  c. 135.05  d. 120.04

Description : A cylinder contains oxygen at a pressure of 10 atm and a temperature of 300 K. The volume of the cylinder is 10 liters. What is the mass of the oxygen in grams? Molecular weight (MW) of oxygen is 32 g/mole?  a. 125.02  b. 130.08  c. 135.05  d. 120.04

Last Answer : 130.08 {(10atm)(10)(32)/(0.0821) (300K)}

Twenty grams of ice at 0˚C melts to water at 0˚C. How much does the entropy of the 20g change in this process?  a. 30.5 J/K  b. 24.6 J/K  c. 21.3 J/K  d. 15.7 J/K

Description : Twenty grams of ice at 0˚C melts to water at 0˚C. How much does the entropy of the 20g change in this process?  a. 30.5 J/K  b. 24.6 J/K  c. 21.3 J/K  d. 15.7 J/K

Last Answer : 24.6 J/K

The compression ratio for Diesel engines is  A. 3 to 6  B. 5 to 8  C. 15 to 20  D. 20 to 30

Description : The compression ratio for Diesel engines is  A. 3 to 6  B. 5 to 8  C. 15 to 20  D. 20 to 30

Last Answer : Answer: C

The compression ratio for petrol engines is  A.3 to 6  B.5 to 8  C.15 to 20  D.20 to 30

Description : The compression ratio for petrol engines is  A.3 to 6  B.5 to 8  C.15 to 20  D.20 to 30

Last Answer : Answer: B

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

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

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

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

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?  A.10.7 ft^3  B.14.7 ft^3  C.15 ft^3  D.24.9 ft^3 Formula: use the ideal gas law pV = mRT T = (100 +460) ˚R V = mRT/p

Description : 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?  A.10.7 ft^3  B.14.7 ft^3  C.15 ft^3  D.24.9 ft^3 Formula: use the ideal gas law pV = mRT T = (100 +460) ˚R V = mRT/p

Last Answer : 24.9 ft^3

Molecular volume of any perfect gas at 600 x 103 N/m2 and 27°C will be  (a) 4.17m3/kgmol  (b) 400 m3/kg mol  (c) 0.15 m3/kg mol  (d) 41.7 m3/kg mol  (e) 417m3/kgmol.

Description : Molecular volume of any perfect gas at 600 x 103 N/m2 and 27°C will be  (a) 4.17m3/kgmol  (b) 400 m3/kg mol  (c) 0.15 m3/kg mol  (d) 41.7 m3/kg mol  (e) 417m3/kgmol.

Last Answer : Answer : a

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

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 SI unit of mass  a. kg  b. g  c. N  d. None of the above

Description : The SI unit of mass  a. kg  b. g  c. N  d. None of the above

Last Answer : kg

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

Two masses, one of the 10kg and the other unknown, are placed on a scale in a region where g = 9.67 m/sec2 . The combined weight of these two masses is 174.06 N. Find the unknown mass in kg.  a. 20 kg  b. 19 kg  c. 18 kg  d. 17 kg formula: m=Fg k / g

Description : Two masses, one of the 10kg and the other unknown, are placed on a scale in a region where g = 9.67 m/sec2 . The combined weight of these two masses is 174.06 N. Find the unknown mass in kg.  a. 20 kg  b. 19 kg  c. 18 kg  d. 17 kg formula: m=Fg k / g

Last Answer : 18 kg

The weight of an object is 50lb. What is its mass at standard condition?  a. 50 lbm  b. 60 lbm  c. 70 lbm  d. 80 lbm formula: m = Fgk /g

Description : The weight of an object is 50lb. What is its mass at standard condition?  a. 50 lbm  b. 60 lbm  c. 70 lbm  d. 80 lbm formula: m = Fgk /g

Last Answer : 50 lbm

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.  a. 11m/s  b. 12m/s  c. 13m/s  d. 10m/s ∫KE = mV2/2gc

Description : 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.  a. 11m/s  b. 12m/s  c. 13m/s  d. 10m/s ∫KE = mV2/2gc

Last Answer : 10m/s

A mass of 5kg is 100m above a given datum where local g = 9.75 m/s2 . Find the gravitational force in newtons. (Formula: Fg= mg/k )  a. 48.75 N  b. 50 N  c. 45 N  d. None of the above

Description : A mass of 5kg is 100m above a given datum where local g = 9.75 m/s2 . Find the gravitational force in newtons. (Formula: Fg= mg/k )  a. 48.75 N  b. 50 N  c. 45 N  d. None of the above

Last Answer : 48.75 N

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

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

Refer to problem # 11. Determine the force that accelerates if to 12 m/s^2. horizontally along frictionless plane.  A. 2474.23 N  B. 2574.23 N  C. 3474.23 N  D. 2374.23 N Formula : M = wk / g F = ma /k

Description : Refer to problem # 11. Determine the force that accelerates if to 12 m/s^2. horizontally along frictionless plane.  A. 2474.23 N  B. 2574.23 N  C. 3474.23 N  D. 2374.23 N Formula : M = wk / g F = ma /k

Last Answer : 2474.23 N

A system weighing 2kN. Determine the force that accelerate if to 12 m/s^2. a. vertically upward when g = 9.7 m/s^2  A. 4474.23 N  B.5484.23 N  C.4495.23 N  D.5488.23 N Formula: F = m/k (a +g)

Description : A system weighing 2kN. Determine the force that accelerate if to 12 m/s^2. a. vertically upward when g = 9.7 m/s^2  A. 4474.23 N  B.5484.23 N  C.4495.23 N  D.5488.23 N Formula: F = m/k (a +g)

Last Answer : 4474.23 N

Who formulated the zeroth law of thermodynamics in 1931?  A. A. Celsuis  B. A. Einstein  C. R.H. Fowler  D. G. Fahrenheit

Description : Who formulated the zeroth law of thermodynamics in 1931?  A. A. Celsuis  B. A. Einstein  C. R.H. Fowler  D. G. Fahrenheit

Last Answer : R.H. Fowler

What is the unit of the total energy of the system?  A. Kj  B. Kj/Kg  C. Kg  D. g

Description : What is the unit of the total energy of the system?  A. Kj  B. Kj/Kg  C. Kg  D. g

Last Answer : Kj

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

While swimming a depth of 13m in a fresh water lake a fish emits an air bubble of volume 2 mm² atmospheric pressure is 100kpa what is the original pressure of the bubble.  a. 217.17 kpa  b. 119 kpa  c. 326.15 kpa  d. 210 kap Pabs = Pg + Patm

Description : While swimming a depth of 13m in a fresh water lake a fish emits an air bubble of volume 2 mm² atmospheric pressure is 100kpa what is the original pressure of the bubble.  a. 217.17 kpa  b. 119 kpa  c. 326.15 kpa  d. 210 kap Pabs = Pg + Patm

Last Answer : 217.17 kpa

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 = 96.2ft/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 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

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

What is defined as the quantity of heat needed to raise the temperature of one kilogram of water by one degrees Celsius?  a. Kilojoule  b. Btu  c. Kilocalorie  d. Latent heat

Description : What is defined as the quantity of heat needed to raise the temperature of one kilogram of water by one degrees Celsius?  a. Kilojoule  b. Btu  c. Kilocalorie  d. Latent heat

Last Answer : Kilocalorie

Which is NOT a correct statement?  A. A superheated vapor will not condense when small amount of heat re removed  B. An ideal gas is a gas that is not a superheated vapor  C. A saturated liquid can absorb as much heat as it can without vaporizing  D. Water at 1 atm and room temperature is subcooled

Description : Which is NOT a correct statement?  A. A superheated vapor will not condense when small amount of heat re removed  B. An ideal gas is a gas that is not a superheated vapor  C. A saturated ... absorb as much heat as it can without vaporizing  D. Water at 1 atm and room temperature is subcooled

Last Answer : An ideal gas is a gas that is not a superheated vapor

What is the specific heat capacity of water in J/kg °C? ∙  A. 4581  B. 4185  C. 4518  D. 4815

Description : What is the specific heat capacity of water in J/kg °C? ∙  A. 4581  B. 4185  C. 4518  D. 4815

Last Answer : 4185

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 of water is  A. 1.817  B. 2512  C. 4.187  D. none of these

Description : The specific heat of water is  A. 1.817  B. 2512  C. 4.187  D. none of these

Last Answer :  C. 4.187

The thermodynamic difference between a Rankine cycle working with saturated steam and the Carnot cycle is that  (a) carnot cycle can’t work with saturated steam  (b) heat is supplied to water at temperature below the maximum temperature of the cycle  (c) a rankine cycle receives heat at two places  (d) rankine cycle is hypothetical  (e) none of the above.

Description : The thermodynamic difference between a Rankine cycle working with saturated steam and the Carnot cycle is that  (a) carnot cycle can't work with saturated steam  (b) heat is supplied to water at temperature ... heat at two places  (d) rankine cycle is hypothetical  (e) none of the above.

Last Answer : Answer : b