61-66
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61. Find the change in internal energy of 5lbm of oxygen when the temperature changes from 100°F to 120°F, Cv = 0.157 Btu/lbm-°R. a. 14.70 Btu c. 16.80 Btu b. 15.70 Btu d. 147 Btu Solution: ΔU = mCvΔT T2 = 120°F = [120°F] + 459.67 = 579.67 °R T1 = 100°C = [100°F] + 459.67 = 559.67 °R m = 5 lbm Cv = 0.157 Btu/lbm-°R = (5lbm)*(0.157 Btu/lbm-°R)*(579.67-559.67 °R) = 15.70 Btu (b)
62. Water (Specific heat, Cv = 4.2 kJ/kg-K) is being heated by 1500-W heater. What is the rate of change in temperature of 1 kg of water? a. 0.043 Kelvin/s c. 0.357 Kelvin/s b. 0.179 Kelvin/s d. 1.50 Kelvin/s Solution:
Q = mCvΔT ΔT = Q/mCv
Q = 1500 watts = 1.5 kw = 1.5 kJ/s Cv = 4.2 kJ/kg-K m = 1 kg ΔT = (1.5 kJ/s)/(1 kg * 4.2 kJ/kg-K) = 0.357 K/s (c)
63. One kilogram of water (Cv = 4.2 kJ/kg-K) is heated by 300 Btu of energy. What is the change in temperature in K? a. 75.36 K c. 73.80 K b. 125.20 K d. 17.96 K Solution: Q = mCvΔT ΔT = Q/mCv m = 1 kg Cv = 4.2 kJ/kg-K Q = 300 Btu (1.055 kJ/Btu) = 316.50 kJ ΔT = 316.50 kJ / (1 kg * 4.2 kJ/kg-K) = 75.36 K (a)
64. Determine the change in enthalpy per lb mass of nitrogen gas as its temperature changes from 500°F to 200°F. (Cp = 0.2483 Btu/lbm-°R) a. – 74.49 Btu/lbm c. – 68.47 Btu/lbm b. – 72.68 Btu/lbm d. – 63.78 Btu/lbm Solution: Q = Δh
Δh = mCp ΔT ; Δh/m = Cp ΔT Cp = 0.2483 Btu/lbm T2 = 200°F = [120°F] + 459.67 = 659.67 °R T1 = 500°C = [100°F] + 459.67 = 959.67 °R Δh/m = (0.2483 Btu/lbm) * (659.67-959.67 °R) = - 74.49 Btu/lbm (a)
65. Calculate the change in enthalpy as 1 kg of nitrogen is heated from 1000 K to 1500 K, assuming the nitrogen is an ideal gas at a constant pressure. The temperature dependent specific heat of nitrogen is Cp = 39.06 – 512 T^(-1.5) + 1072.7 T^(-2) – 820.4 T^(-3) where Cp is in kJ.kg-mol, and T is in K. a. 600 kJ b. 697.27 kJ
c. 800 kJ d. 897.27 kJ
Solution: Cp = δh / δT ; δh = Cp * δT Cp = 39.06 – 512 T^(-1.5) + 1072.7 T^(-2) – 820.4 T^(-3) δT : T2 = 1500 , T1 = 1000 δh = Cp * δT
66. 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 c. 36.4 psia b. 28.6 psia d. 52.1 psia Solution: P1/T1 = P2/T2 P2 = (P1*T2) / T1 P2 = ( 15 psia * (800+460) ) / (200+460) = 28.64 psia (b)
62. Water (Specific heat, Cv = 4.2 kJ/kg-K) is being heated by 1500-W heater. What is the rate of change in temperature of 1 kg of water? a. 0.043 Kelvin/s c. 0.357 Kelvin/s b. 0.179 Kelvin/s d. 1.50 Kelvin/s Solution:
Q = mCvΔT ΔT = Q/mCv
Q = 1500 watts = 1.5 kw = 1.5 kJ/s Cv = 4.2 kJ/kg-K m = 1 kg ΔT = (1.5 kJ/s)/(1 kg * 4.2 kJ/kg-K) = 0.357 K/s (c)
63. One kilogram of water (Cv = 4.2 kJ/kg-K) is heated by 300 Btu of energy. What is the change in temperature in K? a. 75.36 K c. 73.80 K b. 125.20 K d. 17.96 K Solution: Q = mCvΔT ΔT = Q/mCv m = 1 kg Cv = 4.2 kJ/kg-K Q = 300 Btu (1.055 kJ/Btu) = 316.50 kJ ΔT = 316.50 kJ / (1 kg * 4.2 kJ/kg-K) = 75.36 K (a)
64. Determine the change in enthalpy per lb mass of nitrogen gas as its temperature changes from 500°F to 200°F. (Cp = 0.2483 Btu/lbm-°R) a. – 74.49 Btu/lbm c. – 68.47 Btu/lbm b. – 72.68 Btu/lbm d. – 63.78 Btu/lbm Solution: Q = Δh
Δh = mCp ΔT ; Δh/m = Cp ΔT Cp = 0.2483 Btu/lbm T2 = 200°F = [120°F] + 459.67 = 659.67 °R T1 = 500°C = [100°F] + 459.67 = 959.67 °R Δh/m = (0.2483 Btu/lbm) * (659.67-959.67 °R) = - 74.49 Btu/lbm (a)
65. Calculate the change in enthalpy as 1 kg of nitrogen is heated from 1000 K to 1500 K, assuming the nitrogen is an ideal gas at a constant pressure. The temperature dependent specific heat of nitrogen is Cp = 39.06 – 512 T^(-1.5) + 1072.7 T^(-2) – 820.4 T^(-3) where Cp is in kJ.kg-mol, and T is in K. a. 600 kJ b. 697.27 kJ
c. 800 kJ d. 897.27 kJ
Solution: Cp = δh / δT ; δh = Cp * δT Cp = 39.06 – 512 T^(-1.5) + 1072.7 T^(-2) – 820.4 T^(-3) δT : T2 = 1500 , T1 = 1000 δh = Cp * δT
66. 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 c. 36.4 psia b. 28.6 psia d. 52.1 psia Solution: P1/T1 = P2/T2 P2 = (P1*T2) / T1 P2 = ( 15 psia * (800+460) ) / (200+460) = 28.64 psia (b)
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