Application of the Energy Balance to Open Systems

Question 1
Marks : +2 | -2
Pass Ratio : 20%
1 Kg of water (CP = 0.5 J/KgoC) is pumped to a height of 1 m, work done by the motor is 20 J/Kg, and heat liberated by it is 5 J/Kg, what is the final temperature of water if the initial temperature was 20oC?
20oC
30oC
40oC
50oC
Explanation:
∆U = Q + W – m (∆H + ∆PE + ∆KE), => 0 = – 5 + 20 – 1*(∆H + 10*1), => ∆H = 5 J/Kg, => 5 = 0.5(T – 20), => T = 30oC.
Question 2
Marks : +2 | -2
Pass Ratio : 20%
1 Kg of water is pumped to a height of 10 m, the enthalpy is changed by 50 J/Kg, and work done by the motor is 300 J/Kg, assuming a system is insulated what is the change in internal energy of water?
100 J
150 J
200 J
300 J
Explanation:
∆U = 0 + 300*1 – 1*(50 + 10*10) = 150 J.
Question 3
Marks : +2 | -2
Pass Ratio : 20%
5 Kg of water is pumped to a height of 5 m, the enthalpy is changed by 10 J/Kg, and work done by the motor is 100 J/Kg, assuming system is insulated what is the change in internal energy of water?
50 J
100 J
150 J
200 J
Explanation:
∆U = Q + W – m (∆H + ∆PE + ∆KE), => ∆U = 0 + 100*5 – 5 (10 + 10*5) = 500 – 50 – 250 = 200 J.
Question 4
Marks : +2 | -2
Pass Ratio : 20%
What is the energy balance to open system?
Energy in the system is constant
Matter in the system is constant
Matter is constant energy is variable
Both matter and energy are variable
Explanation:
Both matter and energy are variable in an open system.
Question 5
Marks : +2 | -2
Pass Ratio : 20%
1 Kg of water (CP = 0.5 J/KgoC) is pumped to a height of 0.5 m, work done by the motor is 40 J/Kg, and heat liberated by it is 10 J/Kg, what is the final temperature of water if the initial temperature was 10oC?
10oC
30oC
40oC
60oC
Explanation:
∆U = Q + W – m (∆H + ∆PE + ∆KE), => 0 = – 10 + 40 – 1*(∆H + 10*0.4), => ∆H = 25 J/Kg, => 25 = 0.5(T – 10), => T = 60oC.
Question 6
Marks : +2 | -2
Pass Ratio : 20%
2 Kg of water (CP = 0.15 J/KgoC) is pumped to a height of 0.1 m, work done by the motor is 10 J/Kg, and heat liberated by it is 5 J/Kg, what is the final temperature of water if the initial temperature was 27oC?
10oC
19oC
27oC
37oC
Explanation:
∆U = Q + W – m (∆H + ∆PE + ∆KE), => 0 = – 5 + 10 – 2(∆H + 10*0.1), => ∆H = 1.5 J/Kg, => 1.5 = 0.15(T – 27), => T = 37oC.
Question 7
Marks : +2 | -2
Pass Ratio : 20%
1 Kg of water (CP = 0.4 J/KgoC) is pumped to a height of 2 m, work done by the motor is 50 J/Kg, and heat liberated by it is 10 J/Kg, what is the final temperature of water if the initial temperature was 25oC?
25oC
45oC
60oC
75oC
Explanation:
∆U = Q + W – m (∆H + ∆PE + ∆KE), => 0 = – 10 + 50 – 1*(∆H + 10*2), => ∆H = 20 J/Kg, => 20 = 0.4(T – 20), => T = 75oC.
Question 8
Marks : +2 | -2
Pass Ratio : 20%
10 Kg of water is pumped to a height of 2 m, the enthalpy is changed by 15 J/Kg, and work done by the motor is 50 J/Kg, assuming system is insulated what is the change in internal energy of water?
50 J
100 J
150 J
200 J
Explanation:
∆U = Q + W – m (∆H + ∆PE + ∆KE), => ∆U = 0 + 50*10 – 10*(15 + 10*2) = 500 – 350 = 150 J.
Question 9
Marks : +2 | -2
Pass Ratio : 20%
5 Kg of water (CP = 1 J/KgoC) is pumped to a height of 0.5 m, work done by the motor is 50 J/Kg, and heat liberated by it is 5 J/Kg, what is the final temperature of water if the initial temperature was 0oC?
2oC
4oC
6oC
8oC
Explanation:
∆U = Q + W – m (∆H + ∆PE + ∆KE), => 0 = – 5 + 50 – 5*(∆H + 10*0.5), => ∆H = 6 J/Kg, => 6 = 1*(T – 0), => T = 6oC.
Question 10
Marks : +2 | -2
Pass Ratio : 20%
1 Kg of water is pumped to a height of 1 m, the enthalpy is changed by 5 J/Kg, and work done by the motor is 25 J/Kg, assuming a system is insulated what is the change in internal energy of water?
10 J
25 J
35 J
50 J
Explanation:
∆U = Q + W – m (∆H + ∆PE + ∆KE), => ∆U = 0 + 25*1 – 1*(5 + 10*1) = 25 – 15 = 10 J.