250+ TOP MCQs on Application of the Energy Balance to Open Systems and Answers

Chemical Engineering Problems on “Application of the Energy Balance to Open Systems”.

1. What is the energy balance to open system?
a) Energy in the system is constant
b) Matter in the system is constant
c) Matter is constant energy is variable
d) Both matter and energy are variable

Answer: d
Clarification: Both matter and energy are variable in an open system.

2. 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?
a) 50 J
b) 100 J
c) 150 J
d) 200 J

Answer: d
Clarification: ∆U = Q + W – m (∆H + ∆PE + ∆KE), => ∆U = 0 + 100*5 – 5 (10 + 10*5) = 500 – 50 – 250 = 200 J.

3. 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?
a) 50 J
b) 100 J
c) 150 J
d) 200 J

Answer: c
Clarification: ∆U = Q + W – m (∆H + ∆PE + ∆KE), => ∆U = 0 + 50*10 – 10*(15 + 10*2) = 500 – 350 = 150 J.

4. 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?
a) 10 J
b) 25 J
c) 35 J
d) 50 J

Answer: a
Clarification: ∆U = Q + W – m (∆H + ∆PE + ∆KE), => ∆U = 0 + 25*1 – 1*(5 + 10*1) = 25 – 15 = 10 J.

5. 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?
a) 100 J
b) 150 J
c) 200 J
d) 300 J

Answer: b
Clarification: ∆U = 0 + 300*1 – 1*(50 + 10*10) = 150 J.

6. 2 Kg of water (C_P = 0.15 J/Kg^oC) 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 27^oC?
a) 10^oC
b) 19^oC
c) 27^oC
d) 37^oC

Answer: d
Clarification: ∆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 = 37^oC.

7. 1 Kg of water (C_P = 0.5 J/Kg^oC) 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 10^oC?
a) 10^oC
b) 30^oC
c) 40^oC
d) 60^oC

Answer: d
Clarification: ∆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 = 60^oC.

8. 1 Kg of water (C_P = 0.5 J/Kg^oC) 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 20^oC?
a) 20^oC
b) 30^oC
c) 40^oC
d) 50^oC

Answer: b
Clarification: ∆U = Q + W – m (∆H + ∆PE + ∆KE), => 0 = – 5 + 20 – 1*(∆H + 10*1), => ∆H = 5 J/Kg, => 5 = 0.5(T – 20), => T = 30^oC.

9. 1 Kg of water (C_P = 0.4 J/Kg^oC) 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 25^oC?
a) 25^oC
b) 45^oC
c) 60^oC
d) 75^oC

Answer: d
Clarification: ∆U = Q + W – m (∆H + ∆PE + ∆KE), => 0 = – 10 + 50 – 1*(∆H + 10*2), => ∆H = 20 J/Kg, => 20 = 0.4(T – 20), => T = 75^oC.

10. 5 Kg of water (C_P = 1 J/Kg^oC) 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 0^oC?
a) 2^oC
b) 4^oC
c) 6^oC
d) 8^oC

Answer: c
Clarification: ∆U = Q + W – m (∆H + ∆PE + ∆KE), => 0 = – 5 + 50 – 5*(∆H + 10*0.5), => ∆H = 6 J/Kg, => 6 = 1*(T – 0), => T = 6^oC.