Application of the Energy Balance to Closed Systems

Question 1
Marks : +2 | -2
Pass Ratio : 20%
What is the energy balance of a closed system?
Matter in the system is constant
Energy in the system is constant
Neither energy nor matter is constant
Both of energy and matter are constant
Explanation:
For closed system energy balance, matter in the system is constant.
Question 2
Marks : +2 | -2
Pass Ratio : 20%
What is the heat transferred to the system, if a change in internal energy is 10 J?
0
-10 J
10 J
Cannot be determined
Explanation:
Q = ∆U = 10 J.
Question 3
Marks : +2 | -2
Pass Ratio : 20%
What is the heat transferred to the system, if change in enthalpy is 15 J and change in volume is 2 m3 at 5 Pa?
5 J
8 J
10 J
12 J
Explanation:
Q = ∆H – P∆V = 15 – 5*2 = 5 J.
Question 4
Marks : +2 | -2
Pass Ratio : 20%
The specific volume of a gas is 5 m3/Kg, at 25 Pa, H = 40 J/Kg and at 40 Pa, H = 90 J/Kg, what is the heat transferred to the system?
-5 J/Kg
5 J/Kg
-25 J/Kg
25 J/Kg
Explanation:
Q = ∆H – V∆P = (90 – 40) – 5(40 – 25) = -25 J/Kg.
Question 5
Marks : +2 | -2
Pass Ratio : 20%
The specific volume of a gas is 2 m3/Kg, at 12 Pa, H = 30 J/Kg and at 15 Pa, H = 45 J/Kg, what is the heat transferred to the system?
6 J/Kg
9 J/Kg
12 J/Kg
15 J/Kg
Explanation:
Q = ∆H – V∆P = (45 – 30) – 2(15 – 12) = 15 – 6 = 9 J/Kg.
Question 6
Marks : +2 | -2
Pass Ratio : 20%
The specific volume of a gas is 3 m3/Kg, at 6 Pa, H = 4 J/Kg and at 9 Pa, H = 14 J/Kg, what is the heat transferred to the system?
0
1 J/Kg
5 J/Kg
25 J/Kg
Explanation:
Q = ∆H – V∆P = (14 – 4) – 3(9 – 6) = 1 J/Kg.
Question 7
Marks : +2 | -2
Pass Ratio : 20%
What is the heat transferred to the system, if change in enthalpy is 30 J and pressure, volume changed from 8 Pa, 5 m3 to 4 Pa, 12 m3?
12 J
15 J
18 J
22 J
Explanation:
Q = ∆H – V∆P = 30 – (4*12 – 8*5) = 22 J.
Question 8
Marks : +2 | -2
Pass Ratio : 20%
The specific volume of a gas is 5 m3/Kg, at 10 Pa, H = 50 J/Kg and at 20 Pa, H = 100 J/Kg, what is the heat transferred to the system?
0
50 J/Kg
100 J/Kg
150 J/Kg
Explanation:
Q = ∆H – V∆P = (100 – 50) – 5(20 – 10) = 0.
Question 9
Marks : +2 | -2
Pass Ratio : 20%
The specific volume of SO2 is 10 m3/Kg, at 5 Pa, H = 100 J/Kg and at 10 Pa, H = 250 J/Kg, what is the heat transferred to the system?
100 J/Kg
150 J/Kg
250 J/Kg
350 J/Kg
Explanation:
Q = ∆H – V∆P = (250 – 100) – 10(10 – 5) = 100 J/Kg.
Question 10
Marks : +2 | -2
Pass Ratio : 20%
What is the heat transferred to the system, if a change in enthalpy is 25 J and change in pressure is 5 Pa at volume 3 m3?
5 J
10 J
15 J
20 J
Explanation:
Q = ∆H – V∆P = 25 – 3*5 = 10 J.