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A system undergoes a change of state during which 100 kJ of heat is transferred to it, and it does 50 kJ of work. The system is brought back to its original state through a process during which 120 kJ of heat is transferred. The work done by the system is
- a)50 kJ
- b)70 kJ
- c)170 kJ
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A system undergoes a change of state during which 100 kJ of heat is t...
△Q1 =100k Trang≤W1=50kj
△Q2=120kjW2=?
The given process is a cyclic process Hence △U=0
⇒△Q1+△Q2=△W1+△W2
100+120=50+△W2
△W2 = 170kj
Most Upvoted Answer
A system undergoes a change of state during which 100 kJ of heat is t...
Understanding the Thermodynamic Process
The problem describes a system undergoing two distinct processes: one where heat is absorbed, and work is done, and another where the system returns to its original state with heat released.
First Process: Change of State
- Heat transferred to the system: **100 kJ**
- Work done by the system: **50 kJ**
In this first process, the first law of thermodynamics states that:
ΔU = Q - W
Where:
- ΔU = Change in internal energy
- Q = Heat added to the system
- W = Work done by the system
Substituting the values:
ΔU = 100 kJ - 50 kJ = 50 kJ
This indicates that the internal energy of the system increases by **50 kJ**.
Second Process: Returning to Original State
In the second process, the system is brought back to its original state:
- Heat transferred (released) from the system: **120 kJ**
Since the system returns to its original state, the change in internal energy (ΔU) for this process must equal the change from the first process, but in the opposite direction:
ΔU = -50 kJ
Applying the first law again:
-50 kJ = Q - W
Here, Q is the heat released, which is **-120 kJ**:
-50 kJ = -120 kJ - W
Solving for W:
W = -120 kJ + 50 kJ = -70 kJ
This means that **70 kJ** of work is done by the system in this process, leading to the conclusion that the correct answer is option **C) 170 kJ** when considering the total work done in both processes.
The total work done is the sum of work done in both processes:
Total Work = 50 kJ + 70 kJ = 170 kJ
The problem describes a system undergoing two distinct processes: one where heat is absorbed, and work is done, and another where the system returns to its original state with heat released.
First Process: Change of State
- Heat transferred to the system: **100 kJ**
- Work done by the system: **50 kJ**
In this first process, the first law of thermodynamics states that:
ΔU = Q - W
Where:
- ΔU = Change in internal energy
- Q = Heat added to the system
- W = Work done by the system
Substituting the values:
ΔU = 100 kJ - 50 kJ = 50 kJ
This indicates that the internal energy of the system increases by **50 kJ**.
Second Process: Returning to Original State
In the second process, the system is brought back to its original state:
- Heat transferred (released) from the system: **120 kJ**
Since the system returns to its original state, the change in internal energy (ΔU) for this process must equal the change from the first process, but in the opposite direction:
ΔU = -50 kJ
Applying the first law again:
-50 kJ = Q - W
Here, Q is the heat released, which is **-120 kJ**:
-50 kJ = -120 kJ - W
Solving for W:
W = -120 kJ + 50 kJ = -70 kJ
This means that **70 kJ** of work is done by the system in this process, leading to the conclusion that the correct answer is option **C) 170 kJ** when considering the total work done in both processes.
The total work done is the sum of work done in both processes:
Total Work = 50 kJ + 70 kJ = 170 kJ
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A system undergoes a change of state during which 100 kJ of heat is transferred to it, and it does 50 kJ of work. The system is brought back to its original state through a process during which 120 kJ of heat is transferred. The work done by the system isa)50 kJb)70 kJc)170 kJd)None of theseCorrect answer is option 'C'. Can you explain this answer?
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