A closed system undergoes a process `1 to 2` for which the values `W_(1-2)` and `Q_(1-2)` are 50 kJ and -20 kJ respectively. If the system is returned to state 1 and `Q_(2 to 1)` is + 10 kJ the work done `W_(2 to 1)` is:

Options

A. 40 kJ
B. 50 kJ
C. `-60 kJ`
D. `-50 kJ`

Explanation

To solve the problem, we will use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a closed system is equal to the heat added to the system (Q) minus the work done by the system (W).

### Step-by-step Solution:

1. **Identify the given values:**
- Work done from state 1 to state 2: \( W_{1-2} = 50 \, \text{kJ} \)
- Heat transfer from state 1 to state 2: \( Q_{1-2} = -20 \, \text{kJ} \)
- Heat transfer from state 2 to state 1: \( Q_{2-1} = +10 \, \text{kJ} \)
- Work done from state 2 to state 1: \( W_{2-1} = ? \)

2. **Apply the first law of thermodynamics for the process from state 1 to state 2:**
\[ \Delta U_{1-2} = Q_{1-2} - W_{1-2} \]
Substituting the known values:
\[ \Delta U_{1-2} = -20 \, \text{kJ} - 50 \, \text{kJ} = -70 \, \text{kJ} \]

3. **Since the system returns to state 1, the change in internal energy for the complete cycle is zero:**
\[ \Delta U_{total} = 0 \]
Therefore, we can write:
\[ \Delta U_{1-2} + \Delta U_{2-1} = 0 \]
This implies:
\[ \Delta U_{2-1} = +70 \, \text{kJ} \]

4. **Apply the first law of thermodynamics for the process from state 2 to state 1:**
\[ \Delta U_{2-1} = Q_{2-1} - W_{2-1} \]
Substituting the known values:
\[ 70 \, \text{kJ} = 10 \, \text{kJ} - W_{2-1} \]

5. **Rearranging the equation to find \( W_{2-1} \):**
\[ W_{2-1} = 10 \, \text{kJ} - 70 \, \text{kJ} \]
\[ W_{2-1} = -60 \, \text{kJ} \]

### Final Answer:
The work done \( W_{2-1} \) is \( -60 \, \text{kJ} \).

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