Step 1: Understand the cyclic process and apply the first law.
The process is cyclic, so the change in internal energy \( \Delta U = 0 \). The first law of thermodynamics for a cyclic process is:
\[
Q_{\text{net}} = W_{\text{net}},
\]
where \( Q_{\text{net}} \) is the net heat transfer, and \( W_{\text{net}} \) is the net work done.
Step 2: Calculate the net work transfer.
Work interactions:
15 kJ to the fluid (work done on the system, negative work done by the system): \( W_1 = -15 \, \text{kJ} \),
45 kJ from the fluid (work done by the system): \( W_2 = +45 \, \text{kJ} \).
Net work done by the system:
\[
W_{\text{net}} = W_2 + W_1 = 45 + (-15) = 30 \, \text{kJ}.
\]
Step 3: Calculate the known net heat transfer.
Heat interactions:
75 kJ to the fluid (heat added to the system): \( Q_1 = +75 \, \text{kJ} \),
40 kJ from the fluid (heat rejected by the system): \( Q_2 = -40 \, \text{kJ} \),
Third heat transfer \( Q_3 \) (unknown).
Net known heat transfer:
\[
Q_1 + Q_2 = 75 - 40 = 35 \, \text{kJ}.
\]
Step 4: Determine the third heat transfer.
For a cyclic process:
\[
Q_{\text{net}} = W_{\text{net}},
\]
\[
Q_1 + Q_2 + Q_3 = W_{\text{net}},
\]
\[
75 - 40 + Q_3 = 30,
\]
\[
35 + Q_3 = 30,
\]
\[
Q_3 = 30 - 35 = -5 \, \text{kJ}.
\]
- A negative value means 5 kJ of heat is transferred out of the system (from the system).
Step 5: Interpret the magnitude and direction.
Magnitude: 5 kJ,
Direction: From the system (since \( Q_3 \) is negative).
The option "-5 kJ from the system" indicates a heat transfer of 5 kJ out of the system, which matches.
Step 6: Select the correct answer.
The third heat transfer is -5 kJ from the system, matching option (4).