When a system is taken from state i to state f along the path iaf, it is found that \( Q = 50 \, \text{cal} \) and \( W = 20 \, \text{cal} \). Along the path ibf, \( Q = 36 \, \text{cal} \). W along the path ibf is:
Show Hint
For cyclic processes, the change in internal energy is the same along different paths, allowing us to relate heat and work along various paths.
The first law of thermodynamics states that:
\[
\Delta U = Q - W
\]
Since the internal energy change \( \Delta U \) is the same for both paths, we can equate the two expressions for \( \Delta U \) and solve for \( W \) along path \( ibf \). This gives \( W = 16 \, \text{cal} \).
Final Answer:
\[
\boxed{16 \, \text{cal}}
\]
