Step 1: Understand Joule’s experiment and cyclic processes.
Joule’s experiment involves studying the relationship between heat and work, leading to the formulation of the first law of thermodynamics. For a cyclic process, a system returns to its initial state, so the change in internal energy \( \Delta U = 0 \).
Step 2: Apply the first law of thermodynamics to a cyclic process.
The first law of thermodynamics states:
\[
\Delta U = Q - W,
\]
where \( Q \) is the net heat transfer into the system, and \( W \) is the net work done by the system. For a cyclic process:
\[
\Delta U = 0,
\]
\[
Q - W = 0,
\]
\[
Q = W.
\]
This means the net heat transfer \( Q \) (sum of all heat transfers) equals the net work transfer \( W \) (sum of all work transfers). The term "proportional" in the question can be interpreted as equality in this context, as \( Q = W \) implies a proportionality constant of 1.
Step 3: Evaluate the options.
% Option
(1) Change of pressure is proportional to the temperature change: This is not a general result of Joule’s experiment or a cyclic process; it may apply to specific processes (e.g., ideal gas at constant volume), but not universally. Incorrect.
(2) Change of volume is proportional to temperature change: This is not a general result for a cyclic process; it may apply to specific processes (e.g., isobaric), but not always. Incorrect.
(3) Change of internal energy is proportional to temperature change: For a cyclic process, \( \Delta U = 0 \), regardless of temperature changes. This option is incorrect for a cyclic process. Incorrect.
(4) Sum of all heat transfer is proportional to the sum of all work transfer: As derived, \( Q = W \), which matches this statement (with a proportionality constant of 1). Correct.
Step 4: Select the correct answer.
Joule’s experiment for a cyclic process shows that the sum of all heat transfer equals the sum of all work transfer, matching option (4).