Questions: T-S Diagrams: Temperature-Entropy Diagrams

Question 1

A reversible process traces a horizontal line on a T-S diagram from entropy S₁ to S₂ at constant temperature T. What does the area under this line represent?

• A. The work done by the system during the process
• B. The heat transferred during the process
• C. The change in internal energy of the system
• D. The change in enthalpy of the system

Answer: (B)

By δQ_rev = T dS, the area under a T-S curve is heat transfer. For an isothermal process, heat = T × ΔS, which equals the rectangular area under the horizontal line. This is the core utility of T-S diagrams: heat becomes geometrically legible.

Question 2

A student claims: 'A vertical line on a T-S diagram represents an isothermal process because the temperature axis doesn't move.' What is wrong with this reasoning?

• A. Nothing — a vertical line does represent an isothermal process
• B. A vertical line represents an isentropic (constant entropy) process — no heat is transferred, but temperature can change
• C. Vertical lines cannot exist on a T-S diagram
• D. Isothermal processes appear as curved lines, not vertical ones

Answer: (B)

A vertical line means entropy is constant (ΔS = 0), corresponding to a reversible adiabatic (isentropic) process — no heat flows, but temperature can rise or fall. An isothermal process (constant temperature) appears as a HORIZONTAL line, because temperature (the vertical axis) stays fixed while entropy changes as heat is absorbed or released.

Question 3

The Carnot cycle appears as a perfect rectangle on a T-S diagram.

• T. True
• F. False

Answer: True

The Carnot cycle consists of two isothermal processes (horizontal lines at T_H and T_C) connected by two isentropic processes (vertical lines at fixed entropy). These four segments form a rectangle. The efficiency η = (T_H − T_C)/T_H is directly visible as the ratio of the rectangle's height span to its maximum temperature — no algebra required.

Question 4

The area enclosed by a complete thermodynamic cycle on a T-S diagram represents the total heat input to the system.

• T. True
• F. False

Answer: False

The enclosed area represents the NET WORK output of the cycle — that is, heat absorbed minus heat rejected. Heat input alone is the area under the upper portion of the cycle curve (above the baseline); heat rejected is the area under the lower portion. Net work = heat in − heat out = the enclosed area. Confusing 'enclosed area' with 'heat in' is a common error.

Question 5

Why is a T-S diagram more useful than a P-V diagram for analyzing a heat engine's thermal efficiency?

Model answer: A T-S diagram represents heat transfer directly as a geometric area (δQ_rev = T dS), so heat input, heat rejection, and net work appear as readable areas without calculation. On a P-V diagram, the area represents work, but heat is not geometric — it requires knowing the process path and thermodynamic state functions. The T-S diagram makes thermal efficiency visible as the ratio of the enclosed area to the heat-input area.

The key insight is that heat and work play symmetric roles in different diagram spaces: P-V gives work as area, T-S gives heat as area. A Carnot cycle's efficiency η = 1 − T_C/T_H is immediately legible from the rectangle's proportions in T-S space, while extracting the same information from a P-V diagram requires integrating along curved paths.