Question

In a thermodynamic system, Q represents the energy transferred to or from a system by heat and W represents the energy transferred to or from a system by work. I. $Q > 0$ and $W = 0$
II. $Q < 0$ and $W = 0$
III. $W > 0$ and $Q = 0$
IV. $W < 0$ and $Q = 0$ Which of the above will lead to an increase in the internal energy of the system?

• A. I only
• B. II only
• C. I and IV only
• D. II and III only
• E. II and IV only

Hint:

Internal energy increases whenever you "put something in"—either by heating the system ($Q > 0$) or by squashing/compressing it ($W < 0$, work done on the system).

Answer 1

The Correct Option is C

Concept:
The First Law of Thermodynamics is given by: \[ \Delta U = Q - W \] Where: [itemsep=6pt]
• $\Delta U$ is the change in internal energy.
• $Q$ is the heat added to the system ($Q > 0$ if heat is added, $Q < 0$ if heat is removed).
• $W$ is the work done by the system ($W > 0$ if work is done by the system, $W < 0$ if work is done on the system).

Step 1: Analyze each condition for $\Delta U > 0$.
[label=\Roman*., itemsep=8pt]
• $Q > 0, W = 0$: $\Delta U = Q - 0 = Q$. Since $Q > 0$, internal energy increases.
• $Q < 0, W = 0$: $\Delta U = Q - 0 = Q$. Since $Q < 0$, internal energy decreases.
• $W > 0, Q = 0$: $\Delta U = 0 - W = -W$. Since $W > 0$, internal energy decreases (system does work at the expense of internal energy).
• $W < 0, Q = 0$: $\Delta U = 0 - W = -W$. Since $W < 0$, internal energy increases (work done on the system increases its energy).

Step 2: Conclusion.
Internal energy increases in cases I and IV.