Question

Calculate the change in internal energy of the system if \(20\text{ kJ}\) of work is done on the system and it releases \(10\text{ kJ}\) of heat in a particular reaction.

• A. \(30\text{ kJ}\)
• B. \(10\text{ kJ}\)
• C. \(-15\text{ kJ}\)
• D. \(-20\text{ kJ}\)

Hint:

Use sign convention carefully: work on system is positive.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The change in internal energy (\(\Delta U\) or \(\Delta E\)) of a system is governed by the First Law of Thermodynamics, which states that energy cannot be created or destroyed, only transferred. The change in internal energy is equal to the heat added to the system plus the work done on the system.
Step 2: Key Formula or Approach:
The mathematical expression for the first law of thermodynamics is: \[ \Delta U = q + w \] Where:
\(q\) = heat exchanged between the system and surroundings
\(w\) = work done on or by the system
Sign Convention (IUPAC):
- Heat absorbed \textit{by} the system: \(q\) is positive (\(+\)).
- Heat released \textit{from} the system: \(q\) is negative (\(-\)).
- Work done \textit{on} the system: \(w\) is positive (\(+\)).
- Work done \textit{by} the system: \(w\) is negative (\(-\)).
Step 3: Detailed Explanation:
From the problem description:
"\(20\text{ kJ}\) of work is done \textit{on} the system" $\implies w = +20\text{ kJ}$
"it \textit{releases} \(10\text{ kJ}\) of heat" $\implies q = -10\text{ kJ}$
Substitute these values into the formula: \[ \Delta U = q + w \] \[ \Delta U = (-10\text{ kJ}) + (+20\text{ kJ}) \] \[ \Delta U = 10\text{ kJ} \] The change in internal energy is positive \(10\text{ kJ}\).
Step 4: Final Answer:
The correct option is (B).