Question

A system releases 10 kJ of heat and performs 15 kJ of work on the surrounding. Hence the change in internal energy is :

• A. + 5 kJ
• B. -- 5 kJ
• C. + 25 kJ
• D. -- 25 kJ

Hint:

Always look for keywords like "released" or "by the system" to assign a negative sign, and "absorbed" or "on the system" for a positive sign in chemistry thermodynamics.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
According to the first law of thermodynamics, energy can neither be created nor destroyed, only transformed. The change in internal energy (\(\Delta U\)) of a system is the sum of heat exchanged (\(q\)) and work done (\(w\)).
Step 2: Key Formula or Approach:
\[ \Delta U = q + w \]
Sign convention (IUPAC):
Heat released by system: \(q\) is negative.
Work done by system on surroundings: \(w\) is negative.
Step 3: Detailed Explanation:
Given:
Heat released, \(q = -10 \text{ kJ}\)
Work done by system, \(w = -15 \text{ kJ}\)
Substituting the values in the formula:
\[ \Delta U = (-10 \text{ kJ}) + (-15 \text{ kJ}) \]
\[ \Delta U = -25 \text{ kJ} \]
Step 4: Final Answer:
The change in internal energy is \(- 25 \text{ kJ}\).