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, the change in internal energy ($\Delta U$) of a system is the sum of the heat ($q$) exchanged and the work ($w$) done.
Step 2: Key Formula or Approach:
\[ \Delta U = q + w \]
Sign conventions (IUPAC):
Heat released by the system: \( q = -ve \)
Work done by the system on surroundings: \( w = -ve \)
Step 3: Detailed Explanation:
Given:
Heat released, \( q = -10 \text{ kJ} \)
Work performed by the system, \( w = -15 \text{ kJ} \)
Substituting the values:
\[ \Delta U = (-10 \text{ kJ}) + (-15 \text{ kJ}) \]
\[ \Delta U = -25 \text{ kJ} \]
Step 4: Final Answer:
The change in internal energy is -- 25 kJ.