Question

Calculate the change in internal energy of the system if work done by the system is 18 joule and absorbs heat 50 joule in a particular reaction.

• A. 20 J
• B. 32 J
• C. 48 J
• D. 68 J

Hint:

Sign Convention Rule: Anything going INTO the system (absorbing heat, getting compressed by outside work) is POSITIVE. Anything coming OUT of the system (releasing heat, expanding against the surroundings) is NEGATIVE!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks us to calculate the change in internal energy ($\Delta U$) of a thermodynamic system given the magnitude of heat transfer and physical work. We must rigorously apply the standard sign conventions of thermodynamics.

Step 2: Detailed Explanation:
The First Law of Thermodynamics is the law of conservation of energy, stated mathematically as:
$\Delta U = q + w$
where:
$\Delta U$ = Change in internal energy
$q$ = Heat added to or removed from the system
$w$ = Work done on or by the system
Let's carefully apply the IUPAC sign conventions:
- "absorbs heat 50 joule": Heat is flowing INTO the system, increasing its energy. Therefore, $q$ is positive.
$q = +50 \text{ J}$
- "work done by the system is 18 joule": The system is expending its own energy to push against the surroundings (expansion). Therefore, work is negative.
$w = -18 \text{ J}$
Now, substitute these signed values into the First Law equation:
$\Delta U = (+50 \text{ J}) + (-18 \text{ J})$
$\Delta U = 50 - 18$
$\Delta U = +32 \text{ J}$
The internal energy of the system increased by a net total of 32 Joules.

Step 3: Final Answer:
The change in internal energy is 32 J, matching option (b).