Question

A system does 394 J of work on surrounding by absorbing 701 J heat. What is the change in internal energy of the system?

• A. 547 J
• B. 1095 J
• C. 307 J
• D. 394 J

Hint:

Keep a simple arithmetic check in mind: when a system takes in energy as heat ($+701$) and releases energy by performing mechanical work ($-394$), the net change must simply be the subtraction of the two magnitudes ($701 - 394 = 307$).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem asks for the change in internal energy ($\Delta U$) of a thermodynamic system given the amounts of heat absorbed and work performed.

Step 2: Key Formula or Approach:
According to the First Law of Thermodynamics: $$\Delta U = q + W$$ Using standard IUPAC sign conventions: Heat absorbed by the system is positive ($+q$). Work done by the system on its surroundings is negative ($-W$).

Step 3: Detailed Explanation:
From the problem description: 1. Heat absorbed by the system ($q$) = $+701\text{ J}$ 2. Work done by the system on the surroundings ($W$) = $-394\text{ J}$ Substitute these values with their appropriate thermodynamic signs into the first law equation: $$\Delta U = 701\text{ J} + (-394\text{ J})$$ $$\Delta U = 701 - 394 = 307\text{ J}$$ The positive result indicates that the internal energy of the system net increased by 307 Joules.

Step 4: Final Answer:
The change in internal energy of the system is 307 J, which matches option (C).