Question

A system gives out x J of heat and does y J of work on it's surrounding. What is the internal energy change?

• A. $-x - y$ J
• B. $y - x$ J
• C. $x - y$ J
• D. $x + y$ J

Hint:

Think of the system as a bank account. Giving out heat ($x$) means a withdrawal ($-x$), and performing work ($y$) on the surroundings means another withdrawal ($-y$). Therefore, the overall net balance change is simply $-x - y$.

Answer 1

The Correct Option is A

Step 1: Recall the first law.
$\Delta U = q + W$, using the convention that energy entering the system is positive and energy leaving is negative.
Step 2: Assign the heat sign.
The system gives out $x$ J of heat, so heat leaves: $q = -x$ J.
Step 3: Assign the work sign.
The system does $y$ J of work on the surroundings, so energy leaves as work: $W = -y$ J.
Step 4: Substitute into the law.
$\Delta U = (-x) + (-y)$.
Step 5: Simplify.
$\Delta U = -x - y$ J.
Step 6: Interpret and choose.
Both heat loss and work output drain internal energy, so $\Delta U$ is negative - matching option (1).
\[ \boxed{\Delta U = -x - y \text{ J}} \]