Question

When $\mathrm{x}\ \mathrm{kJ}$ heat is provided to a system, work equivalent to $\mathrm{y}\ \mathrm{J}$ is done on it. What is internal energy change during this operation?

• A. $(1000x + y)\ \mathrm{J}$
• B. $1000(x + y)\ \mathrm{J}$
• C. $(x + 1000y)\ \mathrm{J}$
• D. $x + y\ \mathrm{J}$

Hint:

Be alert to unit combinations like $\mathrm{kJ}$ mixed with $\mathrm{J}$! Always expand prefixes like "kilo-" into $1000$ before adding terms together, and remember that any energy added to a system (heat in or work done on it) takes a positive sign.

Answer 1

The Correct Option is A

Step 1: Recall the first law.
$\Delta U = Q + W$, with both $Q$ and $W$ taken as positive when energy enters the system.
Step 2: Read the directions of energy flow.
Heat $x\ kJ$ is provided to the system, so $Q$ is positive. Work $y\ J$ is done on the system, so $W$ is positive too.
Step 3: Match the units.
The answer is wanted in joules, but heat is in kilojoules. Use $1\ kJ = 1000\ J$, so $Q = x\ kJ = 1000x\ J$.
Step 4: Keep work as is.
Work is already in joules, so $W = y\ J$.
Step 5: Add the two inputs.
\[ \Delta U = Q + W = 1000x + y\ (J). \]
Step 6: Identify the option.
The internal energy change is $(1000x + y)\ J$, option (A).
\[ \boxed{(1000x + y)\ J \text{ (option A)}} \]