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: Understanding the Question:
The problem presents a thermodynamic scenario where heat energy equal to $x\ \mathrm{kJ}$ is transferred to a system, and mechanical work equal to $y\ \mathrm{J}$ is simultaneously done on it. We need to express the total change in internal energy ($\Delta U$) in units of Joules ($\mathrm{J}$).

Step 2: Key Formula or Approach:
We apply the first law of thermodynamics: $$\Delta U = Q + W$$ First, verify and synchronize the units into Joules ($\mathrm{J}$). Note that $1\ \mathrm{kJ} = 1000\ \mathrm{J}$.
Next, establish the correct sign conventions:

• Heat provided to the system: $Q = +x\ \mathrm{kJ} = +1000x\ \mathrm{J}$

• Work done on the system: $W = +y\ \mathrm{J}$

Step 3: Detailed Explanation:
Both energy quantities enter the system, which means both parameters must carry a positive thermodynamic sign, increasing the internal energy.
Convert the heat value to Joules: $$Q = x \times 1000 = 1000x\ \mathrm{J}$$ The work component is already in Joules: $$W = y\ \mathrm{J}$$ Summing these inputs together according to the first law: $$\Delta U = 1000x + y$$ This algebraic expression matches the unit formatting in option (A).

Step 4: Final Answer:
The internal energy change is $(1000x + y)\ \mathrm{J}$, which corresponds to option (A).