What is the change in internal energy $\Delta$E of a system when 4.50 *10³J of work is done on the system while it releases 12.0 kJ of energy to the surroundings?

Question

Quizplus · Accepted Answer

The change in internal energy ($\Delta U$) of a system can be calculated using the first law of thermodynamics, which is $\Delta U = Q - W$, where $Q$ is the heat added to the system and $W$ is the work done by the system. In this case, work is done on the system, so $W = -4.50 \,kJ$, and the system releases energy, meaning $Q = -12.0 \,kJ$. Thus, $\Delta U = -12.0 \,kJ - (-4.50 \,kJ) = -7.5 \,kJ$.

What Is the Change in Internal Energy Δ\DeltaΔ E of a System When 4

What Is the Change in Internal Energy $\Delta$ E of a System When 4