Question

homework 2 : 2. in a expansion process, system did 10kj work to surrounding, at the same it released 20kj heat, please calculate the change of internal energy of system.

Explanation:

Step1: Recall First Law of Thermodynamics

The first law of thermodynamics is given by the equation $\Delta U = Q + W$, where $\Delta U$ is the change in internal energy, $Q$ is the heat added to the system, and $W$ is the work done on the system.

Step2: Determine Signs of Q and W

• The system released 20 kJ of heat, so heat is leaving the system. Thus, $Q = - 20\space kJ$ (negative because heat is released by the system).
• The system did 10 kJ of work on the surroundings, which means work is done by the system. In the first law, work done on the system is positive, so work done by the system is negative. Thus, $W=- 10\space kJ$.

Step3: Calculate $\Delta U$

Substitute the values of $Q$ and $W$ into the first law equation:
$$\Delta U=Q + W$$
$$\Delta U=- 20\space kJ+(- 10\space kJ)$$
$$\Delta U=-30\space kJ$$

Answer:

The change in internal energy of the system is $\boldsymbol{-30\space kJ}$ (or the internal energy of the system decreases by 30 kJ).