Question

what is the change in internal energy of a system ($delta u$) if 7.3 kj of heat is absorbed by the system and 2011 j of work is done on the system (by the surroundings) for a certain process? express your final answer in kj. answer:

Explanation:

Step1: Convert work units to kJ

Since 1 kJ = 1000 J, then $2011\ J=\frac{2011}{1000}\ kJ = 2.011\ kJ$.

Step2: Apply the first - law of thermodynamics formula

The first - law of thermodynamics is $\Delta U=Q + W$, where $Q$ is the heat added to the system and $W$ is the work done on the system. Here, $Q = 7.3\ kJ$ and $W=2.011\ kJ$.
So, $\Delta U=7.3\ kJ+2.011\ kJ$.

Step3: Calculate the change in internal energy

$\Delta U=(7.3 + 2.011)\ kJ=9.311\ kJ$.

Answer:

$9.311$