Question

calculate the magnitude of q, in kj, for a system that has 7.53x10³ j of work done on it and for which the change in total internal energy is 10.7 kj. report your answer to one place past the decimal point.

Explanation:

Step1: Convert work from J to kJ

We know that \(1\space kJ = 1000\space J\). So, \(w=7.53\times 10^{3}\space J=\frac{7.53\times 10^{3}}{1000}\space kJ = 7.53\space kJ\)

Step2: Use the first - law of thermodynamics

The first - law of thermodynamics is \(\Delta U=q + w\). We need to solve for \(q\), so \(q=\Delta U - w\). Given \(\Delta U = 10.7\space kJ\) and \(w = 7.53\space kJ\)

$$q=10.7-7.53$$

$$q = 3.17\space kJ$$

Answer:

\(3.2\space kJ\)