Question

calculate the magnitude of q, in kj, for a system that has 6.42×10³ j of work done on it and for which the change in total internal energy is -7.41 kj. report your answer to one place past the decimal point.

Explanation:

Step1: Convert work to kJ

Since \(1\space kJ = 1000\space J\), then \(w=6.42\times 10^{3}\space J=\frac{6.42\times 10^{3}}{1000}\space kJ = 6.42\space kJ\)

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 \(\Delta U=- 7.1\space kJ\) and \(w = 6.42\space kJ\)

$$q=-7.1-6.42$$

$$q=-13.52$$

Answer:

\(-13.5\space kJ\)