Answer:
![Q=-213.84J](https://img.qammunity.org/2021/formulas/chemistry/college/68696lx8la8w5mmepqvzflzb7zrxrv5i63.png)
Step-by-step explanation:
Hello!
In this case, since energy involved when a substance undergoes a temperature change due to heat addition or removal is computed as shown below:
![Q=mC\Delta T](https://img.qammunity.org/2021/formulas/physics/college/ju4johxj2v6sxr0j7d64xf2my4hwldultz.png)
As the aluminum has a specific heat of 0.900 J/g°C and 14.4 g underwent such temperature decrease, the energy change is:
![Q=14.4g*0.900(J)/(g\°C)(23.0\°C-39.5\°C)\\\\Q=-213.84J](https://img.qammunity.org/2021/formulas/chemistry/college/edo5gy3itu38ktfzmk4e2e9ahmzjvj9ntq.png)
Which is an outlet energy flow.
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