The final temperature of the sample, in degrees Celsius is 36.6 °C
From the question, we are to determine the final temperature of the sample
From the formula for heat capacity, we have that
Q = mcΔT
Where Q is the heat energy
m is the mass
c is the specific heat capacity of the subtance
ΔT is the change in temperature ([tex]\Delta T= T_{f} -T_{i}[/tex])
From the given information
m = 687.6 g
Q = 8284.4 J
[tex]T_{i}[/tex] = 23.1 °C
Putting the parameters into the equation, we get
[tex]8284.4 = 687.6 \times 0.89 \times (T_{f} -23.1)[/tex]
NOTE: Specific heat capacity of aluminum = 0.89 J/ g °C
Then,
[tex]8284.4 = 611.964 \times (T_{f} -23.1)[/tex]
[tex](T_{f} -23.1)= \frac{8284.4}{611.964}[/tex]
[tex](T_{f} -23.1)= 13.5[/tex]
[tex]T_{f} = 13.5 + 23.1[/tex]
[tex]T_{f} = 36.6 \ ^\circ C[/tex]
Hence, the final temperature of the sample, in degrees Celsius is 36.6 °C
Learn more on Heat capacity of a substance here: https://brainly.com/question/13439286
