#### Answer

Therefore the final temperature of the aluminum sample is 95.1$^{\circ}$ C

#### Work Step by Step

We will use the equation $q=mc\Delta t$
q = heat transferred
m = mass
c = specific heat
$\Delta t $ = change in temperature
plug in the values.
$2.87 \times 10^{3}$ = (53.2) (0.9) ($\Delta t $)
***We multiply 53.2 times 0.9 and then divide $2.87 \times 10^{3}$ by that answer
*** solve using a calculator
$ \Delta t = 59.9 ^{\circ} $
Therefore the temperature change is $1.82^{\circ}$
The initial temp is 155 so the final temp= initial temp - $ \Delta t$
= 155 - 59.9 = 95.1
Therefore the final temperature is 95.1$^{\circ}$ C