Answer: 28.32 degrees Celsius
Explanation:
For this question, we can use the formula [tex]Q=mc \triangle T[/tex], where [tex]Q[/tex] is the amount of heat absorbed, [tex]m[/tex] is the mass of the sample, [tex]c[/tex] is the specific heat constant, and [tex]\triangle T[/tex] is the change in temperature (final temperature minus initial temperature as stated in the question).
In this question, [tex]m=81.5, Q=1130[/tex], and [tex]c=4.18[/tex] (this is just the specific heat constant for water).
So, [tex]1130=(81.5)(4.18) \triangle T \longrightarrow \triangle T=3.32[/tex]
This means that the temperature increased by 3.32 degrees Celsius, so the final temperature is 28.32 degrees Celsius
