Answer:
The final temperature of the system is 32,2°C
Explanation:
The heat losed by copper tubing is equal to heat gained by water and vessel, thus:
[tex]-Q_{Cu} = Q_{vessel} + Q_{H_{2}O}[/tex] (1)
Qcu = 0,387J/g°C × 285g × (X-89,5°C) = 110,295X - 9871,4
Qvessel = 10,0 J/°C (X-22,8°C) = 10,0X - 228,0
QH₂O = 4,184 J/g°C × 159g × (X-22,8°C) = 665,3X - 15167,8
Where X is the final temperature
Using (1):
-110,295X + 9871,4 = 10,0X - 228,0 + 665,3X - 15167,8
25267,2 = 785,595 X
X = 32,2°C
I hope it helps!
