Answer:
[tex]T_f = 76.46°C[/tex]
Explanation:
Given data:
Mass of mixture = 454 kg
Initial temperature is 10°C
Heat added is Q = 121300 kJ
Heat capacity (Applesuace) at 32.8°C is 4.02kJ/kg K
From heat equation we have
[tex]Q = mCp(T_f -T_i)[/tex]
[tex]\frac{Q}{mCp} = (T_f -T_i)[/tex]
[tex]T_f = T_i + \frac{Q}{mCp}[/tex]
Putting all value to get required final temperature value
[tex]T_f = \frac{121300}{454\times 4.02} + 10[/tex]
[tex]T_f = 76.46°C[/tex]