Answer:
[tex]42.1^{\circ}C[/tex]
Explanation:
The heat extracted from the aluminium is given by:
[tex]Q=mC_s (T_f-T_i)[/tex]
where we have
Q = -1050 J is the heat extracted
m = 65.0 g = 0.065 kg is the mass
Cs = 900 J/kgC is the specific heat of aluminum
[tex]T_i = 60.0^{\circ}[/tex] is the initial temperature
By solving for [tex]T_f[/tex], we find the final temperature:
[tex]T_f = \frac{Q}{m C_s}+T_i=\frac{-1050 J}{(0.065 kg)(900 J/kg C)}+60.0^{\circ}C=42.1^{\circ}C[/tex]
