A sample of aluminum with mass of 53.3 g is initially at 143 ∘C

Question

What is the temperature of the aluminum after it loses 2.76 x 103 J?

Answer 1

heat=mass*c*(Tf-Ti)
-2.75J=53.3g*specific heat*(Tf-143C)
specific heat .9J/gC
Tf=143-2.76e3/(.9*53.3)=85.5 C
check all that logic and math.