A 12.0 gram piece of pure gold originally at 80°C is cooled and in the process releases 72.8 Joules of heat. What is the final temperature of the gold in degrees Celsius?

My answer: 33°C

My work:

Q=mc :delta:T

:delta:T= Q/(mc)

Q= -72.8 J

m= 12 g

c= 0.129 J/(g*K)

:delta:T= ?

Plugged it all in the equation and I get: :delta:T= Q/(mc)= -47 Kelvin

:delta:T=T_{F} -T_{I}

change in T = T final minus T initial where T is Temperature

T_{F}= :delta:T+T_{I}

T_{F}= -47 K + 353 K = 306 K **80°C=353 K**

306 K = 33°C

My answer: 33°C

Correct?