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=TF -TI
change in T = T final minus T initial where T is Temperature
TF= :delta:T+TI
TF= -47 K + 353 K = 306 K **80°C=353 K**
306 K = 33°C
My answer: 33°C
Correct?