J.R. S. answered 03/28/19
Ph.D. University Professor with 10+ years Tutoring Experience
heat lost by copper = heat gained by water
-(mCu)(CCu)(∆T) = (mwater)(Cwater)(∆T)
-(32.2g)(0.385 J/g/deg)(Tf - 99.8 deg) = (215 g)(4.184 J/g/deg)(Tf - 18.5 deg)
-12.4Tf + 1237 = 900Tf - 16650
912.4Tf = 17,887
Tf = 19.6 degrees C