i am studyin\' for a test and trying to \"get\" how to do thistype of equation t

Question

i am studyin' for a test and trying to "get" how to do thistype of equation the book gives the equation and the answer butdoesnt show how they got the answer...what if it's on the testOHNO!! When thermal equalibrium is reached what is the finaltemp?? here is the equation qH20+qmetal=0 [cwater*mwater8(tf-ti)] + [cFe * mFe(tf -ti)] i get this so everything is given except the t final [(4.184J/g*K)(244g)(Tf-292.0K)] +[(0.449J/g*K)(88.5g)(Tf-352.0K)]=0 T final = 295 K BUT HOW?? HOW DO YOU ISOLATE TF AND CALCULATE T FINALW/THIS

Explanation / Answer

[(4.184J/g*K) (244g) (Tf-292.0K)] +[ (0.449J/g*K) (88.5g) (Tf-352.0K) ]=0 Multipy each of the numbers along with Tf to remove thebrackets: 1020.896 Tf -298101.632 + 39.7365Tf - 13987.248 = 0 1060.6325 Tf = 312088.88 Tf = 312088.88 /1060.6325      ˜ 295 K

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