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12. A 25.0 g piece of iron (c 0.451 J/goC ) initially at 21.00C absorbs 3100 J of heat. What is its final temperature? a) 253
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Answer #1

m= 25.09, C=0.4517g.•c Ti = 21.0°C = 294 K , 923100), Tz qe mCAT 3100= 25 X0.451(T2- 21.00 T2 = 295.9°C

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