A chemical engineer studying the properties of fuels placed 1.520 g of hydrocarbon in the bomb of a calorimeter and filled it with 02 gass. The bomb was immersed in 2.550 L of water and the reatcion initiated. The water temp rose from 20 to 23.55 C. If the calorimeter had a heat capcity of 403 J.K, what was the heat of reaction from combustion per gram of the fuel?
...it's for a friend, who's about to give up. I can't help it...It's in my nature to help. If you could show work, too. I'd appreciate it, guys.
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01-28-2009, 10:19 PM #1
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01-28-2009, 10:41 PM #2
Not 100%sure, I haven't done this stuff in a while, but I'll take a shot at it
qrxn=-(qwater + qcalorimeter)
qrxn = -(mcDTwater + mcDTcalorimeter)
qrxn = -[(2.550kg x 4186J/kgK x 3.55K) + (403J/K x 3.55K)]
qrxn = -39324.415
since this is for 1.520 g, divide the answer above by 1.520
final answer = -25.871 kJ/gram
Hope that's right, sorry if it's not but hopefully it will get him on the right track either way