Homework Help: chem 101

Posted by Anonymous on Sunday, April 15, 2012 at 7:44pm.

This is second part is built off the first question. These are all from a practice test.

How much heat is required to raise the temperature of exactly one liter of water fro m 60 °F to 180 °F?

correct answer: 279kj

How much alcohol must be burned to generate enough heat to warm the water as described in problem 3?

correct answer: 10.4

the only mention of an alcohol is ethyl alcohol (C2H6O) in the question before this where you find its heat of combustion.

correct answer: -1238.43 kJ/mol

chem 101 - DrBob222, Sunday, April 15, 2012 at 10:00pm
60F is 15.55 C
180 F is 82.22C
delta T = 82.22-15.55 = 66.67
q = mass x specific heat x delta T
q = 100g x 4.184 x 66.67
q = 278,947 = 279 kJ.

Burning alcohol (46 g ethanol) provides 1238.43 kJ/mol. We want to know how many g ethanol to burn to provide 279 kJ.
46 g x (279/1238.43) = 10.36 g which rounds to 10.4 g.

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