Post a New Question

chem 101

posted by .

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 -

    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.

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question