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

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.

To answer the first question, we need to use the equation for heat transfer: q = m * c * ΔT.

1. First, we need to find the mass of water in grams, since the specific heat (c) of water is often given in J/g°C. One liter of water is equivalent to 1000 grams.

2. Next, we need to convert the temperature difference from Fahrenheit to Celsius. To do this, we use the formula: °C = (°F - 32) * 5/9. So, ΔT = (180 - 60) * 5/9 = 120 * 5/9 = 66.67°C.

3. The specific heat of water, c, is approximately 4.18 J/g°C.

4. Now, we can substitute the values into the equation: q = m * c * ΔT = 1000g * 4.18 J/g°C * 66.67°C.

Calculating this value results in approximately 279,340 J (or 279 kJ).

To answer the second question about the amount of alcohol burned, we can use the concept of the heat of combustion. The heat of combustion is the amount of heat energy released when one mole of a substance is completely burned.

1. We need to find the heat of combustion for ethyl alcohol, which is given as -1238.43 kJ/mol.

2. We know the heat transfer required for the water is 279 kJ.

3. By setting up a proportion, we can calculate the amount of alcohol burned: (279 kJ / 1 L water) = (x kJ / 1 mol alcohol).

4. Solving for x, the amount of alcohol burned, we get:

x = (279 kJ * 1 mol alcohol) / (-1238.43 kJ/mol) ≈ 0.225 mol alcohol.

However, the question asks for the amount of alcohol burned in grams, so we need to convert from moles to grams.

5. The molar mass of ethyl alcohol (C2H6O) is approximately 46.07 g/mol.

6. To convert from moles to grams, we multiply the number of moles by the molar mass:

0.225 mol * 46.07 g/mol ≈ 10.4 g.

Therefore, approximately 10.4 grams of ethyl alcohol must be burned to generate enough heat to warm the water as described.