How many grams of C5H12 must be burned to heat 1.39 kg of water from 21.2 °C to 97.0 °C? Assume that all the heat released during combustion is used to heat the water.

m = 8.9856 g

Explanation:

In order to do this, we need to write the expressions that are to be used. First, to calculate heat:

Q = m*C*ΔT (1)

Where C would be heat capacity of the substance.

The heat can also be relationed with the moles and enthalpy of a compound using the following expression:

Q = n*ΔH (2)

Finally for the mass of any compound, we use the following expression:

m = n*MM (3)

So, in order to calculate the grams of pentane (C5H12), we need to calculate the moles of the compound, and to do that, we need the heat exerted.

So, as we are using water, let's calculate the heat that is been exerted with the water. The C of the water is 4.186 J/g °C so:

Q = (1.39 * 1000) * 4.186 * (21.2 - 97)

Q = - 441,045.33 J

This is the heat neccesary to burn pentane and heat water. Now, with this value, let's calculate the moles used of pentane with expression (2). The ΔH of the pentane is - 3,535 045. kJ/mol or - 3.535x10⁶ J/mol. Solving for n we have:

n = - 441,045.3 / - 3.535x10⁶

n = 0.1248 moles

Finally, we can calculate the grams needed with expression (3). The molar mass of pentane is 72 g/mol

m = 0.1248 * 72

m = 8.9856 g

This is the mass needed to heat 1.39 kg of water