Answer to Question #157568 in Chemistry for James notarte

Question

1.5 grams of ethanol (CH3CH2OH) is burned under a 300-milliliter sample of 

water. This causes the water sample to increase the temperature by 55°C. 

Calculate the heat of combustion of ethanol. Assume that the density of water is 1 

gram/milliliter.

Answer 1

The heat of combustion is usually expressed in energy per mole of the fuel , in our case, of ethanol. Therefore, we must calculate the energy amount that is produced and the number of the moles of ethanol that was burned.

In order to calculate the energy released, we can use the specific heat capacity of water $c = 4.18$ J(g°C)^-1, its mass ( volume times density) and the increase in the temperature of water:

$Q = cm\Delta T = 4.18\cdot300\cdot1\cdot55 = 68970$ J.

In order to calculate the number of the moles of ethanol, we use its mass (1.5 g) and divide it by its molar mass (46.07 g/mol):

$n= \frac{m}{M} = \frac{1.5}{46.07} = 0.03256$ mol.

Finally, the heat of combustion of ethanol is:

$\Delta H = \frac{Q}{n} = \frac{68970}{0.03256} = 2.12\cdot10^6$ J/mol, or 2.12 MJ/mol.

Answer: the heat of combustion of ethanol is 2.12 MJ/mol.