Calculate Heat of Combustion: Grams, Heat Capacity, Temp Change

In summary, the formula for calculating the heat of combustion for a substance is heat capacity multiplied by the temperature difference, divided by the number of grams converted to moles. The temperature difference may not always be a factor, but it is important to take into account the density of the fuel and any energy lost. In the example provided, a sample of napthalene is burned in a calorimeter and the heat of combustion is calculated by determining the number of moles used and the energy received by the calorimeter. This formula allows for the efficient calculation of the heat of combustion for various substances.
  • #1
HeartSoul132
58
0
What's the formula for calculating this, given: grams, heat capacity of substance, and temperature change?

How about heat combustion per kJ/Mol?

Thanks
 
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  • #2
You would write a balanced equation for
Fuel + oxygen -> product (typically water + CO2)

Then account for all the energy needed to break the bonds in the original fuel and oxygen plus the energy released when bonds form in the product.
For real world energy values the water formed is usually steam so you have to account for the heating and vapourisation energy needed.

Or you can lookup the heat of combustion/gram for lots of different fuels on the web or in databooks
 
Last edited:
  • #3
Thanks for the reply.
Don't need an equation, according to solution; just a plug and chug formula. It seems like it is heat capacity * temp diff ALL divided by the grams converted to mols -- wondering why?
 
  • #4
For simply burning the temperature difference shouldn't come into it - if you burn something in liquid oxygen (-200degC) you get just the same energy out!

There is a temperature effect if you have to take into account the density of the fuel (as in liquified gas, or jet fuel at high altitudes) or if you have a non-condensing boiler where some energy is lost as hot steam.
 
  • #5
A .922 sample of napthalene (C10H8) is burned in a calorimeter that has a heat capacity of .944 kJ/K. The temperature of the calorimeter rose from 15.73 C to 19.75 C. Calculate the heat of combustion for this chemical in kJ/mol.

I'm sure there's an equation, wondering how?
 
  • #6
The heat of combustion doesn't depend on the actual temperature - the temperature change of the calorimeter is telling you how much energy was given off.
You get the point - if the fuel burned and heated the calorimeter from 115.73 C to 119.75 C you would get the same result.

You need to work out how many moles of fuel you used, and how much energy the calorimter received that gives you the joules/mole for the fuel.
 

Related to Calculate Heat of Combustion: Grams, Heat Capacity, Temp Change

What is the heat of combustion?

The heat of combustion, also known as the enthalpy of combustion, is the amount of energy released when a substance undergoes complete combustion with oxygen.

How is the heat of combustion calculated?

The heat of combustion is calculated by measuring the mass of the substance, determining its heat capacity, measuring the change in temperature, and then using the formula Q = mCΔT, where Q is the heat of combustion, m is the mass, C is the heat capacity, and ΔT is the change in temperature.

What units are used to measure the heat of combustion?

The heat of combustion is typically measured in joules (J) or kilojoules (kJ) per mole. However, it can also be measured in calories (cal) or kilocalories (kcal) per mole.

What factors can affect the accuracy of the calculated heat of combustion?

The accuracy of the calculated heat of combustion can be affected by various factors, such as incomplete combustion, impurities in the substance, and heat loss to the surroundings during the experiment.

What is the significance of calculating the heat of combustion?

Calculating the heat of combustion is important in understanding the energy content of a substance and its potential as a fuel source. It also allows for the comparison of different substances and their efficiency in releasing energy through combustion.

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