Calorific Value(Heat of combustion)

Main Idea

The Calorific Value of a sample, also known as its Heat of Combustion, is defined as the amount of heat released during the complete combustion of the sample. For combustion to occur, a hydrocarbon is typically put into contact with oxygen and supplied the necessary activation energy. Once the reaction occurs, carbon dioxide, water, and heat are the products. This newfound heat is usually enough to continue the reaction, allowing a flammable substance to burn until there is none left.

The Heat of Combustion is typically measured through experiments using a bomb calorimeter, where the sample is supplied with excess oxygen. This device measures the temperature change. From this, the Heat of Combustion can be computed using the Thermal Energy Equation.

Mathematical Model

A typical combustion reaction looks like this:

[math]\displaystyle{ \text{C}_x\text{H}_y\text{N}_z\text{O}_n + \text{O}_2 \longrightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O} + \frac{z}{2}\text{N}_2 \quad Q = \boldsymbol{\Omega} \ \frac{J}{mol} }[/math], where

[math]\displaystyle{ \bullet \ \text{C}_x = }[/math] [math]\displaystyle{ x }[/math] atoms of Carbon

[math]\displaystyle{ \bullet \ \text{H}_y = }[/math] [math]\displaystyle{ y }[/math] atoms of Hydrogen

[math]\displaystyle{ \bullet \ \text{N}_z = }[/math] [math]\displaystyle{ z }[/math] atoms of Nitrogen

[math]\displaystyle{ \bullet \ \text{O}_n = }[/math] [math]\displaystyle{ n }[/math] atoms of Oxygen gas

[math]\displaystyle{ \bullet \ x\text{CO}_2 = }[/math] [math]\displaystyle{ x }[/math] moles of Carbon Dioxide

[math]\displaystyle{ \bullet \ \frac{y}{2}\text{H}_2\text{O} = }[/math] [math]\displaystyle{ \frac{y}{2} }[/math] moles of Water

[math]\displaystyle{ \bullet \ \frac{z}{2}\text{N}_2 = }[/math] [math]\displaystyle{ \frac{z}{2} }[/math] moles of Nitrogen gas

[math]\displaystyle{ \bullet \ Q = }[/math] the Heat of Combustion

[math]\displaystyle{ \bullet \ \boldsymbol{\Omega} = }[/math] a constant

Let us look at this combustion reaction as an example:

[math]\displaystyle{ \text{CH}_{3}\text{OH} + \text{O}_2 \longrightarrow \text{CO}_2 + 2\text{H}_{2}\text{O} \quad Q = 890 \ \frac{kJ}{mol} }[/math]

Here we see the Heat of Combustion of [math]\displaystyle{ \text{CH}_{3}\text{OH} }[/math], Methanol, is [math]\displaystyle{ 890 \ \frac{kJ}{mol} }[/math].

It is also always important to keep the Thermal Energy Equation in mind when thinking of these things, since it does relate Heat to a Temperature change:

[math]\displaystyle{ \Delta Q = mc \Delta T }[/math]

Computational Model

Insert Computational Model Here

Examples

Simple

A 1.55 gram sample of Ethanol is burned in a bomb calorimeter:

[math]\displaystyle{ \text{C}_{2}\text{H}_{5}\text{OH} + 3\text{O}_2 \longrightarrow 2\text{CO}_2 + 3\text{H}_{2}\text{O} }[/math]

a) If this combustion caused a Temperature increase of 55°C in a 200 gram sample of water, what is the Molar Heat of Combustion of the Ethanol? Water has a Specific Heat Capacity of [math]\displaystyle{ C = 4.18 \ \frac{J}{g \cdot °C} }[/math]. Ethanol has a molar mass of [math]\displaystyle{ Mm = 46.1 \ \frac{g}{mol} }[/math].

Using dimensional analysis, we can find the number of moles of Ethanol in the 1.55 gram sample. Letting [math]\displaystyle{ m }[/math] be the mass of the sample in grams, [math]\displaystyle{ Mm }[/math] be the molar mass of Ethanol in grams per mole, and [math]\displaystyle{ M }[/math] be the number of moles of Ethanol, we see:

[math]\displaystyle{ M = \frac{m}{1} \times \frac{1}{Mm} = \frac{1.55g}{1} \times \frac{1 \ mol}{46.1g} = 0.0336 \ mol }[/math]

Also, from the Thermal Energy equation, we can calculate the Heat transferred to the water:

[math]\displaystyle{ \Delta Q = mC \Delta T }[/math] (1)

The following are known values:

[math]\displaystyle{ m = 200g }[/math]

[math]\displaystyle{ C = 4.18 \ \frac{J}{g \cdot °C} }[/math]

[math]\displaystyle{ \Delta T = 55°C }[/math]

We can plug these values into 1 and get:

[math]\displaystyle{ \Delta Q_{water} = 200 \times 4.18 \times 55 = 45,980 \ J = 45.98 \ kJ }[/math]

By conservation of energy, we can reason that the Heat gained by the water sample is equal to the Heat lost by the Ethanol in the process of combustion. Hence, we make the following assumption:

[math]\displaystyle{ \Delta Q_{water} = Q_{Ethanol} }[/math], where

[math]\displaystyle{ Q_{Ethanol} = }[/math] the Heat of Combustion of this specific Ethanol sample

We can use this to find the Molar Heat of Combustion of Ethanol ([math]\displaystyle{ \boldsymbol{\Omega} }[/math]) as follows:

[math]\displaystyle{ \boldsymbol{\Omega} = \frac{Q_{Ethanol}}{M} = \frac{\Delta Q_{water}}{M} = \frac{45.98 \ kJ}{0.0336 \ mol} = 1,368.4 \ \frac{kJ}{mol} }[/math]

Middling

A bowl has a mass of 100 grams ([math]\displaystyle{ m }[/math]) while there is Ethanol in it. It is known from a previous measurement that the bowl had a mass of 50 grams ([math]\displaystyle{ m_{bowl} }[/math]).

A cup of water with mass 10 grams is heated from 20°C to 50°C by burning some of the sample of Ethanol.

a) Using these two statements, determine how much Ethanol must have combusted.

Let us call the mass of Ethanol combusted [math]\displaystyle{ m_{Ethanol_{c}} }[/math]. From statement one, we can figure out the total mass of the Ethanol sample, [math]\displaystyle{ m_{Ethanol_{t}} }[/math]:

[math]\displaystyle{ m = m_{bowl} + m_{Ethanol_{t}} }[/math]

Therefore:

[math]\displaystyle{ m_{Ethanol_{t}} = m - m_{bowl} = 100 - 50 = 50 }[/math] grams

Now, we want to figure out how much Heat must have been produced to warm the cup of water as much as was stated. We can do this with the Thermal Energy Equation:

[math]\displaystyle{ \Delta Q_{water} = m_{water}C_{water} \Delta T_{water} }[/math]

We know enough of the quantities in the equation to solve for the change in Thermal Energy:

[math]\displaystyle{ \Delta Q_{water} = 10 \times 4.18 \times (50 - 20) = 1254 \ J }[/math] (1)

Referring back to the Simple example, we know the Molar Heat of Combustion of Ethanol ([math]\displaystyle{ \Omega_{Ethanol} }[/math]):

[math]\displaystyle{ \Omega = 1,368.4 \ \frac{J}{mol} }[/math] (2)

We can find the number of moles of Ethanol ([math]\displaystyle{ M_{Ethanol} }[/math]) required to create the amount of Heat in 1, using the Molar Heat of Combustion of Ethanol:

[math]\displaystyle{ M_{Ethanol} = \Delta Q_{water} \frac{1}{\Omega_{Ethanol}} = 1254 \ \left(\frac{J}{1 }\right) \times \frac{1}{1,368.4} \ \left(\frac{mol}{J}\right) = 0.916 \ mol }[/math]

Using the molar mass of Ethanol [math]\displaystyle{ \left(Mm_{Ethanol} = 46.1 \ \frac{g}{mol}\right) }[/math], we can find the number of grams of Ethanol ([math]\displaystyle{ m_{ethanol_{c}} }[/math]) used to cause the necessary combustion:

[math]\displaystyle{ m_{Ethanol_{c}} = M_{Ethanol} \times Mm_{Ethanol} = 0.916 \ \left(\frac{mol}{1}\right) \times 46.1 \ \left(\frac{g}{mol}\right) = 42.23 }[/math] grams

We have found that 42.23 grams of the 50 gram Ethanol sample was combusted to Heat the water.

Difficult

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Connectedness

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History

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See also

Further reading

• Wikipedia: Heat of Combustion
  
• CK-12: Heat of Combustion
  
• CK-12:Heat of Combustion
  
• The Mole in Chemical Reactions
  

References

• Wikipedia: Heat of Combustion
  
• CK-12: Heat of Combustion
  
• CK-12:Heat of Combustion
  
• The Mole in Chemical Reactions