Heat of Combustion

Heat of combustion Matthieu Schaller et Xavier Buffat matthieu.schaller@epfl.ch xavier.buffat@epfl.ch 19 avril 2008 Table des matières 1 Introduction2 2 Theoretical part2 2.1 Gibbs free energy.........................2 2.2 Theorical value..........................2 2.3 Calorimetry............................3 3 Method4 4 Results5 4.1 Constants used..........................5 4.2 Ethanol..............................5 4.3 Gases...............................5 5 Discussion6 6 Conclusion7 7 Appendix8 1 1 INTRODUCTION 2 1 Introduction The first energy to be used by men comes from chemical reactions, in fact, most of the energy we can use still come from chemical reactions. Indeed, we use gazoline to make our car move, different fuels for power plants, even wood to heat our houses up, etc. Each of these reaction is in principle the same is based on the heat created by a chemical reaction called combustion. The heat provided by the combustion depend on the enregy contained in the combustible, this quantity of heat is called calorific value and can be determined by calorimetry. We are going to study the calorific value of some usual combustible using a simple calorimetry method and compare the results with theory. 2 Theoretical part 2.1 Gibbs free energy The Gibbs free energy, also called free enthalpy, is a thermodynamical function that provides a way to determine what is the stable state in which the system is going to go. Indeed, in a spontaneous reaction this function can only be lower in the final state than in the starting state and function is minimal in an equilibrium. In our case, the reaction is happening slowly under constant atmospheric pressure, the Gibbs free energy is G = U − TS + PV and so, by making total differenciation, we have dG = dU − T dS − SdT + P dV + V dP And this, of course, is equal to zero at equilibrium. A consequence of this, which is called Berthelot law is that the reaction that is going to happend is the one which will liberate the most heat, and thus lowing to the minimum its chemical energy. 2.2 Theorical value There is a way to determine the calorific value of a combustible with theory. Indeed, there are ways to determine precisely the energy needed to create a chemical component, which is called enthalpy of formation. There- fore, we can determine the energy it is going to liberate then it is destroyed, using the Hess law. The energy dissipated in heat by a chemical reaction is equal to the difference of the total enthalpy of formation between final and starting componens. In our case we are going to consider the combustion of four different combustibles, one of which is the ethanol, C2H6O. The 3 METHOD 3 reaction is the following : C2H6O + 3O2 −! 2CO2 + 3H2O Therefore we can deternime the heat dissipated by this energy, using stan- dard enthalpy of formation form the table. We have : kJ ∆H = −2 · 394 − 3 · 242 − (−278 + 3 · 0) = −1236 (2.1) mol This calculations can be made in the same way for the other reactions, this give us a way to measure the accuracy of our experiment. 2.3 Calorimetry The experimental device is composed by and heat exchanger between a flow of cooling liquid and a reference liquid. The reference stays in a box that bind around a hole. In this hole goes the suitable burner adapted to the combutible we would like to study. This box is insulated from the outside except for the cooling liquid flowing in. Therefore, we can assume that, if the system is in a stationary state, all the energy produce by the chemical reaction is used to heat up the cooling liquid. Thus, by mesuring the difference of temperature between the in and out going water, and also the quantity of water heated, we can deduce the energy that was produce. Knowing the specific heat of water c, and noting Tin and Tout the temperature of the water and M the mass of water heated, we have : ∆H = c M (Tout − Tin) Therefore, measuring m the quantity of combustible burned, we have the relation : T − T H = c M out in m This is right assuming the water produce during the reaction stays in the state of steam, but this is not the case, this water condensate on the side of the box. Because this state transformation require a lot of energy, we have to take it under consideration (∆Hsteaming(H2O) = 549kcal=kg). Finally we have this formula : c M(T − T ) − M ∆H kcal H = out in c steaming (2.2) m kg where Mc is the mass of condensation water. 3 METHOD 4 Fig. 1 { experimental device schema 3 Method We power up the combustion in the device, while cooling water is flowing through it and wait until the system is in a stationary state, in which the temperatures stay stable. These temperatures are measured by thermo- couple inserted in the water incomming and outgoing. We begin the experiment by opening a valve that direct the cooling water into a box, in order to measure precisely the quantity of water that flew through during the time of the experiment. In the mean time, we do the same with the condensation water. We also need to know precisly the mass burned during this laps of time, in order to do that there is two ways depending on the nature of the combustible. If the combustible is a gaz, we measure the flow, and the time of the experiment, therefore we can deduce the quantity of gaz burned. It is even simplier if the combustible is a liquid, we measure the mass of the burner before and after and then deduce the difference. It is important to insure that there is enough air flowing into the hole of the burner, otherwise the combustion would not be complete. in other words, there will not be enough oxygen to make the complete reaction and therefore the combustion equation (eg. ??) is not the one expected. This generate two problem, the calorific value measured will be lower than it should be, and it is possible that the chemical product are harmful, like CO for instance. We carry out this experiment for liquid ethanol, a fifty-fifty mix of liquid 4 RESULTS 5 butane and propane that is generally used for common burner, the gaz that is provided for Lausanne town, and finally with methan. 4 Results 4.1 Constants used For these experiment we need some physical constants. We used the one that are listed in the table 4.1. Name Symbol Value Value (other units) Specific heat capacity of water c 4180 J=˚kg 1 kcal=˚kg 6 Latent heat of water Hsteaming 2:44 ∗ 10 J=kg 585 kcal=kg Fig. 2 { Physical constants we used 4.2 Ethanol We made three different mesuerement with ethanol. The results are summarized in the following table. Measure Heat of combustion [J/kg] Heat of combustion [kcal/kg] 1 2:16 · 107 5174 2 2:55 · 107 6092 3 3:01 · 107 7211 Fig. 3 { Ethanol's heat of combustion These three measures lead to a mean value of : 6159 ± 833 kcal=kg (4.1) This value is not far from the theoretical value, which is 6460. 4.3 Gases The mesurement for Methane and for (( City Gas ))are a little bit more complex. In order to know the mass of gas we burned, we need to measure the flow of gas and the duration of the experiment. This will give us the volume of gas burned in the burner. A multiplication of this value with the gaz density leads us to the burned mass. The case of the (( City Gas ))is special. This gas is a mixture of ethan, methan, nitrogen and carbon dioxide. the proportion are given in the following table. 5 DISCUSSION 6 Gas Proportion [%] Density [kg=m3] Methan 86 0:72 Ethan 5 1:36 Nitrogen 5 1:25 Carbon dioxide 1:4 1:98 Others 2:6 { Fig. 4 { Composition of the (( City Gas )) The density of this gas is then : 0:77 kg=m3. For each of the gas we made three measures in order to reduces the errors. The values we obtained are summarized in the table5 Gas Heat of combustion [kcal/kg] Precision [%] Theoretical value Methan 13098 ± 212 1:6 14050 City Gas 7505 ± 1004 8:7 9135 Camping burner 11632 ± 1012 13:6 11960 Fig. 5 { Gases heat of combustion The values obtained are correct despite of the little precision we got. The theoretical values are systematicly higher than those we got. 5 Discussion Globally, the values we got are accurate but not very precise. This is often the case in thermodynamics. In order to get better measures there are some improvement that could be done. For example, the measure of the combustible mass we burned, is made with a balance which is not very precise. The last digit of the balance's screen corresponds to a value of 0:1 g. We measured masses from 3 to 10 g. This means that the precision of the measure was not better than 3%. The same problem appears with the measure of the temperautre difference. The precisions of this value is not better than 1%. The amount of water that is changed in steam is also a big source of im- precision. The mass is very small (about 3 to 4 grams) but it requires a huge quantity of energy to get some water to steam. This quantity of energy is nearly the same as the energy needed to heat up the water that flies through the whole system.

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