Data Analysis
In order to determine the total heat energy released by each of the fuel sources, a graph will need to be constructed to demonstrate the change in the temperature of water over the entire duration of the experiment. From this graph, a linear trendline will be created for the last 3 data points where the water was allowed to cool, which can be extrapolated backwards to determine the total decrease in the temperature of the water due to a loss of energy to the surrounding environment during the combustion of the fuel. The difference between the highest and lowest temperature values will then be used to calculate the total change in the temperature of water during the combustion of the fuels. Below, one representative graph and
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Density of water = mass of watervolume of water
mass of water = (0.998 g cm-3)(80.0 mL ± 0.5 1cm31 mL) , given that the density of water at 20.1 °C is 0.998 g cm-3
mass of water = (0.998 g cm-3)(80.0 cm3 ± 1%)
mass of water = 79.8 g ± 1%
mass of water = 79.8 g ± 0.8
q=mcT
q=(79.8 g ± 0.8) (4.18 Jg-1K-1)(71.8 K ± 0.2), given that the change in temperature, in °C, is equivalent to the change in temperature in K.
q=(79.8 g ± 1%) (4.18 Jg-1K-1)(71.8 K ± 0%)
q = 23900 J ± 1%
q = 23900 J ± 200
q = 23900 J ± 200 1kJ1000J
q = 23.9 kJ ± 0.2
Sample calculation 3: calculations performed to determine the total heat energy released per mole of the kerosene during its combustion for the first trial.
Energy per mole = q(initial mass of spirit lamp with kerosene- final mass of spirit lamp with kerosene) MM of kerosene
Energy per mole = 23.9 kJ ± 0.2 (115.93 g ± 0.01 - 114.45 g ± 0.01) 170.38 g mol-1 , given that the main hydrocarbon in kerosene is C12H26
Energy per mole = 23.9 kJ ± 1% 1.48g ± 1% 170.38 g mol-1
Energy per mole = 2750 kJ ± 2%
Energy per mole = 2750 kJ ± 50
The final values for the energy per mole for all trials can be averaged to reduce random experimental error. Once the calculations above are repeated for all fuel sources, the summarized results can be shown, in comparison to literature values for the heat of combustion to determine the most
This graph is identical to experiment 1 graph 1 as it compares the experimental and theoretical molar heats of combustion. Being the same concept this graph shows how the experimental molar heat of combustion is a scale less than the theoretical molar heat of combustion. The significance behind the scale of this graph where the experimental is lower that the theoretical, is how there is a pattern for which every fuel follows the same trend. Also it shows the average heats of combustion and which alkane has the leading heat output rate.
It was desired to compare a theoretical value of enthalpy of combustion to a literature value. To do this, the theoretical value was calculated using a literature value for the heat of sublimation of naphthalene, the heat of vaporization of water and average bond energies, given in Table 1 of the lab packet.1 Equations (1) and (5) were used to calculate the theoretical enthalpy of combustion of gaseous naphthalene, where n was the number of moles, m was the number of bonds, and ΔH was the average bond energy:
Fuel is usually consistent of octane, ethyl benzene, Trimethylpentane, toluene, and brutane. The reactions of 3.51kg of oxygen to one kilogram of fuel causes the production of 1.42kg pf water and 3.09kg of carbon dioxide. Simply written out, it would form an equation similar to C8H18+O2+CO2+H2O —> CO2+H2O. Dismally, this doesn’t work as a proper equation, as one would find it impossible to balance out. The balanced equation would be closer to C8H18+O2 —> CO2+H2O. Now that one is able to balance the equation, we’ll multiply the carbon dioxide and water by eight and nine respectively. Now there are too many oxygen on the right side of the equation (product). We move to the left side of the equation (reactant) and change the O2 to 25/2O2. Balanced, the equation is C8H18+25/2O2 —> 8CO2+9H2O. In terms of heat through energy, this equation can be converted to [(8 x -393.5) + 9 x -241.8)] - 250.1. What this means is for each time the reactant produces the CO2+H2O mixture, -5270kj of energy are produced.
The molar heat of combustion of a substance is the quantity of heat liberated when one mole of that substance is burnt completely in air. In the case of a hydrocarbon, the products are carbon dioxide and water.
The main purpose of this lab is to determine whether or not the number of carbon molecules relate to the amount of energy emitted measured through the temperature change over the course of 2 minutes. The main three tested fuels are fuels methanol (CH3OH), ethanol (C2H5OH), propanol (C3H7OH). Based on those formulas, Propanol alcohol has 3 carbons, Ethanol alcohol has 2 carbons and methanol has 1 carbon. Based on the hypothesis mentioned above “If the number of Carbon molecules in the fuel increases, the amount of energy over 2 minutes is going to increase”. Based on the data shown above, propanol fuel had the most temperature change meaning that it burned with the most energy. Following that is ethanol fuel having the second greatest temperature
The change in temperature is held constant at +20oC. This will be controlled to the best ability (see step 8, method). This is predominantly controlled to allow for consistency within the energy gained by the 100cm3 of water, and to maintain a consistent error in temperature change.
Density of solution = 100 cm3 x 1.00 g/cm3 = 100 g Temperature (K) = Temperature (C) + 273 △T= t final - t initial 304.26667-293.56667= 10.7 K moles(n)=molarity (mol dm3) x volume 2mol dm3 X 50 x 10 ˆ-3
Gas collection bulbs were filled with different proportions of each gas, proportional to the number of moles of gas present, and were tested for relative loudness, which represented power/ explosive energy via a pop test in which a match was struck near the bulb. The loudness was measured via a scale in which the most explosive mixture was given a ten and the least explosive given a zero. The ratios were tested for their ability to launch the bulb a distance during a rocket launch in which the bulb “rocket” was placed on a launch pad and ignited by a piezo sparker. Both dependent variables, distance and loudness represented overall explosion energy and were used to indicate which of the ratios represented the optimum ratio of hydrogen and oxygen to give water in a combustion reaction. To be considered valid, the amount of error in each data point must have fallen into the range of average deviation of the measurements.
The heat of combustion in kJ was -8.372, (Index 1). The heat of combustion in □(kJ/g) is found by (n∆H°rxn)/(mass of fuel burned) , is -20.93□(kJ/g) (Index 1). The heat of combustion in □(kJ/mol) which is found by (n∆H°rxn)/(number of moles), is -5912.4 □(kJ/mol) (Index 1).
Used the formula ‘q = mCΔT’ to calculate the heat/energy produced, (for each trial, then the average was found).
This experiment was dealing with the heating of water with different types and sizes of metals. In it you had to deal with heat capacity. Heat capacity is the amount of heat (normally expressed in calories, kilocalories, or joules) to raise a substances temperature by one (1) degree (normally expressed in Celsius or Kelvin). With the answer expressed in units of thermal energy per degree temperature. With substances having certain specific dimensions you can use to types of heat capacity, Molar Heat Capacity and Specific Heat Capacity.
After a further analysis of the results it has been seen that the averages of the results have proven the hypothesis to be correct. Hence the constructed calorimeter has been more efficient in reducing the enthalpy number upon burning ethanol compared to the copper calorimeters that have been previously used. Its has become known that due to more heat being absorbed the enthalpy number has significantly been improved by almost 10%.
The literature value for the enthalpy change of the last reaction which was provided by our teacher is -97 kJ mol-1. Agreeing this value, our result can be considered accurate.
Figure 1. The Dependency of Water on Temperature. From 12.0 °C - 20.0 °C the equation of best fit line was found to be y=-0.00162(g/((mL* °C)) )x+1.0015(g/mL) and the equation from 20.0 °C – 30.0 °C was found to be y=-0.000256 (g/((mL* °C)))x+1.0034 (g/mL).
The purpose of this experiment was to calculate the specific heat of an unknown object. To perform this experiment one must use a calorimeter to measure the heat transferred from a metal object, of known mass, to a known mass of water by measuring the temperature change. First one must apply heat to unknown object, then put object into water (record temperature of water before inserting object) and record the heat being released from object to find change in temperature. The purpose is to find specific heat, which is the amount of heat (J) applied to unknown object, then divide by mass (g) multiplied by change in temperature. Done correctly you should concluded specific heat to be: 0.456 J/g Celsius. The unknown object base on specific heat