Example: 1.40 g of vegetable oil is placed in a bomb calorimeter with excess oxygen and ignited with a spark. If the calorimeter temperature changes from 20.000 °C to 21.195 °C, find the energy released per gram of oil . The calorimeter contains 2.50 kg of water. The calorimeter without water has a heat capacity of 1.00 kJ°C^{1}.
q = nC_{p}T, where C_{p} is the molar heat capacity at constant pressure (= 75.3 J C^{1}mol^{1} for water).
q= q_{water} + q_{calorimeter}
q_{water}= {(2.50 kg H_{2}O)(1000g/kg) / (18.02 g H_{2}O/mole)}{75.3 J C^{1}mol^{1}}{1.195 °C} = 1.25 x 10^{4}J
q_{calorimeter} = CT = (1.00 x 10^{3}J°C^{1})(1.195 °C) = 1.195 x 10^{3}J
q_{tot}= 1.25 x 10^{4}J + 1.195 x 10^{3}J = 1.369 x 10^{4}J
E/g = (1.369 x 10^{4}J) / 1.40 g = 9.78 kJ/g Notice that this is now the energy released, and it will also be the energy you could potentially get from consuming this much oil, since we are working with state functions, and the pathway (fire or metabolism) doesn't matter.
Hess's Law
Hess's Law states that changes in enthalpy in any process depends only on the nature of the reactants and products, and is independent of the number of steps in the process or the pathway taken. Hess's Law is thus a result of the fact that enthalpy is a state function.
Hess's Law turns out to be extremely useful for determining the energy of various processes, and thus the conditions necessary for reactions to proceed. The pathway independence is particularly nice, because we can look at processes that have never been observed to occur in a laboratory, and reasonably discuss the thermochemistry of processes that might occur at the Earth's core or the heart of a comet etc.
In order to use Hess's Law we need to keep some properties of enthalpy in mind.
When a reaction is written in reverse, the sign of H is reversed. The magnitude of H is directly proportional to the amount of reactants. Thus if the coefficients of a reaction are multiplied, then H is multiplied by the same amount.
Formation Reactions and Standard Enthalpies of Formation: If we consider a reaction in which compounds are formed from elements in their standard states then the value of H is the standard enthalpy (heat) of formation.
Standard States:
With this information we can now find H for any chemical reaction!
Example: Find the value of H for the reaction:
2 CO_{2} + 7 H_{2} C_{2}H_{6} + 4 H_{2}O_{(g)} From Table find H values:

now we can put the reactions together, adding the enthalpies for products, and subtracting enthalpies for reactants (since the reaction directions are reversed), and multiplying enthalpy values by the coefficients of the balanced equation (since all of the formation reactions were based on coefficients of one).
2 (H_{CO2}) 7 (H_{H2}) + (H_{C2H6}) + 4 (H_{H2O}) = H_{rxn}
Electromagnetic Radiation comprises the various types of forms of radiation which propagate through space not associated with mass. The visible spectrum encompasses a very narrow region of the overall electromagnetic spectrum as seen below and on figure 7.2 on p 276 of your text.
Electromagnetic radiation behaves in most circumstances as waves [Figure 7.1 p 276] and can thus be characterized as waves.
Three parameters determine a wave:
These parameters are related by the the expression:
For electromagnetic radiation (light) the speed is defined in a vacuum: v = c = 2.9979 x 10^{8}m/s
Syllabus / Schedule 
© R A Paselk
Last modified 4 March 2011