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Page history last edited by Douglas Martin 10 years, 6 months ago

Comments by Dr. Cooper: Good work. Make sure each figure is individually titled and labeled. You also need to number each equation, and preferably, don't include an equation in a sentence.



Black (Svetlana Makarova) Purple (Katharine Mehalic) Maroon (Doug Martin)




Heat transactions


As mentioned in the previous lecture, the internal energy of a system cannot be measured directly. However, the change in internal energy may be calculated: 



dU = dq +dwexp + dwe          (1)


The above equation incorporates the energy supplied (q > 0) or obtained (q <0) as heat when a system undergoes a change of state under constant volume conditions, taking into account the expansion work (wexp) and any other additional work done by the system (we). Under constant volume (such as in a rigid container), the gas cannot expand, thus the system cannot do any expansion work (wexp = 0). In addition, if the system cannot do any other work under these conditions (such as electric work,), we can assume wex = 0. Thus, the above equation becomes: 


U = dq                          (2)

Formulaq                      (3) 


In other words, in order to obtain the change in the internal energy of a system, we can measure the energy as heat given off or absorbed by a system under restricted volume. By eliminating the variables of work from equation (1), this gives us a more direct way of measuring the change in internal energy by measuring the amount of heat transferred. 






Calorimetry is a quantitative method of measuring heat transfer that takes place due to physical and chemical changes of a reaction. The most common way to measure heat transfer, and thus a change in internal energy, is to use a calorimeter, a device used to measure the transfer of energy as heat by observing changes in temperature (FormulaT). This change in temperature reflects the heat released or absorbed during a chemical reaction. It is important to note that before any measurements can be made, the device is calibrated and the calorimeter constant (C) is obtained in order to accurately quantify the heat of a reaction. 


How do we obtain a calorimeter constant? 


Although multiple different types of calorimeters exist, the most common device used in calorimetry is a bomb calorimeter. In order to obtain a calorimeter constant, a constant electrical current (I) is sent through the device for a known period of time (t) with a source of known voltage (V), which results in electrical heating. The temperature changes are then measured in order to obtain the calorimeter constant. Thus, in order to obtain C: 


C = energy supplied to the calorimeter / increase in temperature            (4)


Example 1. An electrical current supplies 10 kJ of energy to the bomb calorimeter, which results in 2 °K increase in temperature. Calculate the calorimeter constant. 


Using these values:


C = 10 kJ / 2 °K = 5 kJ °K-1


This calorimeter constant suggests that in order to increase the temperature of the water bath by 1 °K, we must supply 5 kJ of energy to the system.



The changes in temperature of a calorimeter is proportional to the changes in heat of a reaction. Note, the calorimeter constant is specific to a particular experimental set-up and may vary from one device to another, although the calorimeters might be exactly the same. For this reason, calorimeter constant becomes an important parameter in measurements of heat (which acts as a proportionality constant in calculating heat transfer) :


q = C FormulaT                      (5)


The electrical heating method for obtaining the calorimeter constant involves calculation of heat, using known electrical current, voltage and time (which can be controlled by using a multimeter and a stop watch):


q = I V t                         (6)


Thus, by using these known parameters we can quantify heat transferred. We can then use equation (5) to solve for calorimeter constant (see example below).


Example 2. If a 10 A current is passed to the calorimeter, which comes from a 12 V source for 300 s, how much energy is supplied as heat? What is the calorimeter constant if an increase of 5.5 °K in temperature is observed?


                        Use equation (6) to solve for heat:


                        q = I V t = (10 A) (12 V) (300 s) = 36 x 104 A V s = 36 kJ


     Use equation (5) to solve for calorimeter constant (by rearranging the equation): 


                        C = q / FormulaT  = (36 kJ) / (5.5 °K) = 6.5 kJ °-1



 Figure 1. Bomb calorimeter                        

 Another common method for obtaining calorimeter constant involves using a known substance or a standard (such as benzoic acid standard), which is placed into the calorimeter (or the bomb), along with a small amount of water onto the sample plate. The device is then filled with excess oxygen through the oxygen inlet (see Figure 1) and then immersed into a water bath (that is precisely measured) that contains a thermometer. In the bomb calorimeter, the sample is exploided which results in a combustion reaction that gives off heat. The heat is transferred to the water surrounding the bomb, which is then measured as temperature. It is important to note that heat may be also absorbed by the sample, accordingly temperature for this particular heat transfer may also be measured and quantified as heat (q).




 In Summary


Since internal energy cannot be measured directly, the changes in internal energy may be determined. In order to quantify the changes in the internal energy, we restrict the system to a constant volume where it cannot expand or do any additional work, such as in a calorimeter. A bomb calorimeter is often used to measure heat transfer, where temperature is measured and multiplied by the calorimeter constant (which acts as a proportionality constant and is specific to the experimental set-up, see equation 5). 


Calorimetry is often used to measure the amount of energy content in foods, which we see on nutritional labels. The way we often think of calories is the energy intake that allows us to do work (such as driving, writing, lifting items, etc.) Similarly, measuring the heat transfer for the reaction of interest gives us a way to quantify the changes in internal energy, which we can then relate to the amount of work that particular system can do.



Heat Capacity


As more heat is transferred to an object or substance, the temperature of the object or substance will rise, however the magnitude of temperature change produced by a certain amount of heat can vary widely from substance to substance.  The temperature change in an object or substance created by a certain amount of heat is known as the heat capacity and is symbolized by a capital C.  Basically, the greater the heat capacity, the greater the amount of heat required to produce the given increase in temperature.  To illustrate heat capacity it is best to plot the internal energy (U) against temperature (T) (see figure 2)




As you slide the tangent line along the curve, the slope of the tangent is the heat capacity at that temperature.  Formally defined heat capacity at a constant volume is:


Formula (7)



This value, CV, tells us how much the internal energy needs to be changed to change the temperature. 





For the above example, FIGURE 3, you must make the volume constant in order to get the tangent line which depicts heat capacity. 

As you increase the internal energy, U, the temperature, T, will also increase.  The equation proves this as it is in direct correlation (to maintain the same value for CV you must increase ∂U the same amount that you increase ∂T. 

Heat capacity is an extensive property, for example, 1g H2O has 1/100 of the heat capacity of 100g H2O.  When working with problems which involve heat capacity they may be presented in a variety of different ways.  Molar heat capacity is just the heat capacity divided by the number of moles.  As you already know, all molar properties are intensive and thus so is molar heat capacity.  Finally, another variant of heat capacity which you may run into is specific heat capacity is just the heat capacity divided by the mass. 

Sometimes when dealing with heat capacities you can even consider the process independent of temperature when you are looking for approximate values.  You can do this when dealing with a small range of temperature which is at or above room temperature. 

You can manipulate the following equation:




to a more usable one through a few steps.  First multiply both sides by ∂T to get the following:


∂U = CV ∂T       (8)

This equation shows that because the volume remains constant, a small change in temperature will lead to a small change in energy.  As well we can say that such a change could be depicted by the following:

∆U = CV ∆T          (9)

This shows that a change in temperature will bring about a proportionate change in energy required.  This leads us to use the following equation which uses the symbol for required energy instead of internal energy:

qV = CV ∆T           (10)

This equation indicates that a large heat capacity will result in a small increase in temperature of a substance.  It also indicates that a substance with infinite heat capacity will have no increase in temperature.  This could be if the energy is being used to carry out a phase change, and thus there will be no energy put into the system which will result in a stable temperature. 




Enthalpy is, in the simpliest terms, when the internal enrgey and the output of energy, as heat, are not equal in an open  system (a system that has constant pressure, but can change in volume). The heat that in a closed system that would have contributed to the internal energy would go to work in an open system.  The image below shows how this looks, where the piston in the system is not fixed, and allows for the expansion of the system. Also, the image also notes that ΔU is less than the work (q).


Figure 1:  Comparison of Enthalpy versus Internal Energy











However, dH= dq when the pressure is at a constant and the only work is expansion. The enthalpy of an open system at constant pressure is equal to the energy, as heat, given to the system.

 Measuring Enthalpy Change

To measure enthalpy change, a calorimeter can be used. To determine the enthalpy change calorimetrically, one just needs to monitor the temperature change and have the sample stay at a constant pressure. An isobaric calorimeter can be used, which can be as simple as a styrofoam cup with a thermometer stuck through it. Also, a normal bomb calorimeter can be used to measure the change in enthalpy.

For liquids and solids, the molar volumes tend to be small when compared to gases, that pV or pVm (pressure x molar Volume) can be neglected from the molar enthalpy. The equation for molar enthalpy is Hm = Um +pVm ≈ Um.

In gases, the ideal gas law (pV= nRT) can be incorporated into the enthalpy equation (H = U +pV).  The resulting equation is H= U + nRT and change of enthalpy equation is ΔH= ΔU + ΔngRT, where ng is the stoichiometric change in number of moles of gas.



The molar enthalpy of vaporization of methanol is 43.5 kJ/mol at its boiling point (176 K). What is the molar internal energy change? How long would a 12 V source need to supply a 0.5 A current in order to vaporize a 10.0 g sample?



CH3OH(liq)à CH3OH(g)

ΔHm= ΔUm + ΔngRT

43.5 = ΔUm + (1 x 8.314 x 10-3 x 176)

ΔUm= 29.7 kJ/mol

For the second half of the problem:

ΔHm= 43500 J/mol

ΔHm= 43500 J/mol x (10 g / 32.05 g/mol) = 13572.5 J

ΔH= qp

13572.5= 12 x 0.5 x t

t= 2262.1 s





Comments (1)

Svetlana Makarova said

at 1:03 am on Sep 16, 2009

Hey Doug and Katharine, I hope that you guys received my emails regarding our assigned lecture. Feel free to email me about any of the information. Also, I have yesterday's lecture recorded, if you guys wanted to use it.


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