From UCDavis Chemwiki link 1 and link 2

Remember from a previous discussion:

The 1st Law of Thermodynamics

The First Law of Thermodynamics states that energy can be converted from one form to another with the interaction of heat, work and internal energy, but it cannot be created nor destroyed, under any circumstances. The equation that supports the First Law of Thermodynamics mathematically is:

`DeltaU=q+w`

Here `DeltaU` is the total change in internal energy of a system, `q` is the heat exchanged between a system and its surroundings, and `w` is the work done by or on the system. Work is also equal to the negative external pressure on the system multiplied by the change in volume:

`w=-pDeltaV`

where `P` is the external pressure on the system, and `DeltaV` is the change in volume.

The internal energy of a system would decrease if the system gives off heat or does work. Therefore, internal energy of a system increases when the heat increases (this would be done by adding heat into a system). The internal energy would also increase if work were done onto a system. Any work or heat that goes into or out of a system changes the internal energy. However, since energy is never created nor destroyed (thus, the first law of thermodynamics), the change in internal energy always equals zero. If energy is lost by the system, then it is absorbed by the surroundings. If energy is absorbed into a system, then that energy was released by the surroundings:

`DeltaU_"system"=-DeltaU"surroundings"`

where  `DeltaU_"system"` is the total internal energy in a system, and `DeltaU_"surroundings"` is the total energy of the surroundings.

 

 Table 1

Process	Sign of heat (q)	Sign of Work (w)
Work done by the system	N/A	-
Work done onto the system	N/A	+
Heat released from the system- exothermic (absorbed by surroundings)	-	N/A

 

Table 2

The Process	Internal Emergy Change	Heat transfer of thermal energy (q)	Work `w=PDeltaV`	Example
q=0 Adiabatic	`+`	`0`	`+`	Isolated system in which heat does not enter or leave similar to styrofoam
DeltaV=0 Constant Volume	`+`	`+`	`0`	A hard, pressure isolated system like a bomb calorimeter
Constant Pressure	`+` or `-`	enthalpy (`DeltaH`)	`-PDeltaV`	Most processes occur are constant external pressure
DeltaT=0 Isothermal	`0`	`+`	`-`	There is no change in temperature like in a temperature bath

 

Measurement of Internal Energy

Equation (1) tells us how to detect and measure changes in the internal energy of a system. If we carry out any process in a closed container the volume remains constant), the quantity of heat absorbed by the system equals the increase in internal energy.

`q_v = DeltaU` (1)

A convenient device for making such measurements is a bomb calorimeter (Fig. 6), which contains a steel-walled vessel (bomb) with a screw-on gas-tight lid. In the bomb can be placed a weighed sample of a combustible substance together with `O_2 (g)` at about 3 MPa (30 atm) pressure. When the substance is ignited by momentarily passing electrical current through a heating wire, the heat energy released by its combustion raises the temperature of water surrounding the bomb. Measurement of the change in temperature of the water permits calculation of `q_v` (and thus `DeltaU`), provided the heat capacity of the calorimeter is known.

The heat capacity can be determined as in Example 1 from Heat Capacities or by igniting a substance for which `DeltaU` is already known.

Figure 6 A schematic of a bomb calorimeter.

EXAMPLE 11 When 0.7943g of glucose, `C_6 H_(12) O_6`, is ignited in a bomb calorimeter, the temperature rise is found to be 1.841 K. The heat capacity of the calorimeter is `6.746 kJ K^(-1)`. Find `DeltaU_m` for the reaction

`C_6 H_(12) O_6(s) + 6 O_2(g) -> 6CO_2 (g) + 6H_2 O(l)`

under the prevailing conditions. 

Solution:

The heat energy absorbed by the calorimeter in increasing its temperature by 1.841 K is given by

`q = CDeltaT = 6.745kJ K^(-1) & 1.841K = 12.42kJ`

Since this heat energy was released by the reaction system, we must regard it as negative. Accordingly,

`q_v = -12.42kJ = DeltaU`

We need now only to calculate the change in internal energy per mole, that is, `DeltaU_m`. Now

`n_"glucose" = 0.7953g * (1mol)/(180.16g) = 4.409 * 10^(-3) mol`

Thus

`DeltaU_m = (-12.42kJ)/(4.409) * 10^(-3)mol = -2817kJ mol^(-1)`

Subpages (1): Example 11