# Intro to Thermodynamics

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Intro to Thermodynamics
Mr Nelson – 2010

Energy Energy is the ability to do work or transfer heat.
Energy used to cause an object to move is called work. Energy used to cause the temperature of an object to rise is called heat.

Potential Energy Potential energy is energy an object possesses by virtue of its position or chemical composition.

Kinetic Energy Kinetic energy is energy an object possesses by virtue of its motion. 1 2 KE =  mv2

Units of Energy The SI unit of energy is the joule (J). kg m2
An older, non-SI unit is still in widespread use: the calorie (cal). 1 cal = J 1 J = 1  kg m2 s2

Energy in Foods Most of the fuel in the food we eat comes from carbohydrates and fats.

System and Surroundings
The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). The surroundings are everything else (here, the cylinder, piston and the universe).

Work Energy used to move an object over some distance is work.
w = F  d where w is work, F is the force, and d is the distance over which the force is exerted.

Heat Energy can also be transferred as heat.
Heat flows from warmer objects to cooler objects.

Conversion of Energy Energy can be converted from one type to another.
For example, the cyclist above has potential energy as she sits on top of the hill.

Conversion of Energy As she coasts down the hill, her potential energy is converted to kinetic energy. At the bottom, all the potential energy she had at the top of the hill is now kinetic energy.

First Law of Thermodynamics
Energy is neither created nor destroyed. The total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

Exchange of Heat between System and Surroundings
When heat is absorbed by the system from the surroundings, the process is endothermic.

Exchange of Heat between System and Surroundings
When heat is released by the system into the surroundings, the process is exothermic.

Heat Capacity and Specific Heat
The amount of energy required to raise the temperature of a substance by 1C is its heat capacity.

Heat Capacity and Specific Heat
We define specific heat as the amount of energy required to raise the temperature of 1 g of a substance by 1 C .

Heat Capacity and Specific Heat
Specific heat, then, is Or the way we usually use the equation: q = m  c  T Specific heat = heat transferred mass  temperature change c = q m  T

Constant Pressure Calorimetry
By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

Constant Pressure Calorimetry
Because the specific heat for water is well known (4.184 J/g-K), we can measure H for the reaction with this equation: q = m  c  T

Enthalpy Enthalpy is the heat content of a system at constant pressure and volume: Enthalpy = H

Enthalpy The absolute amount of Enthalpy cannot be measured, so it is indirectly measured with temperature. When the system changes at constant pressure, the change in enthalpy, H, is H = Hfinal - Hinital

Enthalpy So, at constant pressure, the change in enthalpy is the heat gained or lost. H = q

Bomb Calorimetry

Different H’s Every delta H has the same type of definition: Hfusion
The heat released or absorbed when: Hfusion Hsolid Hformation Hreaction Hvaporization Hcombustion

Endo vs Exo A process is endothermic when H is positive.

Endo vs Exo A process is endothermic when H is positive.
A process is exothermic when H is negative.

Heat Transfer Problem 1 Calculate the heat that would be required an aluminum cooking pan whose mass is 400 grams, from 20oC to 200oC. The specific heat of aluminum is J g-1 oC-1. Solution DQ = mCDT = (400 g) (0.902 J g-1 oC-1)(200oC – 20oC) = 64,944 J

Heat Transfer Problem 2 What is the final temperature when 50 grams of water at 20oC is added to 80 grams water at 60oC? Assume that the loss of heat to the surroundings is negligible. The specific heat of water is J g-1 oC-1 Solution: DQ (Cold) = DQ (hot) mCDT= mCDT Let T = final temperature (50 g) (4.184 J g-1 oC-1)(T- 20oC) = (80 g) (4.184 J g-1 oC-1)(60oC- T) (50 g)(T- 20oC) = (80 g)(60oC- T) 50T = – 80T 130T =5800 T = oC

Chemical Reactions Chemical bonds are broken Atoms are rearranged
In a chemical reaction Chemical bonds are broken Atoms are rearranged New chemical bonds are formed These processes always involve energy changes

Energy Changes Breaking chemical bonds requires energy
Forming new chemical bonds releases energy

Exothermic and Endothermic Processes
Exothermic processes release energy C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4H2O (g) kJ Endothermic processes absorb energy C(s) + H2O (g) +113 kJ  CO(g) + H2 (g)

Energy Changes in Chemical Reactions
In an endothermic reaction there is more energy required to break bonds than is released when bonds are formed. The opposite is true in an exothermic reaction.

Enthalpy of Reaction The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = Hproducts − Hreactants

Enthalpy of Reaction This quantity, H, is called the enthalpy of reaction, or the heat of reaction.

Enthalpy is an extensive property. H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction. H for a reaction depends on the state of the products and the state of the reactants.

Hess’s Law Hess’s law states that “[i]f a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.” We can estimate H using published H values and the properties of enthalpy.

Hess’s Law

Hess’s Law The total enthalpy change depends only on the initial state of the reactants and the final state of the products.

Standard Enthalpies of Formation
Standard enthalpies of formation, Hf°, are measured under standard conditions (25 °C and 1.00 atm pressure).

Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring in three steps: C3H8 (g)  3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g)  3 CO2 (g) 4 H2 (g) + 2 O2 (g)  4 H2O (l)

Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring in three steps: C3H8 (g)  3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g)  3 CO2 (g) 4 H2 (g) + 2 O2 (g)  4 H2O (l)

Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring in three steps: C3H8 (g)  3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g)  3 CO2 (g) 4 H2 (g) + 2 O2 (g)  4 H2O (l)

Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
The sum of these equations is: C3H8 (g)  3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g)  3 CO2 (g) 4 H2 (g) + 2 O2 (g)  4 H2O (l) C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)

Calculation of H We can also use Hess’s law in this way:
H = nHf°products – mHf° reactants where n and m are the stoichiometric coefficients.

Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
H = [3( kJ) + 4( kJ)] – [1( kJ) + 5(0 kJ)] = [( kJ) + ( kJ)] – [( kJ) + (0 kJ)] = ( kJ) – ( kJ) = kJ

Energy in Fuels The vast majority of the energy consumed in the U.S. comes from fossil fuels.