Entry Task: Block 2 Sept 27

Entry Task: Block 2 Sept 27 Question: The specific heat of copper metal is 0.385 J/g-K. How many joules of heat energy are necessary to raise the temperature of a 5.0 g block of copper from 25.0oCto 88.5oC? You have 5 minutes!!

Agenda: • Sign off on Ch. 5 sec 1-5 notes • HW: Calorimetric problems

I can… • Describe the relationship between energy (E), work (w) and heat (q). • Define the state of function and how it relates to enthalpy • Calculate the enthalpy change in a reaction • Use the calorimetry equation to solve for q. • Solve for q for solutions reactions.

Energy • The ability to do work or transfer heat. • Work: Energy used to cause an object that has mass to move. • Heat: Energy used to cause the temperature of an object to rise.

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

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

kg m2 1 J = 1  s2 Units of Energy • The SI unit of energy is the joule (J). • An older, non-SI unit is still in widespread use: The calorie (cal). 1 cal = 4.184 J

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 and piston).

Work • Energy used to move an object over some distance. • 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.

Transferal of Energy • The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.

Transferal of Energy • The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall. • As the ball falls, its potential energy is converted to kinetic energy.

Transferal of Energy • The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall. • As the ball falls, its potential energy is converted to kinetic energy. • When it hits the ground, its kinetic energy falls to zero (since it is no longer moving); some of the energy does work on the ball, the rest is dissipated as heat.

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

Internal Energy The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E.

Internal Energy By definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system: E = Efinal−Einitial

Changes in Internal Energy • If E > 0, Efinal > Einitial • Therefore, the system absorbed energy from the surroundings. • This energy change is called endothermic

Changes in Internal Energy • If E < 0, Efinal < Einitial • Therefore, the system released energy to the surroundings. • This energy change is called exothermic.

Changes in Internal Energy • When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). • That is, E = q + w.

E, q, w, and Their Signs

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 absorbed by the system from the surroundings, the process is endothermic. • When heat is released by the system to the surroundings, the process is exothermic.

State Functions Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

State Functions • However, we do know that the internal energy of a system is independent of the path by which the system achieved that state. • In the system below, the water could have reached room temperature from either direction.

State Functions • Therefore, internal energy is a state function. • It depends only on the present state of the system, not on the path by which the system arrived at that state. • And so, E depends only on Einitial and Efinal.

State Functions • However, q and w are not state functions. • Whether the battery is shorted out or is discharged by running the fan, its E is the same. • But q and w are different in the two cases.

Heat (Enthalpy) change DH • Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure

Enthalpy Hproducts< Hreactants Hproducts> Hreactants DH < 0 DH > 0

H2O (s) H2O (l) DH = 6.01 kJ Endothermic and Exothermic • A process is endothermic when H is positive. • 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.

DH = -890.4 kJ CH4(g) + 2O2(g) CO2(g) + 2H2O (l) Endothermic and Exothermic • A process is exothermic when H is negative. • 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.

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 ofreaction, or the heat ofreaction.

The Truth about Enthalpy • Enthalpy is an extensive property. • CH4(g) + 2O2(g) CO2(g) + 2H2O(l)∆H of -890kJ • 2CH4(g) + 4O2(g) 2CO2(g) + 4H2O(l)∆H of -1780 kJ • H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction. • CO2(g) + 2H2O (l) 2O2(g) + CH4(g)∆H of +890kJ • H for a reaction depends on the state of the products and the state of the reactants. • CH4(g) + 2O2(g) CO2(g) + 2H2O(l)∆H of -890kJ • CH4(g) + 2O2(g) CO2(g) + 2H2O(g)∆H of -802kJ

5.3 problem Hydrogen peroxide can decompose to water and oxygen by the reaction 2 H2O2(l) 2 H2O(l) + O2(g) H = –196 kJ Calculate the quantity of heat released when 5.00 g of H2O2(l) decomposes at constant pressure. We have to find out how many moles of H2O2 are in 5.00g. 1 mole H2O2 5.00 g = 0.0147 mol H2O2 34 g H2O2 And there are 2 moles of H2O2 to release -196 kJ of energy. -196 kJ 0.0147 mol H2O2 = -14.4 kJ 2 mol H2O2

5.28 a) Exothermic or endothermic? Its endothermic (pos ∆H) Consider the reaction: CH3OH(g)  CO(g) + 2H2(g) ∆H = +90.7 b) Calculate amount of energy when 45.0 g is transferred. = 1.40 molCH3OH 1 mole CH3OH 45.0 g 32.04 g CH3OH +90.7 kJ 1.40 mol CH3OH = 127 kJ 1 mol CH3OH

5.28 c) If the enthalpy is changed to 16.5 kJ, how many grams of hydrogen gas are produced? Consider the reaction: CH3OH(g)  CO(g) + 2H2(g) ∆H = +90.7 =0.364mol H2 2 mole H2 16.5 kJ +90.7 kJ 2 g H2 0.364 molH2 = 0.727g 1 molH2

5.28 d)How many joules of heat are released when 10.0g of CO reacts completely with hydrogen to form CH3OH at constant pressure Consider the reaction: CH3OH(g)  CO(g) + 2H2(g) ∆H = +90.7 =0.357mol CO 1 mole Co 10.0 g CO 28 g CO +90.7 kJ 0.357 molCO = -32.4 kJ 1 molCO

Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow.

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

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

heat transferred Specific heat= mass  temperature change q Specific Heat (c)= m T Heat Capacity and Specific Heat Specific heat, then, is

Since we know the specific heat of most substances, we will rearrange the equation to solve for the amount of heat (q) released or absorbed which is what we don’t know.

q = c x m x ΔT • q = the heat absorbed or released • c= the specific heat of water = 4.184 Joules • m= mass of the sample in grams • ΔT= the temperature difference (Tf- Ti)

If the temperature of 34.4g of ethanol increases from 25.0˚C to 78.8˚C, how much heat has been absorbed by ethanol? Specific heat of ethanol is 2.44 J/g C q=cmT • q = the heat absorbed or released • c= the specific heat of ethanol= 2.44 J/g C • m= 34.4 g • ΔT= 53.8 4515 J or 4.515 x 103 J

A 4.50 g nugget of pure gold absorbs 276 J of heat. What was the final temperature of gold if the initial temperature was 25.0˚C? q=cmT • q = 276 J • c= the specific heat of gold= 0.129 Joules • m= 4.5 g • ΔT= 25-X T = q/cm = 276J = 475°C 0.129 J/(gC) x 4.5 g Since the temperature started at 25.0°C, the final temperature is 500°C.

A 155 g sample of an unknown substance was heated from 25.0˚C to 40.0˚C. In the process, the substance absorbs 5696 J of energy. What is the specific heat of the substance? What is this substance? q=cmT • q = 5696J • c= the specific heat = X Joules • m= 155 g • ΔT= 25-40=15 c= q/Tm = 5696J = 2.44 J/(gC) 15 C x 155g

5.4 problem q=cmT (a) Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of the rocks is 0.82 J/g–K. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0 C. (b) What temperature change would these rocks undergo if they emitted 450 kJ of heat? a. (0.82J/g-K)(50000g)(12K) = 4.92 x 105 J b. (0.00082 kJ/g-K)(50000g)(X) = 450 kJ = 10.9 or 11.0 temp increase (X) = 450 kJ (41)

WHAT ABOUT REACTIONS? 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.

Reactions in Solutions 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 = c m T

Entry Task: Block 2 Sept 27