I. Energy. - power which may be translated into mo8on, overcoming resistance, or effec8ng physical change; the ability to do work.

Transcription

1 I. Energy - power which may be translated into mo8on, overcoming resistance, or effec8ng physical change; the ability to do work.

2 A. Work response to a force moving an object some distance W=F x D Joules = Newtons (Kg x m/s 2 ) x m =

3 B. Law of Conserva8on of Energy- Energy is neither created nor destroyed during chemical change. - energy and mass can be converted, but it is only observable during nuclear reac8on.

4 C. Poten&al Energy stored energy, not doing work. Ex. Chemical E, Magne8c E, Electrical E D. Kine&c Energy- Energy of mo8on, work being done Ex. Heat E, Sound E, Light E

5 E. Energy is measured by the amount of work work it can do or the amount of heat it can become. 1. Joules (the SI base unit) is the unit of work. derived from kg x m 2 /s 2 or Nm. W=FxD J = N(kg x m/s 2 ) x m

6 2. Calorie- commonly used measurement of heat. - 1 calorie is the heat required to raise 1 g of water by 1 C. - 1 calorie = J - 1 Calorie ( kilocalorie) = 1000 calories

7 F. Energy of Chem. Reac8ons- every chemical reac8on involves a change in energy, but the total amount of energy is always conserved. 1. Exergonic reac&on- chemical change involving the release of kine8c E. * Exothermic releases heat - Reactants à Products + Kine&c E - More poten8al E in reactants than products. Pot.E is converted to Kin. E. - Some exergonic reac8ons require Ac&va&on E start the reac8on

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9 2. Endergonic (Endothermic) Reac&on- reac8on that absorbs Kin. E h\p:// v=lrlonfvjmlu - Reactants + Kin E à Products - Products have more Pot. E than the reactants. - Requires con8nuous input of E

10 G. Kine&c Theory of Heat and Temperature- heat is transferred to cooler bodies un8l temperature is equal throughout the system. 1. Par8cles of ma\er contain Kin E and are constantly colliding. - Higher Kin E = Faster mo8on= Higher Temp - Faster moving par8cles collide with slower ones transferring energy thus raising the temperature un8l thermal equilibrium is reached. h\p:// n90pfibje

11 2. Heat- quan8ty of Kin E measured in Joules h\p:// 3. Temperature- concentra8on of heat measured in degrees. - measures the average Kin E of a substances par8cles. a. Absolute zero is the theore8cal lowest possible temperature. Ma\er has no Kin E. - Basis of the Kelvin Scale - 0 K = absolute zero = C = F

12 b. Celcius scale- Andreas Celcius, based on the freezing and boiling points of water. - O C = 273 K - K= C C = K 273 c. Farenheit scale- Daniel Farenheit 0 0 F = ice,water, and ammoium chloride 32 0 F = ice and water 0 F = (9/5 x 0 C) F = body heat 0 C = ( 0 F - 32) x 5/ F = boiling water

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14 H. Measurement of heat. 1. Calorimetry- measure of heat absorbed or released during a chemical or physical change. 2. Calorimeter- Device used to measure the change in temp. of water surrounding a reac8on chamber, or measures the direct change of temp of a water sample during heat transfer.

15 Chemical reac8on calorimeter

16 Heat transfer calorimeter (physical)

17 3. Measuring heat transferred. a. Heat transferred (J) = mass (g) x change of temp ( O C) x specific heat (J/(gX 0 C)) * * specific heat is the heat required to raise 1 g of a substance by 1 0 C ** if we are using a calorimeter then the measurements of the equa&on are from the water in the calorimeter. Specific Heat of Water is J/ (g x 0 C)

18 EX. How much heat is required to raise the temperature of 1000 g of water from 20.0 C to 80.0 C? EX. A sample of ice goes from 0 C to 25.0 C while gaining 2.50 x 10 4 J of heat. What is its mass?

19 EX. How much heat is required to raise the temperature of 1000 g of water from 20.0 C to 80.0 C? Q = 1000g x (80.0 C- 20.0C) x 4.18J/gxc Q= 2.51 x 10 5 EX. A sample of ice goes from 0 C to 25.0 C while gaining 2.50 x 10 4 J of heat. What is its mass? 2.50 x 10 4 J = M x (25.0-0C) x 4.18 J/gxc 2.50 x 10 4 J = M x J/g 2.50 x 10 4 J/104.5 j/g = M 2.39 x10 2 g =m

20 Heat of solu8on Example Problem 2: A chemistry student dissolves 4.51 grams of sodium hydroxide in ml of water at 19.5 C (in a calorimeter cup). As the sodium hydroxide dissolves, the temperature of the surrounding water increases to 31.7 C. Determine the heat of solu8on of the sodium hydroxide in J/g. Once more, the solu8on to this problem is based on the recogni8on that the quan8ty of energy released when sodium hydroxide dissolves is equal to the quan8ty of energy absorbed by the water in the calorimeter. In equa8on form, this could be stated as Q NaOH dissolving = - Q calorimeter

21 Q calorimeter = m C ΔT Q calorimeter = (100.0 g) (4.18 J/g/ C) (31.7 C C) Q calorimeter = J The assump8on is that this energy gained by the water is equal to the quan8ty of energy released by the sodium hydroxide when dissolving. So Q NaOH- dissolving = J. (The nega8ve sign indicates an energy lost.) This quan8ty is the amount of heat released when dissolving 4.51 grams of the sodium hydroxide. When the heat of solu8on is determined on a per gram basis, this J of energy must be divided by the mass of sodium hydroxide that is being dissolved. ΔH solu8on = Q NaOH- dissolving / m NaOH ΔH solu8on = ( J) / (4.51 g) ΔH solu8on = J/g ΔH solu8on = x10 3 J/g (rounded to three significant figures)

22 Heat capacity Amount of heat required to raise a sample of a substance by one degree celsius. Extensive property. Calculated by dividing heat (J) divided by temperature change (C). Heat (J)/ Temp. change (C)

23 Ex. A sample of a metal at 95.0 C is placed in calorimeter with ml of water at 25.0 C. The metal causes the temperature of the water to rise to 28.5 C. What is the heat capacity?

24 Solu8on HT H2O = g x (28.5 C- 25.0C)x 4.18J/gC HT H2O = 1463 J so HT metal = J HC = 1463J/ (95.0C 28.5C) = 1463J/ 66.5C HC = 22.0 J/C