1. HEAT AND ENERGY ….. “It’s Smokin”

2. Thermometry and Temperature Conversions  The temperature of a system is defined as simply the average energy of microscopic motions (Average Kinetic Energy) of a single particle in the system per degree of freedom. For a solid, these microscopic motions are principally the vibrations of the constituent atoms about their sites in the solid. For an ideal monatomic gas, the microscopic motions are the translational motions of the constituent gas particles. For multiatomic gases, vibrational and rotational motion should be included too.  Fahrenheit is a temperature scale named after the German-Dutch physicist Daniel Gabriel Fahrenheit (1686–1736), who proposed it in 1724. Where did his scale come from? There are many different version, take a look. http://en.wikipedia.org/wiki/Fahrenheit http://en.wikipedia.org/wiki/Fahrenheit  In 1742, Anders Celsius (1701 – 1744) created a "reversed" version of the modern Celsius temperature scale whereby 100 represented the freezing point of water and zero represented the boiling point of water.

3. Thermometry and Temperature Conversions  The Kelvin scale and the Kelvin are named after the Irish-born physicist and engineer William Thomson, 1st Baron Kelvin (1824 – 1907), who wrote of the need for an “absolute thermometric scale.”  The Kelvin unit and its scale, by international agreement, are defined by two points: absolute zero, and the triple point of specially prepared water (VSMOW). This definition also precisely relates the Kelvin scale to the Celsius scale. Absolute zero— the temperature at which nothing could be colder and minimal heat energy remains in a substance—is defined as being precisely 0 K and -273.15 °C. The triple point of water is defined as being precisely 273.16 K and 0.01 °C. This definition does three things:  It fixes the magnitude of the Kelvin unit as being precisely 1 part in 273.16 parts the difference between absolute zero and the triple point of water;  It establishes that one Kelvin has precisely the same magnitude as a one-degree increment on the Celsius scale;  and It establishes the difference between the two scales’ null points as being precisely 273.15 Kelvins (0 K = -273.15 °C and 273.16 K = 0.01 °C). Temperatures in Kelvin can be converted to other units per the table at the bottom.

4. Thermometry and Temperature Conversions

5. ***Note any change in Celsius is the same "change" in Kelvin. The Kelvin scale is 273 units greater than Celsius.

6. Convert These Temperatures ProblemHighlight to Reveal Answers 100 o C-->F(100 * 9/5) + 32 = 212 o F 212 o F--> K(212 - 32) * 5/9 + 273= 373K 40C --> K40 + 273= 313K 550 o C-->F(550 * 9/5) + 32 = 1022 o F -40 o F--> K(-40 - 32) * 5/9 + 273= 233K 2000K --> C2000 - 273= 1727 o C 500K --> F(500 - 273) * 9/5 + 32= 440 o F

7. Exothermic and Endothermic  Exothermic- the word describes a process that releases energy in the form of heat. Forming a chemical bond releases energy and therefore is an exothermic process. Exothermic reactions usually feel hot because it is giving heat to you.  Endothermic - a process or reaction that absorbs energy in the form of heat. Breaking a chemical bond requires energy and therefore is Endothermic. Endothermic reactions usually feel cold because it is taking heat away from you.

8. Exothermic and Endothermic

9. Heat Conversions  Energy is defined as the ability to do work. It is measured primarily in Joules (J), British Thermal Units (BTU) and calories (cal).  The Système International d'Unités (SI) unit for energy is the joule (J), after James Joule, who demonstrated that work can be converted into heat. Lifting a medium-sized potato a distance of 1 m (3.28 ft) would require approximately one joule of energy. Energy is often expressed as the calorie (cal), which is the amount of heat needed to raise the temperature of one gram of water by one degree Celsius. One calorie is equal to 4.184 joules. The Calorie (Cal), which is used to express the energy in food, is 1,000 calories. Note** it uses a capital "C" vs. the lower case "c".  1 cal =4.184J  1 kJ = 1,000 J  1kcal= 4.184 kJ  1kcal = 1,000 cal = 1 Calorie (food)

10. Specific Heat  Specific Heat Capacity (C or S ) - The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity of the substance. The quantity of heat is frequently measured in units of Joules(J). Another property, the specific heat, is the heat capacity of the substance per gram of the substance. The specific heat of water is 4.18 J/g° C.

11. Specific Heat  q = m x C p x Δ T  q = m x C p x (T f - T i )  q = amount of heat energy gained or lost by substance  m = mass of sample  C p = heat capacity (J/ °C g or J /K g) T f = final temperature T i = initial temperature

12. Sample Heat Questions Sample QuestionsAnswers 1. Calculate the amount of heat needed to increase the temperature of 250g of water from 20ºC to 46ºC. q = m x C x ∆T q = 250g x 4.18J/gºC x 26ºC q = 37,620J or 38kJ 2. Calculate the specific heat capacity of copper given that 204.75 J of energy raises the temperature of 15g of copper from 25º to 60º. q = m x C x ∆T C= q/m x ∆T C = 204.75J /(15g x 35ºC ) C= 0.39 J/gºC 3. 216 J of energy is required to raise the temperature of aluminum from 15º to 35ºC. Calculate the mass of aluminum. (Specific Heat Capacity of aluminum is 0.90 JºC -1 g -1 ). q = m x C x ∆T m= q/C x ∆T m= 216J/(0.90J/gºC x 20ºC ) m= 12g 4. The initial temperature of 150g of ethanol was 22ºC. What will be the final temperature of the ethanol if 3240 J was needed to raise the temperature of the ethanol? (Specific heat capacity of ethanol is 2.44 JºC -1 g -1 ). q = m x C x ∆T ∆T = q/m x C ∆T = 3240J/(150g x 2.44J/goC) ∆T = 8.85ºC T final = 22ºC +8.85ºC = 30.9ºC

13. Heat of Fusion of Water  Heat of Fusion-the amount of heat required to convert unit mass of a solid into the liquid without a change in temperature. (or released for freezing)  For water at its normal freezing point of 0 ºC, the specific heat of Fusion is 334 J/g. This means that to convert 1 g of ice at 0 ºC to 1 g of water at 0 ºC, 334 J of heat must be absorbed by the water. Conversely, when 1 g of water at 0 ºC freezes to give 1 g of ice at 0 ºC, 334 J of heat will be released to the surroundings.  Heat of Fusion of Water (H f = 334 J /g)  q= m H f  Note- The Heat of Fusion equation is used only at the melting/freezing transition, where the temperature remains the same only and that is why there is no temperature change ( Δ T) in this formula. It stays at 0 Celsius for water.  Note #2-Energy is required to melt and released when it freezes

14. Heat of Fusion of Water The diagram on the left shows the uptake of heat by 1 kg of water, as it passes from ice at -50 ºC to steam at temperatures above 100 ºC, affects the temperature of the sample. A: Rise in temperature as ice absorbs heat. B: Absorption of latent heat of fusion. C: Rise in temperature as liquid water absorbs heat. D: Water boils and absorbs latent heat of vaporization. E: Steam absorbs heat and thus increases its temperature. from-http://www.physchem.co.za/Heat/Latent.htm

15. Sample Problems Sample QuestionsAnswers 1. How much energy is required to melt 10.g of ice at its melting point? q= m H f q = 10.g x 334 J/g = 3340J or 3.34kJ 2. How much energy is released when 20. g of water is frozen at 0ºC? q= m H f q = 20.g x 334 J/g = 6680J or 6.68kJ

16. Heat of Vaporization of Water  Heat of Vaporization-the amount of heat required to convert unit mass of a liquid into the vapor without a change in temperature.  For water at its normal boiling point of 100 ºC, the heat of vaporization is 2260 J/g. This means that to convert 1 g of water at 100 ºC to 1 g of steam at 100 ºC,  2260 J of heat must be absorbed by the water. Conversely, when 1 g of steam at 100 ºC condenses to give 1 g of water at 100 ºC, 2260 J of heat will be released to the surroundings.  Heat of Vaporization of Water H v = 2260 J /g  q= m H v

17. Heat of Vaporization The diagram on the left shows the uptake of heat by 1 kg of water, as it passes from ice at -50 ºC to steam at temperatures above 100 ºC, affects the temperature of the sample. A: Rise in temperature as ice absorbs heat. B: Absorption of latent heat of fusion. C: Rise in temperature as liquid water absorbs heat. D: Water boils and absorbs latent heat of vaporization. E: Steam absorbs heat and thus increases its temperature. from- http://www.physchem.co.za/Heat/Latent.ht m

18. Sample Questions Answers 1. How much energy is required to vaporize 10.g of water at its boiling point? q= m H v q = 10.g x 2260 J/g = 22600J or 22.6kJ 2. How much energy is released when 20. g of steam is condensed at 100ºC? q= m H v q = 20.g x 2260 J/g = 45200J or 45.2kJ