Internal Energy (also known as Heat)

Conversions between scales: O F O C K Water boils 212o 100o 373 Human body temperature 98o 37o 310 Room Temperature 68o 20o 293 Water freezes 32o 0o 273 Absolute zero -459.7o -273o Absolute Zero- avg. KE of substance is a minimum; lowest possible temperature

Conversions between scales: (cont.) Celsius Kelvin K = oC + 273 Examples: 200 oC = ? K K = 200 oC + 273 = 473 K 400 K = ? oC 400 = oC + 273 = 127 oC -30 oC = ? K K = -30oC + 273 = 243 K 100 K = ? oC 100 = oC + 273 = -173 oC

B. Temperature change and its affect on matter Heat is energy, so when it is added to or removed from a substance, the mass of that substance does not change 2. When objects are heated, they expand 3. When objects are cooled, they contract

Water contracts after it melts until it reaches 4o, then it expands. Generally, gases expand and contract the most with changes in temperature, followed by liquids. Solids expand and contract the least Exception: Water Temperature oC Volume of 1g of water (ml) 4 10 1.0001 1.0000 Water contracts after it melts until it reaches 4o, then it expands.

Since the volume of water changes while its mass doesn’t, another property must change as well- Its Density Temperature oC Density (g/ml) 4 10 1.0000 0.99985 Water contracts after it melts until it reaches 4o, then it expands.

Hydrogen bonding in ice

Hydrogen bonding in ice

II. Heat and Energy A. Heat- the energy flowing between 2 objects because of a difference in temperature. Heat flows from the warmer object to the cooler object until the objects reach the same temperature 50o F 40o F 42o F The unit of heat is the Joule or calorie. 1 calorie = 4.19 J

Phase Change Graphs: Cooling Curve Gas Condensing Liquid Freezing Solid

The PE of the molecules in the substance changes 1. Melting and Freezing The temperature of most materials remains constant during melting or freezing (avg. KE remains constant) When a substance melts, it must gain energy; when a substance freezes, it must lose energy. But wait: If the Avg. KE remains constant, where does the energy go or come from when the object is melting or freezing? The PE of the molecules in the substance changes

q = mHf Melting and Freezing (cont.) HEAT OF FUSION: Amount of heat needed to melt 1 gram of a particular substance or the amount of heat lost when 1 gram of a substance freezes. q = mHf q = heat (in J) m = mass (in g) Hf = Heat of fusion (J/g) (334 J/g for water)

q = mHf q = mHf Heat of Fusion Examples: 1) How much heat is needed to melt 200 g of ice? q = mHf = (200g) (334 J/g) = 66,800 J 2) If 50,100 J of heat is released when a sample of water is frozen, how much water was in the sample? q = mHf 50,100 J = m (334 J/g) m = 150 g

Melting and Freezing (cont.) Factors affecting melting/freezing points 1. A substance dissolved in a liquid decreases its freezing/melting point. Salt “melts” ice by lowering its freezing temperature. Antifreeze prevents water from freezing

Melting and Freezing (cont.) 2. Pressure The pressure of a skate blade on the ice melts the ice directly beneath it. The ice refreezes after the blade is removed. Applying pressure on the surface of ice lowers the melting point of the ice

2. Boiling (or Vaporizing) and Condensing The temperature of most materials remains constant during boiling or condensing (avg. KE remains constant) When a substance vaporizes, it must gain energy; when a substance condenses, it must lose energy. While a substance boils or condenses, there is a change in the PE of the molecules

q = mHv Boiling and Condensing (cont.) HEAT OF VAPORIZATION: Amount of heat needed to vaporize 1 gram of a particular substance or the amount of heat lost when 1 gram of a substance condenses. q = mHv q = heat (J) m = mass (in g) Hv = Heat of vaporization (J/g) (2260 J/g for water)

Why does it take more energy to vaporize a substance than it does to melt it? 1. Molecules need more energy to be in the gas phase than they do the liquid phase. 2. Molecules need energy to overcome attractive forces (hydrogen bonding, van der Walls, cohesion/adhesion)

q = mHv q = mHv Heat of Vaporization Examples: 1) How much heat is released when 400 g of steam condenses? q = mHv = (400g) (2260 J/g) = 904,000 J 2) If 565,000 J of heat is absorbed when a sample of water is vaporized, how much water was in the sample? q = mHv 565,000 J = m (2260 J/g) m = 250 g

Boiling and Condensing (cont.) Factors affecting boiling/condensation points 1. A substance dissolved in a liquid increases its boiling/condensation point. Antifreeze not only prevents freezing but also prevents water in the cooling system from boiling away.

Boiling and Condensing (cont.) 2. Pressure Increasing the pressure on a liquid raises the boiling point of that liquid Example: Pressure cooker We say that water boils at 100 oC (32 oF, 373 K). This (and other temp’s on a reference chart) is known as the normal boiling point. The normal boiling point occurs at Standard Pressure (101.3 kPa, 1 atm, 760 mmHg, or sea level pressure)

Boiling and Condensing (cont.) Technically water (or other liquids) can boil at any temperature. Boiling occurs when the pressure of the vapor of the substance is equal to atmospheric (air) pressure. See Table H in the Chemistry Reference Tables. Examples: 1) What is the normal boiling point of propanone? 56 oC 2) At what pressure will water boil at 70 oC 30 kPa 3) At a pressure of 80 kPa, ethanoic acid boils at 110 oC

3. Evaporation: Process where the faster moving (more KE) molecules at a liquid’s surface escape into the gas phase. When this happens, the temperature of the substance decreases. Perspiration Examples: Dog panting

C. Specific Heat Capacity (or Specific Heat, or Heat Capacity) The amount of heat needed to raise 1 gram of a substance 1 oC ( or how much needs to be lost to lower the temperature 1 oC) 40oC 41oC Amount of heat equal to specific heat added to substance **The calorie is defined by how much heat is needed to raise the temperature of 1g of water 1 oC

Specific heat capacities of some substances: Phase J/g oC Ammonia liquid 2.43 Aluminum solid 0.90 Copper 0.39 Iron 0.45 Lead 0.13 Titanium 0.52 ice 2.05 water 4.18 steam gas 2.01

Understanding Specific Heat Capacity Assuming that each of the following substances have equal masses, start at the same temperature, and have the same amount of heat added to them: Iron, copper, aluminum, lead 1. Which substance will heat up the quickest? Lead- smallest specific heat capacity 2. Which substance will heat up the slowest? Aluminum- largest specific heat capacity 3. Which substance will cool off the quickest? Lead- smallest specific heat capacity

Q = m c DT Q = heat (in J) m = mass (in g) The heat added to (or removed from) a substance can be found by the following formula: Q = heat (in J) m = mass (in g) C = specific heat capacity (in J/g oC) DT = temperature change (in oC or K) Q = m c DT

Sample questions How much heat is needed to raise the temperature of 100 g of iron from 20 oC to 40 oC? Q = m C DT = (100g) (0.45 J/g oC) (20 oC) = 900 J If 3150 J of heat is added to 50g of an unknown substance to raise its temperature from 10 oC to 80 oC, what is the substance? Hint- find specific heat Q = m C DT C =0.9 J/ g oC 3150 J = (50g) C (70oC) Aluminum 3150 J = (3500goC) C

Heat Lost = Heat Gained D. Conservation of Energy Neglecting heat energy lost to the surroundings, when 2 objects/ substances at different temperatures are brought into contact, the heat energy lost by the warmer object/ substance is gained by the cooler object/ substance. Heat Lost = Heat Gained

Sample Problem 1a. When a 1000 g piece of iron is placed in 500 g of 20 oC water, the water’s temperature rises to 25 oC. How much heat is lost by the iron? Heat lost by iron = heat gained by water Q = m C DT = (500g) (4.18 J/ g oC) (5 oC) = 10450 J If the water gains 10450 J, then iron lost 10450 J. 1b. What is the temperature change of the iron? Q = m C DT 10450 J = (1000g) (0.45J/g oC) DT 10450 J= 450 DT DT = 23.2oC

E. Calorimetry Every substance has a unique set of characteristics that can help us identify it. One of those characteristics is a substance’s specific heat capacity. If we can determine how much heat an unknown substance absorbs, we can identify it by looking up its specific heat.

A calorimeter is a device that allows us to measure the heat exchange between 2 substances. thermometer Insulated calorimeter cup Unknown object water

Sample Problem An unknown metal with a mass of 500 g is heated to 100 oC and then is placed in 500 g of water at a temperature of 20oC. If the final temperature of the water is 27.8 oC, what is the identity of the metal? Unknown m1C1DT1= m2C2DT2 Water (500g)(C1)(72.2oC) = (500g)(4.18 J/goC)(7.8oC) 36100 C1 = 16302 = 0.45 J/goC Iron

III Laws of Thermodynamics 1. First Law Whenever heat is added to a system, it transforms to an equal amount of some other form(s) of energy. OR Heat Added = increase in + Work done internal energy by system

2. Second Law of Thermodynamics An object or system will naturally proceed from a state of order to disorder (will break down, become unorganized). Disorder is also known as ENTROPY

Entropy (cont.) To make a system more ordered, energy or work must be inputted into the system. Increased Entropy Decreased Work done to order system

Entropy (cont.) Entropy and Phases of Matter Solid Liquid Gas Increasing Entropy Decreasing Entropy 3. Third Law It is impossible to reach the temperature of absolute zero (0 K)