Conservation of Mass Energy Energy and Chemical and Physical Change Temperature Energy and Heat Capacity Calculations Law of Conservation of Matter Energy (Law of Conservation of Energy) Units of Energy Exothermic/Endothermic Reactions Temperature Calculations Energy and Heat Capacity Calculations

Conservation of Mass Matter is neither created nor destroyed in a chemical reaction. Example: 58g of butane reacts with 208 g of oxygen to form 176 g of carbon dioxide and 90g of water Reactants Products Butane and oxygen carbon dioxide and water 58g + 208g 176 g + 90g 266g 266g

Example 3.4 Conservation of Mass A chemist forms 16.6 g of potassium iodide by combining 3.9 g of potassium with 12.7 g of iodine. Show that these results are consistent with the law of conservation of mass. SOLUTION The sum of the masses of the potassium and iodine is: 3.9 g + 12.7 g = 16.6 g The sum of the masses of potassium and iodine equals the mass of the product, potassium iodide. The results are consistent with the law of conservation of mass. SKILLBUILDER 3.4 Conservation of Mass Suppose 12 g of natural gas combines with 48 g of oxygen in a flame. The chemical change produces 33 g of carbon dioxide. How many grams of water form? Answer: 27 g For More Practice Example 3.15; Problems 45, 46, 47, 48, 49, 50.

Example: Are these data sets on chemical changes consistent with the law of conservation of mass? (a) A 7.5-g sample of hydrogen gas completely reacts with 60 g of oxygen to form 67.5 g of water. (b) A 60.5-g sample of gasoline completely reacts with 243g of oxygen to form 206g of carbon dioxide gas and 88g of water.

Energy Energy is defined as the capacity to do work. Work is defined as the result of a force acting on a distance. Like matter energy is conserved. Law of Conservation of Energy – Energy is neither created nor destroyed. The total amount of energy remains constant. It is transferred from one form of energy to another form of energy

Kinetic energy is energy of motion. Example: a recoiling spring moving water Potential energy is energy associated with position or composition. Example: a stretched spring water behind a damn Electrical energy is energy associated with the flow of electrical charge. Thermal energy is energy associated with random motion of atoms and molecules. Chemical energy is energy related to the making and breaking of bonds in a chemical reaction.

Units of Energy joule (J) – SI unit of energy calorie (cal) – the amount of energy required to raise the temperature of 1 g of water by 1oC Calorie(Cal) – kcal, used for the “Calories” in food Kilowatt-hour (kWh) – used to measure electricity use in homes

Example 3.5 Conversion of Energy Units Continued SOLVE Follow the solution map to solve the problem. Begin with 225 Cal and multiply by the appropriate conversion factors to arrive at J. Round the answer to the correct number of significant figures (in this case, three because of the three significant figures in 225 Cal). CHECK Check your answer. Are the units correct? Does the answer make physical sense? SOLUTION The units of the answer (J) are the desired units. The magnitude of the answer makes sense because the J is a smaller unit than the Cal; therefore, the quantity of energy in J should be greater than the quantity in Cal. SKILLBUILDER 3.5 Conversion of Energy Units The complete combustion of a small wooden match produces approximately 512 cal of heat. How many kilojoules are produced? Answer: 2.14 kJ SKILLBUILDER PLUS Convert 2.75  104 kJ to calories Answer: 6.57  106 cal For More Practice Example 3.16; Problems 51, 52, 53, 54, 55, 56, 57, 58.

Energy and Chemical and Physical Changes Exothermic Reaction – releases energy Endothermic Reaction – absorbs energy

Example 3.6 Exothermic and Endothermic Processes Identify each change as exothermic or endothermic. (a) wood burning in a fire (b) ice melting SOLUTION (a) When wood burns, it emits heat into the surroundings. Therefore, the process is exothermic. (b) When ice melts, it absorbs heat from the surroundings. For example, when ice melts in a glass of water, it cools the water as the melting ice absorbs heat from the water. Therefore, the process is endothermic. SKILLBUILDER 3.6 Exothermic and Endothermic Processes (a) water freezing into ice (b) natural gas burning Answers: (a) exothermic (b) exothermic For More Practice Problems 61, 62, 63, 64.

Temperature Temperature measure of its thermal energy. Heat is energy and has units of energy. It is considered the transfer or exchange of thermal energy. Three temperature scales: Fahrenheit (oF) scale, Celsius (oC) scale, and Kelvin (K) scale. oF = oC(1.8) + 32 oC = (oF – 32)÷1.8 K = oC + 273.15

Example 3.7 Converting between Celsius and Kelvin Temperature Scales Convert 25 C to kelvins. SORT You are given a temperature in degrees Celsius and asked to find the value of the temperature in kelvins. GIVEN: 25 C FIND: K STRATEGIZE Draw a solution map. Use the equation that relates the temperature in kelvins to the temperature in Celsius to convert from the given quantity to the quantity you are asked to find. SOLVE Follow the solution map to solve the problem by substituting the correct value for C and calculating the answer to the correct number of significant figures. SOLUTION MAP RELATIONSHIPS USED K = C + 273.15 (presented in this section) SOLUTION K = C + 273.15 K = 25 C + 273.15 = 248 K

Example 3.7 Converting between Celsius and Kelvin Temperature Scales Continued CHECK Check your answer. Are the units correct? Does the answer make physical sense? The units (K) are correct. The answer makes sense because the value in kelvins should be a more positive number than the value in degrees Celsius. SKILLBUILDER 3.7 Converting between Celsius and Kelvin Temperature Scales Convert 358 K to Celsius. Answer: 85 C For More Practice Example 3.17; Problems 65c, 66d.

Example 3.8 Converting between Fahrenheit and Celsius Temperature Scales Convert 55 F to Celsius. SORT You are given a temperature in degrees Fahrenheit and asked to find the value of the temperature in degrees Celsius. STRATEGIZE Draw the solution map. Use the equation that shows the relationship between the given quantity (F) and the find quantity (C). SOLVE Substitute the given value into the equation and calculate the answer to the correct number of significant figures. GIVEN: 55 F FIND: C SOLUTION MAP RELATIONSHIPS USED (presented in this section) SOLUTION

Example 3.8 Converting between Fahrenheit and Celsius Temperature Scales Continued Check Check your answer. Are the units correct? Does the answer make physical sense? The units (C) are correct. The value of the answer (13 C) is smaller than the value in degrees Fahrenheit. For positive temperatures, the value of a temperature in degrees Celsius will always be smaller than the value in degrees Fahrenheit because the Fahrenheit degree is smaller than the Celsius degree and the Fahrenheit scale is offset by 32 degrees (see Figure 3.17). FIGURE 3.17 Comparison of the Fahrenheit, Celsius, and Kelvin temperature scales.

Example 3.8 Converting between Fahrenheit and Celsius Temperature Scales Continued SKILLBUILDER 3.8 Converting between Fahrenheit and Celsius Temperature Scales Convert 139 C to Fahrenheit. Answer: 282 F For More Practice Example 3.18; Problems 65a, 66a, c.

Example 3.9 Converting between Fahrenheit and Kelvin Temperature Scales Convert 310 K to Fahrenheit. SORT You are given a temperature in kelvins and asked to find the value of the temperature in degrees Fahrenheit. STRATEGIZE Build the solution map, which requires two steps: one to convert kelvins to degrees Celsius and one to convert degrees Celsius to degrees Fahrenheit. GIVEN: 310 K FIND: F SOLUTION MAP RELATIONSHIPS USED

Example 3.9 Converting between Fahrenheit and Kelvin Temperature Scales Continued SOLVE Solve the first equation for C and substitute the given quantity in K to convert it to C. Solve the second equation for F. Substitute the value of the temperature in C (from the previous step) to convert it to F and round the answer to the correct number of significant figures. SOLUTION CHECK Check your answer. Are the units correct? Does the answer make physical sense? The units (F) are correct. The magnitude of the answer is a bit trickier to judge. In this temperature range, a temperature in Fahrenheit should indeed be smaller than the temperature in kelvins. However, because the Fahrenheit degree is smaller, temperatures in Fahrenheit become larger than temperatures in kelvins above 575 F.

Example 3.9 Converting between Fahrenheit and Kelvin Temperature Scales Continued SKILLBUILDER 3.9 Converting between Fahrenheit and Kelvin Temperature Scales Convert – 321 F to kelvins Answer: 77K For More Practice Problems 65b, d, 66b.

Specific Heat Amount of heat needed to change 1 g of a substance 1oC. Units of J/goC (or cal/goC)

Energy and Heat Capacity Calculations When a substance absorbs(releases) heat its temperature change is in direct proportion to the amount of heat absorbed(released) and the amount of substance present. Heat = Mass x Specific Heat x Temperature Change q = m x C x ΔT q is in units of joules (or cals), m is in grams, C is in J/goC (or cal/goC), Δ means change in, T is temperature in oC.

Example 3.10 Relating Heat Energy to Temperature Changes Gallium is a solid metal at room temperature but melts at 29.9 C. If you hold gallium in your hand, it melts from your body heat. How much heat must 2.5 g of gallium absorb from your hand to raise its temperature from 25.0 C to 29.9 C? The specific heat capacity of gallium is 0.372 J/g C. SORT You are given the mass of gallium, its initial and final temperatures, and its specific heat capacity, and are asked to find the amount of heat absorbed. STRATEGIZE The equation that relates the given and find quantities is the specific heat capacity equation. The solution map indicates that this equation takes you from the given quantities to the quantity you are asked to find. GIVEN: 2.5 g gallium (m) Ti = 25.0 C Tf = 29.9 C C = 0.372 J/g C FIND: q SOLUTION MAP RELATIONSHIPS USED

Example 3.10 Relating Heat Energy to Temperature Changes Continued SOLVE Before solving the problem, you must gather the necessary quantities—C, m, and T —in the correct units. Substitute C, m, and T into the equation, canceling units, and calculate the answer to the correct number of significant figures. CHECK Check your answer. Are the units correct? Does the answer make physical sense? SOLUTION * This is the amount of heat required to raise the temperature to the melting point. Actually melting the gallium requires additional heat. The units (J) are correct. The magnitude of the answer makes sense because it takes almost 1 J to heat the 2.5 g sample of the metal by 1 C; therefore, it should take about 5 J to heat the sample by 5 C.

Example 3.10 Relating Heat Energy to Temperature Changes Continued SKILLBUILDER 3.10 Relating Heat Energy to Temperature Changes You find a copper penny (pre-1982) in the snow and pick it up. How much heat is absorbed by the penny as it warms from the temperature of the snow, –5.0 C, to the temperature of your body, 37.0 C? Assume the penny is pure copper and has a mass of 3.10 g. You can find the heat capacity of copper in Table 3.4 (p. 74). Answer: 50.1 J SKILLBUILDER PLUS The temperature of a lead fishing weight rises from 26 C to 38 C as it absorbs 11.3 J of heat. What is the mass of the fishing weight in grams? Answer: 7.4 g For More Practice Example 3.19; Problems 75, 76, 77, 78.

Example 3.11 Relating Heat Capacity to Temperature Changes A chemistry student finds a shiny rock that she suspects is gold. She weighs the rock on a balance and obtains the mass, 14.3 g. She then finds that the temperature of the rock rises from 25 C to 52 C upon absorption of 174 J of heat. Find the heat capacity of the rock and determine whether the value is consistent with the heat capacity of gold. SORT You are given the mass of the “gold” rock, the amount of heat absorbed, and the initial and final temperature. You are asked to find the heat capacity. STRATEGIZE The solution map shows how the heat capacity equation relates the given and find quantities. GIVEN: 14.3 g 174 J of heat absorbed Ti = 25 C Tf = 52 C FIND: C SOLUTION MAP RELATIONSHIPS USED q= m · C · T (presented in this section)

Example 3.11 Relating Heat Capacity to Temperature Changes continued SOLVE First, gather the necessary quantities—m, q, and T —in the correct units. Then solve the equation for C and substitute the correct variables into the equation. Finally, calculate the answer to the right number of significant figures. SOLUTION m = 14.3 g q = 174 J T = 52 C – 25 C = 27 C By comparing the calculated value of the specific heat capacity (0.45 J/g C) with the specific heat capacity of gold from Table 3.4 (0.128 J/g C), we conclude that the rock could not be pure gold.