Hydrocarbons and Heat Most hydrocarbons are used as fuels. Knowing how much energy a fuel provides, can tell us if it is useful for a certain application. For example, the amount of energy a food releases when burned, can tell us about its caloric content (fats release lots of energy). Heat energy released during combustion can be measured with a calorimeter. A bomb calorimeter is shown. It includes water in a heavily insulated container, a stirrer, valve, bomb chamber, ignition wires, & a thermometer (pg 577).

Exothermic and Endothermic changes An alternative to the bomb calorimeter is a coffee cup calorimeter, where two nested polystyrene cups take the place of the container In either case, the change in heat of the water tells us about the reaction of the chemicals. An increase in water temperature indicates that the chemicals released energy when they reacted. This is called an exothermic reaction. In an endothermic reaction, water temperature decreases as the chemicals absorb energy. We will see that heat is measured in Joules (J) or kiloJoules (kJ). Before we do any heat calculations, you should know several terms …

Specific heat capacity balloon demo The heat needed to the temperature of 1 g of a substance by 1 C. Symbol: c, units: J/(g C). Heat capacity The heat needed to the temperature of an object by 1 C. Symbol: C (=c x m), units: J/ C Heat of reaction The heat released during a chemical reaction. Symbol: none, units: J. Specific heat (of reaction) The heat released during a chemical reaction per gram of reactant. Symbol: h, units: J/g. Molar heat of reaction The heat released during a chemical reaction per mole of reactant. Symbol: H, units: J/mol.

Heat Calculations To determine the amount of heat a substance produces or absorbs we often use q = cm T q: heat in J, c: specific heat capacity in J/(g C), m: mass in g, T: temperature change in C, This equation makes sense if you consider units J = J g C x g x C For a list of c values see page 568 (table 3) Sample problem: (must know water = 4.18 J/g C) When 12 g of a food was burned in a calorimeter, the 100 mL of water in the calorimeter changed from 20 C to 33 C. Calculate the heat released. = 4.18 J/(g C) x 100 g x 13 C = 5.4 kJ q=cm T

More practice 1.5 g of copper was heated from 20 C to 80 C. How much energy was used to heat the Cu? 2. If a 3.1 g ring is heated using 10.0 J, its temp. rises by 17.9 C. Calculate the specific heat capacity of the ring. Is the ring pure gold? 3. Do questions 5, 6 on pg. 596 Do questions 7, 8 on pg. 570 q=cm T= 0.38 J/(g C) x 5 g x 60 C = 114 J q=cm T 10.0 J 3.1 g x 17.9 C = 0.18 J/(g°C) c = q/m T= The ring is not pure. Gold is 0.13 J/(g C) -pg. 568

5. q=cm T= 4.18 J/(g C) x 2570 g x 92 C = J = 988 kJ = MJ q=cm T J 4.18 J/(g C) x g = 33.5°C T = q/cm = 6. Since the temperature started at 5.0°C, the final temperature is 38.5°C. 7. q=cm T= 0.86 J/(g C) x 2500 g x 335 C = 720 kJ or 0.72 MJ 8.Total heat = water heat + pot heat = 4.18 J/(g C) x 1200 g x 53.0 C = J = J/(g C) x 450 g x 53.0 C = J = 278 kJ

Heat Capacity Calculations Recall that heat capacity (J/ C) is different from specific heat capacity (J/g C). Heat capacity is sometimes a more useful value For example, because a calorimeter includes wires, the stirrer, thermometer, etc. some heat will be transferred to these other materials. Rather than having to calculate q for each material (like question 8) a J/ C value is used. Sample problem: A calorimeter has a heat capacity of 2.05 kJ/ C. How much heat is released if the temperature change in the calorimeter is 11.6 C? x 11.6 C = 23.8 kJ q = cm T q= 2.05 kJ/ C q=cm T

Thermochemical Equations Thermochemical equations are chemical equations with an added heat term. KBrO 3 (s) + 42 kJ KBrO 3 (aq) This is endothermic (heat is absorbed/used) 2 Mg(s) + O 2 (g) 2 MgO(s) kJ This is exothermic (heat is produced/released) Read 12.3 (pages 582 – 585). Do 2-6 (pg. 585). Sample problem: 3.00 g of octane was burned in a calorimeter with excess oxygen, the 1000 mL of water in the calorimeter rose from 23.0 C to 57.6 C. Write the thermochemical equation for octane, representing the molar heat of combustion.

2.Specific refers to mass in grams. 3.Specific heat of reaction: J/g or kJ/g, etc. Molar heat of reaction: J/mol or kJ/mol, etc. 4.J/mol = J/g x g/mol (molar mass is used) 5. Exothermic. E.g. CH 4 + 2O 2 CO 2 + 2H 2 O + x kJ Other examples include propane, octane, Mg kJ g = 1299 kJ/mol g mol x C 2 H O 2 2CO 2 + H 2 O kJ Page 585 or 2C 2 H 2 + 5O 2 4CO 2 + 2H 2 O kJ

Sample: First, calculate the heat released by the combustion reaction via q=cm T … = 4.18 J/(g C) x 1000 g x 34.6 C = J h = We can determine the molar heat of reaction 1) via the specific heat of reaction or 2) directly J g J 3.00 g = J/g = 48.2 kJ/g = = 5508 kJ/mol 48.2 kJ g g mol x Octane (C 8 H 18 ) has a molar mass of g/mol H = C 8 H O 2 8CO 2 + 9H 2 O kJ = 5508 kJ/mol J mol J mol = H = or For more lessons, visit