Temperature and Heat

Transcription

1 Temperature and Heat Energy: capacity to do work Kinetic energy (associated with motion) Thermal, atoms, molecules, and ions are in motion all matter has thermal energy Mechanical, moving tennis ball, automobiles Electrical, e moving through a conductor Sound, corresponds to compression and expansion of the space between molecules Potential energy (results from object s position) Gravitational, possessed by a ball held above floor Electrostatic, associated with the separation of two dissimilar electrical charges Chemical, burning coal heat work Thermodynamics: part of chemistry that deals with the energy changes involved in chemical rxns and changes in the physical state of substances. Thermochemistry: branch of Thermodynamics that deals with the evolution or absorption of energy as heat in chemical processes Work: w = F d (recall F = ma) and the units of work are: kg m s 2 m = kg m 2 s 2 Units of energy: Joule (J): kg m 2 s 2 Calorie (cal): energy required to raise the temperature of 1.0 g of water by 1 o C (from 14.5 to 15.5 o C) 1 cal = J and 1 J = cal 1 kj = 1000 J Nutritional calorie (Cal): energy available from food (1 Cal = 1000 cal) The temperature of an object is a measure of its heat content and of its ability to transfer heat. We measure temperature by thermometers. HEAT IS NOT THE SAME AS TEMPERATURE! The more thermal energy a substance has, the greater the motion of atoms and molecules. 1

2 System: object or collection of objects under study Surroundings: everything outside the system can exchange energy with the system Thermal equilibrium: associated with directionality of heat transfer Calorimetry: measures flow of heat across boundaries Heat Capacity: measures the ability of the substance to pick up heat Heat capacity under constant pressure, C p equals the amount of heat required to raise the q temperature of a system by 1 o C (1 K) under constant pressure C P = (J K -1 ) T CP q Molar hat capacity: cp = = q = ncp T (units: J K -1 mol -1 ) n n T Heat capacity under constant volume, C v c Specific heat capacity: c = P s (J K -1 kg -1 ) q mcs T m =, so c P = Mcs the molar heat capacity equals its specific heat capacity multiplied by its molar mass. Heat Transfer The quantity of heat transferred from or to an object depends on a) the quantity of material b) the size of temperature change c) the identity of the material gaining or losing heat See Table 10.1 How to find the temperature at thermal equilibrium amt of heat gained by the cooler body = amt of heat lost by the warmer body means: final - initial T = + q = + heat transferred from surroundings to system T = - q = - heat transferred from system to surroundings The heat content changes within a given system is zero. q + q + q = 2

3 Example 55.0 g of a hot metal at 98.8 o C is immersed in a beaker containing g of cool water at 21.0 o C. The final temperature is 23.1 o C. Find the specific heat capacity of the metal and using Table 10.1 identify the metal. q q = 0 assuming no loss in the surroundings water + metal c s, mh ( ) +, ( ) = 0 2 O Tf T water i cs m metal metal Tf Ti (4.184 J K -1 g -1 )(225.0 g)( )K + c s, metal (55.0 g)( )K c, =0.469 J K -1 g -1 Fe is the closest in Table 10.1 s metal Example Calculate the minimum amt of heat required to raise the temperature of 45 kg of granite by 15.0 K The heat absorbed by the granite is: 3 5 q = mcs T = (45x10 g)(0.82jk g )(15.0K) = 5.5x10 J = 550kJ This is the minimum heat lost to surroundings as the rock heats up Now, Calculate the temperature change of 45.0 kg of water if it stores the same amt of thermal energy as the 45 kg of granite T 5 q 5.5x10 J = = 1 3 c m (4.184JK g )(45x10 g) s = 2.9K Calorimeters Containers that have an interior space thermally insulated from the surroundings and a means to measure the temperature in the interior. The idea is to measure the temperature at thermal equilibrium among well-defined systems on the inside while preventing heat flow to or from outside q = C CAL T C CAL is the calorimeter constant and it tells us how much heat the interior part absorbs Combustion (bomb) calorimeter confines the rxn into a fixed volume 3

4 Example A stryrofoam cup calorimeter is filled with g of water at room temperature. Adding 1312 J of heat to the contents raises the interior temperature by 1.93 K. Calculate the calorimeter constant. Assume that the calorimeter does not leak heat. q + q = J water CAL 1312 cs, mh O T + CCAL T = 1312J (4.184JK -1 g -1 )(150.0g)(1.93K)+C CAL (1.93K)=1312 J, and C CAL =52.2 JK -1 water Now we dissolve some LiCl in the g of water contained in the same calorimeter and we observe a temperature rise by 3.46 K. Calculate the amt of heat evolved in the dissolution of LiCl if the heat capacity and mass of LiCl are the same as of water. q solution + qcal + qrxn = 0 m solutioncs, solution T + CCAL T + qrxn = 0 (150.0 g)(4.184 JK -1 g -1 )(3.46K) + (52.2 JK -1 )(3.46K) + q rxn = 0 q rxn = J, the dissolution rxn absorbs J and the heat that evolves is the opposite of it, the reverse of absorbing it, J EXAMPLE A manufacturer claims that its new dietetic dessert has fewer than 10 Calories per serving. To test the claim a chemist at the Department of Consumer Affairs places one serving in a bomb calorimeter and burns it in oxygen. The heat capacity of the calorimeter is kj K -1. The temperature increases o C. Is the manufacturer s claim correct? qsample + qcal = 0 qcal = qsample q CAL = (8.151 kj K -1 )(4.937K)=40.24 kj The heat released upon burning is gained by the calorimeter 10 Calories = (10 kcal)(4.184 kj/kcal) = kj because 1 Calorie = 1 kcal 2 4

5 And kj > kj and the manufacturer is honest!!! EXAMPLE A chemist burns g. of graphite in a new bomb calorimeter and CO 2 is formed. If T increases K and kj of heat is released per mole of graphite, what is the heat capacity of the calorimeter? q + q = 0 1molC (0.865g)( )( 393.5kJmol 12.0gC C sample CAL CAL = 10.85kJK ) + (2.613K) C Enthalpy The heat content of a substance under constant pressure, H = q P. It is a state property. H= H f - H i Enthalpy of a rxn is the change in enthalpy when a chemical rxn occurs H= H products H reactants CAL = 0 CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (l) H rxn = q p < 0 exothermic N 2 (g) + O 2 (g) 2 NO 2 (g) H rxn = q p > 0 endothermic Enthalpies of rxn are written in association with a balanced equation CO (g) + ½ O 2 (g) CO 2 (g) H rxn = kj mol -1 2 CO (g) + O 2 (g) 2 CO 2 (g) H rxn = kj mol -1 EXAMPLE The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by Al 2 O 3 (s) 2Al (s) + 3/2 O 2 (g) H rxn = 1676 kj. How many grams of Al can be formed when x 10 3 kj of heat is transferred? (1.000 x 10 3 kj) x (2 mol Al/1676 kj)(26.98 g Al/1 mol Al) = g Al Changes in State (Phase diagrams) Common phase transitions involve a transfer of heat between system and surroundings at constant temperature and pressure. 5

6 GAS Condensation Vaporization Sublimation Deposition LIQUID Fusion to m.p. Freezing SOLID EXAMPLE Heating of g of water from -50 o C to 200 o C (Heat of fusion: 333 J/g, heat of vaporization: 2256 J/g). How much is needed for this process to occur? Step 1: warm water from -50 o C to 0 o C (c s of ice = 2.06 J K -1 g -1 ) q 1 = c s,ice m ice T = (2.06 J K -1 g -1 )(500.0 g)( )K = 5.15 x 10 4 J Step 2: melt ice at 0 o C q 2 = (500.0 g)(333 J g -1 ) = 1.67 x 10 5 J Step 3: provide heat to raise the temperature of water from 0 o C to 100 o C q 3 = cs, mh O T = (4.184 J K -1 g -1 )(500.0 g)( )k = 1.13 x 10 5 J water 2 Step 4: provide heat to evaporate water at 100 o C q 4 = (500.0 g)(2256 J g -1 ) = 1.13 x 10 6 J Step 5: raise the temperature of steam from 100 o C to 200 o C q 5 = cs, mwater T = (1.92 J g -1 )(500.0 g)( )k = 9.60 x 10 4 J steam q TOTAL = q 1 + q 2 + q 3 + q 4 + q 5 = 1.60 x 10 6 J = 1600 kj 6

7 Determining the heat of a reaction You place 50.0 ml of M NaOH in a coffee-cup calorimeter at o C and carefully add 25.0 ml of M HCl, also at o C. After stirring, the final temperature is o C. Calculate q soln (in J) and H rxn (in kj/mol). (Assume the total volume is the sum of the individual volumes and that the final solution has the same density and specific heat capacity as water: d = 1.00 g/ml and c = J/g K) 1. We find the mass of the solution m soln = (25.0 ml ml)(1.00 g/ml) = 75.0 g 2. We find T soln, T soln = o C o C = 2.21 o C = 2.21K 3. Now we find q soln qso ln = mso lncs, so ln Tso ln = (75.0g)(4.184JK g )(2.21K ) = 693J 4. Now we find H rxn We have a neutralization rxn HCl + NaOH NaCl + H 2 O Let us write the net ionic equation: H + (aq) + OH (aq) H 2 O (l) We find the moles of reactants and products n H = ( 0.500mol / L)( ) = L n OH = ( 0.500mol / L)(0.050L) = H + is the limiting reactant and only mol of H 2 O will be formed. The heat gained by water was lost from the rxn: q soln = - q rxn q rxn = J H rxn = qrxn molh 2 O 1kJ 693J 1kJ x 1000J = x J (to get the units requested) = 55.4kJmol ENTHALPY CHANGES ACCOMPANY CHEMICAL REACTIONS 7

8 Summary of the relationship between amount (mol) of substance and the heat (kj) transferred during a reaction. 8