Calorietry: easures flow of heat across boundaries Heat Capacity: easures the ability of the substance to pick up heat Heat capacity under constant pressure, C p euals the aount of heat reuired to raise the teperature of a syste by 1 o C (1 K) under constant pressure C P T (J K -1 ) CP Molar heat capacity: cp ncp T (units: J K -1 ol -1 ) n n T Heat capacity under constant volue, C v c Specific heat capacity: c P s (J K -1 kg -1 ) cs T, so c P Mcs the olar heat capacity euals its specific heat capacity ultiplied by its olar ass. When energy is transferred fro one object to another, it appears as work and/or as heat. Heat Transfer The uantity of heat transferred fro or to an object depends on a) the uantity of aterial b) the size of teperature change c) the identity of the aterial gaining or losing heat How to find the teperature at theral euilibriu At of heat gained by the cooler body At of heat lost by the warer body eans: final - initial T + + heat transferred fro surroundings to syste T - - heat transferred fro syste to surroundings The heat content changes within a given syste is zero. + + +... 0 1 3 1
Exaple 55.0 g of a hot etal at 98.8 o C is iersed in a beaker containing 5.0 g of cool at 1.0 o C. The final teperature is 3.1 o C. Find the specific heat capacity of the etal and identify the etal. 0 assuing no loss in the surroundings c + etal ( T T ) + c, ( T T ) s, H O f i s etal etal f i c, (4.184 J K -1 g -1 )(5.0 g)(96.3 94.)K + s etal c, 0.469 J K -1 g -1 Fe is the closest s etal 0 (55.0 g)(96.3 371.95)K Calorieters Containers having an interior space therally insulated fro the surroundings and a eans to easure the teperature in the interior. The idea is to easure the teperature at theral euilibriu aong well-defined systes on the inside while preventing heat flow to or fro outside C T C CAL is the calorieter constant and it tells us how uch heat the CAL interior part absorbs A stryrofoa cup calorieter is filled with 150.0 g of at roo teperature. Adding 131 J of heat to the contents raises the interior teperature by 1.93 K. Calculate the calorieter constant. Assue that the calorieter does not leak heat. + CAL 131J cs, H O T + CCAL T 131J (4.184JK -1 g -1 )(150.0g)(1.93K)+C CAL (1.93K)131 J, and C CAL 5. JK -1 Now we dissolve soe LiCl in the 150.0 g of contained in the sae calorieter and we observe a teperature rise by 3.46 K. Calculate the at of heat evolved in the dissolution of LiCl if the heat capacity and ass of LiCl are the sae as of.
solution + CAL + rxn 0 solutioncs, solution T + CCAL T + rxn 0 (150.0 g)(4.184 JK -1 g -1 )(3.46K) + (5. JK -1 )(3.46K) + rxn 0 rxn - 35 J, the dissolution rxn absorbs - 35 J and the heat that evolves is the opposite of it, the reverse of absorbing it, + 35 J A anufacturer clais that its new dietetic dessert has fewer than 10 Calories per serving. To test the clai a cheist at the Departent of Consuer Affairs places one serving in a bob calorieter and burns it in oxygen. The heat capacity of the calorieter is 8.151 kj K -1. The teperature increases 4.937 o C. Is the anufacturer s clai correct? saple + CAL 0 CAL saple CAL (8.151 kj K -1 )(4.937K)40.4 kj The heat released upon burning is gained by the calorieter 10 Calories (10 kcal)(4.184 kj/kcal) 41.84 kj because 1 Calorie 1 kcal And 41.84 kj > 40.4 kj and the anufacturer is honest!!! A cheist burns 0.865 g. of graphite in a new bob calorieter and CO is fored. If T increases by.613 K and 393.5 kj of heat is released per ole of graphite, what is the heat capacity of the calorieter? saple + CAL 0 1olC (0.865g)( )( 393.5kJol 1.0gC ) + (.613K) C CCAL 10.85kJK Heating of 500.0 g of fro -50 o C to 00 o C (Heat of fusion: 333 J/g, heat of vaporization: 56 J/g). How uch is needed for this process to occur? CAL 0 3
Step 1: war fro -50 o C to 0 o C (c s of ice.06 J K -1 g -1 ) 1 c s,ice ice T (.06 J K -1 g -1 )(500.0 g)(73.15 3.15)K 5.15 x 10 4 J Step : elt ice at 0 o C (500.0 g)(333 J g -1 ) 1.67 x 10 5 J Step 3: provide heat to raise the teperature of fro 0 o C to 100 o C 3 cs, H O T (4.184 J K -1 g -1 )(500.0 g)(373.15-73.15)k.09 x 10 5 J Step 4: provide heat to evaporate at 100 o C 4 (500.0 g)(56 J g -1 ) 1.13 x 10 6 J Step 5: raise the teperature of stea fro 100 o C to 00 o C 5 cs, T (1.9 J g -1 )(500.0 g)(473.15-373.15)k 9.60 x 10 4 J stea TOTAL 1 + + 3 + 4 + 5 1.60 x 10 6 J 1600 kj ENTHALPY CHANGES ACCOMPANY CHEMICAL REACTIONS You place 50.0 L of 0.500 M NaOH in a coffee-cup calorieter at 5.00 o C and carefully add 5.0 L of 0.500 M HCl, also at 5.00 o C. After stirring, the final teperature is 7.1 o C. Calculate soln (in J) and H rxn (in kj/ol). (Assue the total volue is the su of the individual volues and that the final solution has the sae density and specific heat capacity as : d 1.00 g/l and c 4.184 J/g K) 1. We find the ass of the solution soln (5.0 L + 50.0 L)(1.00 g/l) 75.0 g. We find T soln, T soln 7.1 o C - 5.00 o C.1 o C.1K 3. We find soln 4
so ln so lncs, so ln Tso ln (75.0g)(4.184JK g )(.1K ) 693J 4. We find H rxn We have a neutralization rxn HCl + NaOH NaCl + H O Let us write the net ionic euation: H + (a) + OH (a) H O (l) We find the oles of reactants and products n H + ( 0.500ol / L)(0.050L) 0.015 n OH ( 0.500ol / L)(0.050L) 0.050 H + is the liiting reactant and only 0.015 ol of H O will be fored. The heat gained by was lost fro the rxn: soln - rxn rxn - 693 J H rxn rxn olh O 1kJ 693J 1kJ x 1000J x 0.015 1000J (to get the units reuested) 55.4kJol 5