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Chapter 12. Section 4 – Calorimetry. Calorimetry. the study of heat flow and heat measurements. How to determine the enthalpy change?. C 6 H 12 O 6 + 6O 2 6CO 2 + 6H 2 O Δ H o =-2803 kJ. Calorimeter. A completely insulated container – no energy can’t go in or come out.
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Chapter 12 Section 4 – Calorimetry
Calorimetry • the study of heat flow and heat measurements
How to determine the enthalpy change? • C6H12O6+ 6O2 6CO2 + 6H2O ΔHo=-2803 kJ
Calorimeter • A completely insulated container – no energy can’t go in or come out
Conservation of Energy • amount of energy released by a chemical reaction = amount of energy absorbed by water • qreaction = ‒qwater • water = surrounding
Heat Transfer (Q) Calculation (Example) A 5.10-kg cast-iron skillet is heated on the stove from 295 K to 450 K. How much heat had to be transferred to the iron? CFe = 0.45 J/g·˚C
Specific Heat, C • The amount of energy needed to raise 1˚C (or 1 K) of 1 g (or 1 kg) of a substance • Unit: J/kg∙K; J/g ∙˚C • Unusually high specific heat of ice and water
Sample Problem, Pg 396 • When a 4.25-g sample of solid NH4NO3 dissolves in 60.0 g of water in a calorimeter, the temperature drops form 21.0 °C to 16.9 °C. Calculate ΔH for the solution process. NH4NO3 (s) NH4+(aq) + NO3‒(aq) ΔH=? (Ans: +19.4 kJ)
Practice Problem 2, Pg 397 • What is the specific heat of nickel if the temperature of a 32.2-g sample of nickel is increased by 3.5 °C when 50. J of heat is added? (Ans = 0.44 J/g·°C)
Practice Problem 1, Pg 397 • When a 13.7-g sample of solid Pb(NO3)2 dissolves in 85.0 g of water in calorimeter, the temperature drops from 23.4°C to 19.7 °C. Calculate ΔH for the solution process. (Ans = 32 kJ) Pb(NO3)2(s) Pb2+(aq) + 2NO3-(aq) ΔH = ?
Heat Needed to Melt a Solid, Hf • No temperature change while a solid is melting • Q = mHf • Hf= heat of fusion = heat needed to melt 1 g or 1 kg of a solid (Ex) Hf of copper = 2.05×105 J/kg Hf of ice = 3.34×105 J/kg
Heat Needed to Boil a Liquid, Hv • No temperature change while a liquid is boiling • Q = mHv • Hv= heat of vaporization = heat needed to boil off 1 g or 1 kg of a liquid (Ex) Hv of copper = 5.07×106 J/kg Hv of water = 2.26×106 J/kg
Example • You need to melt 1.50 kg of snow at 0.0 ˚C and heat it to 70.0 ˚C to make hot cocoa. How much heat will be needed? Hf of ice = 3.34×105 J/kg; Hv of water = 2.26×106 J/kg
Energy Transfer Between Two Objects • Energy transfer (=heat) occurs until two objects reach the same final temperature • One object loses heat; another gains • Qgiven offbyA = ‒ QabsorbedbyB mCA (Tf– Ti, A) = –mCB (Tf– Ti, B)
Example • A calorimeter contains 0.50 kg of water at 15 ˚C. A 0.040-kg block of zinc at 115 ˚C is placed in the water. What is the final temperature of the system? Cwater = 4180 J/Kg∙˚C; Czn = 388 J/Kg∙˚C;
Three Laws of Thermodynamics • First law • Conservation of energy • ∆U = Q + W ∆U = Change in thermal energy (= internal energy) in a system +Q = Heat absorbed by the system ‒Q = Heat given off by the system +W = work done on the system ‒W = work done by the system
Second law • Entropy, S, (randomness, disorder, or chaos) in Universe is always increasing (= spontaneity) • Heat added increases entropy (Ex) Which will add more disorder? a) 10 J heat added to a system at 100 ˚C b) 10 J heat added to a system at -10 ˚C (3) (T must be in K)
Third law: The entropy of a “perfect” crystal is zero. • http://www.allaboutscience.org/common/printable-third-law-of-thermodynamics-faq.htm