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HEAT EXCHANGE

HEAT EXCHANGE. The exchange of thermal energy is simply referred to as heat. If an object rises/falls in temperature it has gained/lost thermal energy. Heat from here on in refers to the exchange of thermal energy . therefore, heat refers to either “warming up” or “cooling down”.

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HEAT EXCHANGE

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  1. HEAT EXCHANGE The exchange of thermal energy is simply referred to as heat. If an object rises/falls in temperature it has gained/lost thermal energy. Heat from here on in refers to the exchange of thermal energy therefore, heat refers to either“warming up”or “cooling down”

  2. How much heat energy an object can gain or lose depends upon 3 factors: 2 kg of iron 1) Mass 1 kg of iron 2) Temperature Change 1 kg of Al 3) Type of Substance 1 kg of Fe

  3. Heat Exchange Different objects will gain/lose thermal energy at different rates: • 1 kg of aluminum will melt more ice than 1 kg of iron • 1 kg of iron will melt more ice than 1 kg of lead The amount of heat that an object will absorb in order to raise its temperature 1˚C is called the heat capacity of the object.

  4. Heat Capacity refers to how much heat an object can gain/lose. • 1 kg of Al will have a higher heat capacity than .5 kg of Al because of the larger mass even though they gain and lose heat at the same rate. Specific Heat (c) is the amount of heat needed to raise 1 g of a substance 1˚C: units: Q m∆T c = J g˚C

  5. Specific heats at 1 atm :

  6. Using this definition, the amount of heat needed to warm/cool a substance can be calculated: Q = mc∆T units: J The Law of Heat Exchange:  When 2 or more objects of different temperatures are mixed together, they will exchange heat until reaching a final, equilibrium temperature. Q lost = Q gained

  7. A copper calorimeter with a mass of 150.0 g contains 350.0 g of water. The temperature of both the calorimeter and the water is 20.0˚C. A metal cylinder, with a mass of 200.0 g and a temperature of 99.5˚C, is placed into the calorimeter. The final temperature of the calorimeter, water, and cylinder is 26.7˚C. Find the specific heat of the metal cylinder. mm =200.0 g cm = ? mc = 150.0 g mw = 350.0 g Tm = 99.5˚C cc = .387 J/g˚C Tc/w = 20.0˚C cw= 4.19 J/g˚C Tc/w/m = 26.7˚C

  8. Qlost = Qgained (mc∆T)m = (mc∆T)c + (mc∆T)w (200.0g)(cm)(72.8˚C) = (150.0g)(.387 J/g˚C)(6.7˚C) + (350.0 g)(4.19 J/g˚C)(6.7 ˚C) 14,560(cm) = 10,214 cm = .702 J/g˚C

  9. A 115 g piece of copper is heated to 99.8ºC and then placed in a 65.0 g Aluminum calorimeter that contains 200.0 g of water at 13.5ºC. What will the equilibrium temperature be? mc= 115g mw = 200.0 g cc = .387 J/gºC cw = 4.19 J/gºC Tc = 99.8ºC Ta/w = 13.5 ºC ma = 65.0 g T = ? ca = .909 J/gºC NOTE: Temperature changes in heat exchange problems are always Higher minus Lower!

  10. Qlost = Qgained (mcT)c= (mcT)w + (mcT)a (115 g)(.387 J/g˚C)(99.8 - T) = (200.0 g)(4.19 J/g˚C)(T - 13.5) + (65.0 g)(.909 J/g˚C)(T - 13.5) 4440 - 44.5(T)= 838(T) - 11,300 + 59.1(T) - 798 16,538 = 942(T) T = 17.6 ˚C

  11. 1) A 75.0 g aluminum calorimeter contains 100.0 g of water at 20.0 ˚C. A student wishes to heat the water to 100.0 ˚C by immersing a piece of iron heated in a furnace to 350.0 ˚C. How much iron will he need to use? 2) What will be the equilibrium temperature when a 125 g copper block at 99.0 ˚C is dropped into a 50.0 g aluminum calorimeter containing 145 g of water at 15.0 ˚C? 3) A 170.0 g copper calorimeter containing water at 19.5 ˚C is raised to 42.1 ˚C when 75.0 g of water at 90.5 ˚C is poured in. How much cool water was originally in the cup?

  12. Heat of Fusion/Vaporization When an object gains/loses heat, one of two things must happen: 1) It changes temperature. OR 2) It changes phase. It cannot do both- it is one or the other! When a substance gain/loses heat without changing temperature, it must be changing potential energy and therefore changing phase!

  13. When a substance changes phase, it must gain or lose heat to do so it stores or gives up stored energy: The amount of heat needed to melt/freeze one gram of a substance is called the Heat of Fusion (Latent Heat-- Lf): Q m Lf = Q = mLf It requires 334 J to turn one gram of ice at 0˚C into water at 0˚C. For water: Lf = 334 J/g

  14. The amount needed to boil (vaporize) one gram of a substance is called the Heat of Vaporization (Lv): Q m Lv = Q = mLv For water: Lv = 2260 J/g How much heat would be needed to raise 10.0 g of ice at -20.0˚C and turn it into steam?

  15. 1) “warm” the ice: Q=mc∆T =(10.0g)(2.22 J/g˚C)(20.0˚C)= 444 J 2) melt the ice: Q = mLf = (10.0 g)(334 J/g) = 3340 J 3) warm the water: Q = mc∆T = (10.0 g)(4.19 J/g˚C)(100.0˚C) = 4190 J 4) boil the water: Q = mLv = (10.0 g)(2260 J/g) = 22,600 J

  16. The total amount of heat energy needed: 444 J + 3340 J + 4190 J + 22,600 J = 30, 574 J Approximately 74% of the thermal energy gained is acquired while changing phase from liquid to gas. This is why steam burns so much worse than boiling water even though they are both examples of H2O at 100˚C!

  17. An aluminum calorimeter cup of 50.0 g is filled with 120.0 g of water at 5.0 ˚C when 11.7 g of steam is pumped into the calorimeter. The final temperature of the calorimeter, water and condensed steam is measured to be 55.0 ˚C. What is the experimental heat of vaporization of water for this trial? cw = 4.19 J/(g˚C) mc = 50.0 g Tw+c = 5.0˚ C cc = .909 J/(g˚C) Lv = ? ms = 11.7 g mw = 120.0 g T = 55.0 ˚ C

  18. An aluminum calorimeter, mass of 150.0 g, contains 420.0 g of water at 35.0˚C. An 83.5 g amount of ice at 0˚C is placed into the calorimeter and it melts completely. The final temperature is 17.0 ˚C. Calculate the heat of fusion for water. Ta+w = 35.0˚C ca = .909 J/g˚C Ti = 0 ma = 150.0 g mi = 83.5 g mw = 420.0 g Tf = 17.0˚C cw = 4.19 J/g˚C Lf = ?

  19. Qlost = Qgained (mc∆T)w + (mc∆T)a = miLf + micw∆T (420.0 g)(4.19 J/g˚C)(18.0˚C) + (150.0 g)(.909 J/g˚C)(18.0˚C) = (83.5 g)(Lf) + (83.5 g)(4.19 J/g˚C)(17.0˚C) 31,700 + 2450 = 83.5Lf + 5950 Lf = 28,200 83.5 = 338 J/g

  20. What is the final temperature when 25.0 g of steam is pumped into a 150.0 g aluminum calorimeter that contains 200.0 g of water at 35.0˚C? Tf = ? cw = 4.19 J/g˚C ms = 25.0 g Tw+a = 35.0˚C Lv = 2260 J/g ma = 150.0 g Remember: When steam (or ice) is involved you must account for heat gained/lost while changing phase! ca = .909 J/g˚C mw = 200.0 g

  21. Q lost = Q gained msLv + mscw∆T = maca∆T + mwcw∆T (25.0 g)(2260 J/g) + (25.0 g)(4.19 J/g˚C)(100˚C - Tf) = (150.0 g)(.909 J/g˚C)(Tf - 35.0˚C) + (200.0 g)(4.19 J/g˚C)(Tf - 35.0˚C) 56,500 + 10,500 - 105 Tf = 136 Tf - 4770 + 838 Tf - 29,300 101,000 = 1080 Tf Tf = 93.5 ˚C

  22. A copper ball of mass 4.54 kg is removed from a furnace and dropped into a bucket with 1.36 kg of water at 22.0˚C. After the water stops boiling, the combined mass of the water and the copper ball is 5.45 kg. What was the temperature of the furnace? (No heat was absorbed by the bucket.) A mixture of ice and water, mass 200.0 g, is in a 100.0 g calorimeter with a specific heat of .800 J/(g˚C). When 40.0 g of steam is pumped into that mixture, the final temperature is 60.0˚C. How many grams of ice were in the calorimeter originally?

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