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7.1 The Principle of Heat Exchange. December 6, 2010. Mixing Water Lab. Mixed different amounts of water, at different temperatures, and found their final temperature What factors determined the final temperature of your mixtures? Temperature of cold and hot water
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7.1 The Principle of Heat Exchange December 6, 2010
Mixing Water Lab • Mixed different amounts of water, at different temperatures, and found their final temperature • What factors determined the final temperature of your mixtures? • Temperature of cold and hot water • Mass of the cold and hot water
Quantity of Thermal Energy • The amount of thermal energy transferred to an object resulting in a change in temperature • This quantity was based on 3 factors: • Mass of an object (kg) • Change in temperature of an object (K) • Specific Heat Capacity of an object (J/kg•K) Q = mc ΔT
Let’s do an Example! Calculate the amount of heat when 3.8 kg of ethelyne glycol cools from 35°C to 20°C. (Hint: the specific heat capacity of ethylene glycol is 2200 J/kg•K) (Addison Wesley, p. 203) Given: • m = 3.8 kg • Ti = 35°C • Tf = 20°C • c = 2200 J/kg•K N.T.F.: • ΔT= ? • Q = ? Solution: The ethelyne glycol lost 1.3 x 105 J of thermal energy by cooling from 35°C to 20°C. 308 K • ΔT = Tf -Ti • =293 K – 308 K • = -15 K • Q = mcΔT • = (3.8 kg)(2200 J/kg•K)(-15 K) 293 K = -1.3 x 105 J Addison Wesley. Physics 11. Toronto: 2002.
The principle of Thermal Energy Exchange • “When two substances at different temperatures are mixed, the amount of thermal energy lost by the hotter substance in cooling is equal to the amount of thermal energy gained by the colder substance in warming” (Pearson, 214) • In other words, the heat lost from object 2 is equal to the heat gained in object 1. • We can write this as an equation: Q1 = -Q2 m1c1ΔT1 = -m2c2ΔT2 Pearson. PhysicsSource 11. Toronto: 2011.
Friday’s Lab • We conducted one trial where: • m1 = m2 • c1 = c2 = c • In theory what should our data look like for: • ΔT1, ΔT2? • Q1, Q2? • Why do you think we had different results?
Let’s do Another Example! When 2.0 kg of a cold metal at a temperature of 248 K is immersed in 3.0 kg of water at a temperature of 313 K, the final temperature of the mixture is 309 K. What is the specific heat capacity of the metal? (Addison Wesley, p. 203) Given: • m1 = 2.0 kg • Ti1 = 248K • m2 = 3.0 kg • Ti2 = 313 K • c2 = 4190 J/kg•K • Tf = 309 K N.T.F.: • C1 = ? Solution: Q1 = -Q2 m1c1ΔT1 = -m2c2ΔT2 m1c1(Tf -Ti1) = -m2c2(Tf –Ti2) (2.0 kg) c1(309 K –248 K) = -(3.0 kg)(4190 J/kg•K)(309 K –313 K) (2.0 kg) c1(61 K) = -(3.0 kg)(4190 J/kg•K)(-4K) (122 kg•K) c1 = 50280 J c1 = 412 J/kg•K Addison Wesley. Physics 11. Toronto: 2002. The specific heat capacity of the metal is 412 J/kg•K.
Activity! • Set-Up • Clear your desks • Find a partner • Partner A: get 1 ZipLock bag of squares • Partner B: get 1 whiteboard, 1 marker, and 1 cloth • Once seated, you can separate the squares into piles based on their colours • Activity • Pick 1 square from each pile. • You will use these givens to solve for your unknown • Partner A: solve problem on whiteboard • Partner B: solve problems in notebook (or GOOS paper) • After problem is solved: switch! • Exceptions • Unknown: Ti1, Ti2, Tf What to do: remove the corresponding given value • Unknown: c1, c2, m1, m2 What to do: let your given c/m equal the • remaining value (ex: Unknown: m1, • then m (given) = m2)
Consolidation Question A sample of iron is heated to 80.0°C and placed in 100mL of water at 20.0°C. What is the mass of the iron? The final temperature of the mixture is 22.0°C. (hint: 1mL = 1 g for water, and the specific heat capacity for iron is 4.5 x 10² J/kg•K). Answer: miron = 3.2 x 10-2 kg Nelson. Physics 11: University Preperation. Toronto: 2010.
