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Energy changes in reactions Text chapter 3, 4, 5, 6 & 7
Thermodynamics Energy changes in reactions Text chapter 3, 4, 5, 6 & 7
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Terminology: Please write Thermodynamics: Energy (J):
study of heat changes that occur during chemical reactions. Energy (J): Cannot be seen, touched, smelled, or weighed. Cannot be created or destroyed (1st law of thermodynamics) it is recognized by its effects; heat, light, electricity etc Useful forms of energy in chemistry: Radiant (sun) Thermal (associated with the temperature Q=mc∆T) Chemical (stored in molecular bonds) Kinetic (movement EK = ½ mv2) & Potential (EP=mgh)
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Heat Please write is a transfer of thermal energy.
Eg. A change in temperature between a system & its surroundings. System: The part we are observing. Heat will flow from a warmer body to a cooler body, until both bodies have the same temperature. (2nd law of thermodynamics) Surrounding. Everything else! ENERGY
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Types of systems: Please write Open system: Closed system:
Energy & matter can easily enter or leave. Eg. Beaker Closed system: Energy can enter or leave, but not matter. Eg. Balloon Isolated system: Neither energy nor matter can enter or leave. Eg. Insulated container or calorimeter.
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Bomb Calorimetry Please write
Is a device used to measure heat transfer. An inner chamber (AKA “bomb”) where the rxn happens. A thermometer . An outer chamber that holds water (or another substance of known specific heat capacity) Insulation to prevent loss of heat to the surroundings. When the water changes temperature, thermal energy has been absorbed or released. Qsystem vs Qsurroundings
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Please write Exothermic reactions: Endothermic reaction:
Release heat into the surroundings Qsystem = neg Eg campfire Endothermic reaction: Absorbs heat from the surroundings. Qsystem = pos Boiling water (liquid to gas)
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ENDOTHERMIC E + NH4NO3(s) NH4 (aq) + NO3 EXOTHERMIC CaCl2 (s) Ca Cl- (aq) + E Sodium Acetate. Supersaturated solution.
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Please write Q=mc∆T When an object absorbs heat, its temperature increases. Q = thermal energy (joules) m = mass (g) c = specific heat capacity (J/gºC) (how well it absorbs heat) ∆T = change in temperature (in ºC) = Tfinal – Tinitial Q gained by the surrounding = Q lost by the system (rxn)
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Specific heat capacity=
Amount of energy required to raise the temperature of 1g of a substance by 1°C. Substance Specific Heat Capacity Liquid Water 4.19 J/(g∙°C) Water vapour 1.41 J/(g∙°C) Ice (solid water) 2.05 J/(g∙°C) Ethylene glycol 2.20 J/(g∙°C) Aluminum 0.90 J/(g∙°C) Copper 0.39 J/(g∙°C) Glass 0.84J/(g∙°C) Air (dry) 1.02 J/(g∙°C)
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Ex.1 Please write If 125g of copper forms a layer on the bottom of a frying pan, how much heat is need to raise the temp of the copper from 25oC to 300oC. The specific heat capacity for Cu is 0.39J/goC. Q = m c ΔT m = 125g Ti = 25oC Tf = 300oC c = 0.39J/goC QCu = (125g)(0.39J/goC)( 300oC - 25oC) QCu = J or 1.3 x 104J Heat is absorbed by the frying pan.
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Recall H2O density of 1g/mL
Please write Ex.2 HCl was added to of NaOH in a calorimeter, producing 50mL of water. At the start, the solutions were at 25oC. During the reaction, the highest temp observed was 32oC. What is the heat of the reaction? m = 50g Ti = 25oC Tf = 32oC c = 4.19 J/goC Recall H2O density of 1g/mL So, 50mL x 1g/mL = 50g
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Q = m c ΔT QH2O = (50g)(4.19J/goC)( 32oC - 25oC) QH2O = J or 1.5 x 103J Heat is absorbed by the water.
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Heat is given off by the iron nail
Please write Ex.3 Calculate the heat involved when a 5.5g iron nail is cooled from 37oC to 25oC. The heat capacity of iron is 0.45J/goC. Q = m c ΔT QFe = (5.5g)(0.45J/goC)( 25oC - 37oC) QFe = J m = 5.5g Ti = 37oC Tf = 25oC c = 0.45 J/goC Heat is given off by the iron nail
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Laws of thermodynamics
1st law: Energy cannot be created or destroyed. 2nd law: Heat will flow from a warmer body to a cooler body. Entropy always increases. 3rd law: Entropy of a system approaches a constant value as the temperature approaches absolute zero. Entropy is the measure of a system’s thermal energy per unit of temperature that is unavailable for doing useful work. Measure of disorder.
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Heat transfer calculations
Please write The heat lost by a system is equal numerically but opposite in sign to the heat gained by the surroundings. Consider adding milk to coffee. 1=milk 2=coffee 𝑸 𝟏 =− 𝑸 𝟐 The coffee cools down the milk heats up. 𝑸 𝟏 =− 𝑸 𝟐 𝒎 𝟏 𝒄 𝟏 ∆𝑻 𝟏 =− 𝒎 𝟐 𝒄 𝟐 ∆𝑻 𝟐 𝒎 𝟏 𝒄 𝟏 (𝑻 𝟏𝒇 − 𝑻 𝟏𝒊 )=− 𝒎 𝟐 𝒄 𝟐 (𝑻 𝟐𝒇 − 𝑻 𝟐𝒊 ) But what is the final temperature?
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75 mL Ex.1 𝒎 𝟏 𝒄 𝟏 (𝑻 𝟏𝒇 − 𝑻 𝟏𝒊 )=− 𝒎 𝟐 𝒄 𝟐 (𝑻 𝟐𝒇 − 𝑻 𝟐𝒊 ) / /
Please write Ex.1 What volume of milk at 2.0oC, must be added to 160.0mL of coffee in order to obtain the perfect cup of cappuccino at 55.5oC? The initial coffee temperature is 80.5oC. Assume milk & coffee have the same density as water 1.00g/mL 𝒎 𝟏 𝒄 𝟏 (𝑻 𝟏𝒇 − 𝑻 𝟏𝒊 )=− 𝒎 𝟐 𝒄 𝟐 (𝑻 𝟐𝒇 − 𝑻 𝟐𝒊 ) / / 160 𝑚𝐿(55.5℃−80.5℃)=− 𝑚 2 (55.5℃−2.0℃) 160 𝑚𝐿(−25.0℃)=− 𝑚 2 (53.5℃) −4000 𝑚𝐿∙℃ 53.5℃ =− 𝑚 2 / / / 75 mL / 𝑚 2 = 𝑚𝐿
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ex2 Please write 500ml of 10oC water is mixed with 250ml of 60oC water. What is the final temperature of the water “mixture”?
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What is the final temperature of the water “mixture”?
Q = -Q 500ml of 10oC water + 250ml of 60oC water. What is the final temperature of the water “mixture”? mc∆T = -mc∆T (500)(4.19)(x-10) = -(250)(4.19)(x-60) 500x = -250x 750x = 20000 Tf =27oC x =26.67
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Please do! P137 2, 3 &6 p141 4, 6 & 8 Due next class.
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