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Thermochemistry: The study of heat changes that occur during chemical reactions and physical changes of state.

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Presentation on theme: "Thermochemistry: The study of heat changes that occur during chemical reactions and physical changes of state."— Presentation transcript:

1 Thermochemistry: The study of heat changes that occur during chemical reactions and physical changes of state

2 Energy Heat (q) = Form of energy always flows from a warmer object to a cooler object Law of Conservation of Energy – Energy is neither created nor destroyed

3 Heat (q) The unit is the joule (J) or calorie (cal)
1 calorie = J 1 Calorie (food) = 1000 cal (science) = 1 kcal

4 Specific Heat (C) Specific Heat (C) – the amount of heat required to raise 1 gram of a substance by 1C Specific heat is an intensive property, Every substance has its own specific heat Ex. Water = 4.18 J/(g ∙ ºC) Glass = 0.50 J/(g ∙ ºC)

5 Specific Heat Units for Specific Heat ( C ) = J / g ● ºC (joules per gram degree Celsius) Equation for Specific Heat: C = q / (m Δ T) This equation can be rearranged to solve for heat (q) q= CmΔT C = specific heat; q = heat; m = mass and ΔT = change in temperature (Tf – Ti)

6 Specific Heat A 10.0 g sample of iron changes temperature from 25.0C to 50.4 C while releasing 114 joules of heat. Calculate the specific heat of iron.

7 Example C= q/ (m∆T) C=114 J/ (10.0 g x 25.4°C) c = J/g C

8 Another example If the temperature of 34.4 g of ethanol increases from 25.0 C to 78.8 C how much heat will be absorbed if the specific heat of the ethanol is 2.44 J/g C

9 Another example First, rearrange the specific heat formula to solve for heat q = CmT q = (2.44 J/g°C)(34.4g)(78.8°C – 25.0°C) q = 4520 J

10 Yet another example 4.50 g of a gold nugget absorbs 276 J of heat. What is the final temperature of the gold if the initial temperature was 25.0 C & the specific heat of the gold is 0.129J/g C

11 Yet another example C= q/ (m∆T); rearrange to find ∆T = q / (C x m)
∆T = 276 J / (.129 J/g°C x 4.50 g) T = 475C T = Tf-Ti 475 = Tf - 25 Tf = 500 C

12 Phase Changes How many joules are required to raise the temperature of g of ice at -10.0˚C to 50.0˚C? Cice = 20.9 J/g ˚C Cwater = J/g ˚C ΔHfusion = 6.01 kJ/mol Answer: kJ

13 How much energy is required to bring 45 g of benzene from - 10˚C to 70˚C?
(hint: draw a heating curve to help)


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