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Energy

Thermochemistry: The study of heat changes that occur during chemical reactions and physical changes of state. Energy. Energy – the ability to do work or produce heat Heat (q) = the energy that transfers from one object to another because of a temperature difference between them

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Energy

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

  2. Energy • Energy – the ability to do work or produce heat • Heat (q) = the energy that transfers from one object to another because of a temperature difference between them • Heat always flows from a warmer object to a cooler object

  3. Energy • Kinetic energy – in a chemical reaction temperature is the determining factor • The higher the temperature…the faster the particles move…the higher the average kinetic energy • Temperature is a measure of the average kinetic energy • Kelvin scale: 0 K = -273 °C • Potential energy – in a chemical reaction deals with the types of atoms & what bonds they form

  4. Law of Conservation of Energy • Law of Conservation of Energy – Energy is neither created nor destroyed

  5. Heat (q) • Heat or energy can be in joules, calories, kilocalories, or kilojoules • The SI unit is the joule • 1 Cal = 1000 cal = 1 kcal • 1 cal = 4.186 J

  6. 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, and therefore does not depend on size • Every substance has its own specific heat • Ex. Water = 4.18 J/(g x ºC) Glass = 0.50 J/(g x ºC)

  7. Specific Heat • Units for 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

  8. 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.

  9. Example • C= q/ (m∆T) • C=114 J/ (10.0 g x 25.4°C) • c = 0.449 J/g C

  10. 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

  11. 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

  12. 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

  13. 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

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