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CALORIMETRY

CALORIMETRY. The Measurement of Heat. Calorimetry is. the science of measuring the heat of chemical reactions or physical changes http://en.wikipedia.org/wiki/Calorimeter. Things to Know. Heat is the energy that is passed from one object to another

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CALORIMETRY

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  1. CALORIMETRY The Measurement of Heat

  2. Calorimetry is • the science of measuring the heat of chemical reactions or physical changes • http://en.wikipedia.org/wiki/Calorimeter

  3. Things to Know • Heat is the energy that is passed from one object to another • Net flow of heat flows from the warmer object to the cooler object • Temperature is the average kinetic energy of all the particles in a substance • Specific heat is the amount of heat needed to raise 1 gram of a substance 1°C

  4. Things to Know • The unit for heat energy is either the calorie or the Joule • A calorie is defined as the amount of heat needed to raise 1 gram of water 1°C • 1 calorie = 4.18 Joules • NOTE: a food calorie (C) is actually a kilocalorie or 1000 calories

  5. How to Calculate the Heat Gained or Lost by a Substance • Q = m c ΔT , where • Q = amount of heat • m = mass of substance in grams • c = specific heat of substance in Joules/gram ∙ °C • ΔT = change in temperature substance experience; Final T – Initial T

  6. Law of Conservation of Energy • Just as we have learned about Matter, Energy also cannot be created nor destroyed but can be changed from one form to another • So, that means if one substance loses heat, another substance must gain that same amount of heat lost

  7. Example Problem • A 50.0 g sample of an unknown metal is heated to 115°C and then placed into a calorimeter containing 125 g of water at 25.60°C. The final temperature 29.30°C. • What is the specific heat of the unknown metal? • What is the metal?

  8. Find Out How Much Heat was Gained • Since the hot metal was placed into the cooler water, heat was transferred from the metal to the water. • Heat Gained by the Water: • Q=mcΔT=(125g)(4.18J/g∙°C)(29.30°C-25.60°C)=+1933.25 J • The (+) in front of the amount of energy tells us that heat was absorbed or added

  9. Find Out How Much Heat was Lost • As stated in previous slide, heat was transferred from the hot metal to the cooler water • Heat Lost by the Metal: • Q=mcΔT=(50.0g)(c)(29.30°C-115°C)= -4285c J [since we do not know the specific heat of the metal, c is used] • The (-) in front of the amount of energy tells us that heat was lost

  10. Determining the Metal’s Specific Heat • 1. We know that energy is conserved, therefore: • Heat Gained by Water = Heat Lost by Metal • 2.So, Qwater = - QMetal [(-) because heat is lost] • Substituting what we know: +1933.25 J = - (-4285c) J

  11. Solving for Specific Heat +1933.25 J = - (-4285c) J c = 1933.25 / 4285 = 0.451 J/g∙°C

  12. Determining Metal’s Identity • Using Table 15.2 in our Chemistry book, we find that Iron is the closest metal to the one in the example with a specific heat of 0.449 J/g∙°C.

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