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Published byDayna Goodman Modified over 8 years ago
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Review Endothermic reactions _________________ energy causing the q and ∆H to be ______________. Exothermic reactions ___________________ energy causing the q and ∆H to be _____________________. absorb positive release negative
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Energy – the ability to do work or produce heat Exists in 2 forms: Kinetic energy – energy of ___________ Potential energy – energy at rest or energy of ____________ HEAT -- the energy that transfers from one object to another because of a temperature difference between them Heat always flows from a ___________ object to a _______________ object motion position warmer cooler
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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 Therefore, the lower the temperature the ________ the kinetic energy. Temperature is a measure of the average kinetic energy Kelvin scale : 0 K = -273 °C, °C + 273 = K lower
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Law of Conservation of Energy Law of Conservation of Energy – Energy is neither created nor destroyed What else have we seen conserved in a chemical reaction? Mass
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Units for Measuring Heat Flow Heat or energy can be in joules, calories, kilocalories, or kilojoules The SI unit is the joule 1000 cal = 1 kcal 1 cal = 4.186 J 1kcal = 4186J 1 J = 0.239 cal
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Heat Capacity The amount of heat it takes to change an object’s temperature by 1ºC Depends on an object’s mass and chemical composition Ex. A cup of water has a greater heat capacity than a drop of water.
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Specific Heat (C) Specific Heat (C) – the amount of heat required to raise the temperature of 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 (look at your CRM) Ex. Water = 4.184 J/(gºC) Glass = 0.500 J/(gºC) *The higher the specific heat, the greater amount of energy! The lower the specific heat, the larger the increase in temperature!
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Specific Heat Units for C = J/gºC (joules per gram degree Celsius) Equation for Specific Heat: C = q / (m Δ T) C = specific heat; q = heat; m = mass ΔT = change in temperature This equation can be rearranged to solve for heat (q) q= mCΔT
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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.
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Example C= q/ (m∆T) C=114 J/ (10.0 g x 25.4°C) C = 0.45 J/g C
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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
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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 = 475.45 C T = Tf-Ti 475.45 = Tf-25 Tf = 500.45 C
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