Specific Heat Capacity. ANNOUNCEMENTS Unit 9 test tomorrow!
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Transcript of Specific Heat Capacity. ANNOUNCEMENTS Unit 9 test tomorrow!
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Specific Heat Capacity
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ANNOUNCEMENTS
• Unit 9 test tomorrow!
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Take 1
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Things heat/cool at different rates
• Everything changes temperature when it is heated or cooled, but certain things change temperature faster/slower than others.
• For example, put a metal skillet on a stove and it heats up quickly, so fast that it can burn you.
• Put water in a skillet on the stove, and the metal skillet heats faster than the water.
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Why?
• Water is more resistant to a change in temperature than metal.
• In other words, water has a higher heat capacity.
• Heat capacity- The quantity of heat energy to change the temperature of a substance by 1°C.
• Every different element has a specific heat capacity.
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Some heat capacities Substance Specific Heat Capacity
Lead .128 J/g °C
Gold .128 J/g °C
Silver .235 J/g °C
Copper .385 J/g °C
Iron .449 J/g °C
Aluminum .903 J/g °C
Water 4.18 J/g °C
Do you notice a difference in heat capacity between water and the metals?
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So?
• Water has a much higher heat capacity than the metals in the table.
• What does that mean? (Review of the definition of heat capacity)
Water takes more energy to heat 1 °C than the metals.
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Apply
• If water has a higher specific heat capacity than the metals that make up the soil and rocks around us, why might a city by the water have lower temperature than a city that is surrounded by land?
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Solving problems with heat capacity
• Often in chemistry we want to know how much heat must be absorbed or lost in order for a compound to change temperatures.
• This is why we have an easy formula to find heat gained and heat lost.
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Formula
• q= m*C*ΔT• q=heat • m= mass of substance• C= specific heat capacity • ΔT=change in temperature
• SIDENOTE: Δ means change in any situation
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Problem
• q=mCΔT• How much heat must be absorbed for 2
grams of copper to heat from 25 °C to 1084 °C? The specific heat of copper is .385 J/g °C.
• q=• m=• C=• ΔT=
?3g
.3851084-25=1059 °C
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Solved
• q=m x C x ΔT
• ?=(3) x (.385) x (1059) =
• In other words, it takes 1223.14 Joules of energy to heat copper 1059 °C
1223.14
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Guided
• Now lets work a few more problems on the board together.
• Next- Practice work to really get it down.
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Remastery time!
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Last minutes
• What did we learn?
• Top ten: how’d we do?