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Heat. Lab 31 Terms. Δ (delta) means change Subtract FINAL – INITIAL. Temperature = average kinetic energy of particles in a substance Measured with a thermometer. Q means HEAT measured in calories or Joules ΔQ means change in heat. HEAT

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Lab 31 terms
Lab 31 Terms

  • Δ (delta) means change

    • Subtract FINAL – INITIAL

  • Temperature = average kinetic energy of particles in a substance

  • Measured with a thermometer


  • Q means HEAT

    measured in calories or Joules

  • ΔQ means change in heat.

  • HEAT

    The flow of thermal energy from HIGH to LOW temperature

    Must be calculated


  • calorie = amount of heat required to raise the temperature of 1 g of water 1oC

  • 1 calorie = 4.18 Joules


Calorimeter
Calorimeter

Styrofoam cup calorimeter

an insulated container where a reaction can be performed and heat can be calculated

http://www.chem.purdue.edu/gchelp/howtosolveit/Thermodynamics/ThermoArt/Calorimeter.JPG


Flip page over to ii heat equation q m c t
Flip page over to II. Heat EquationΔQ = m · c · ΔT

Heat = mass x heat capacity x change in temp

cal = g X X ºC

c for water = 1

density of H2O = 1 g/ml

  • Find by taking T2 – T1


Try this

m

T1

T2

?

1

Try this:

100 ml of water warms from 10 ºC to 30 ºC. What is the heat gained?

m =

ΔT =

c =

ΔQ = m · c · ΔT

ΔQ = 100g · 1 · (20 ºC)

ΔQ = 2000 cal

100 g

30oC – 10oC = 20oC


m

T1

T2

?

Lab 31 Pre Lab Questions

If 10.00 g of water is changed from 70ºC to 85ºC. What is the ΔQ (change in heat)?

ΔQ = m · c · ΔT

ΔQ = 10g · 1 · (85 ºC - 70 ºC)

ΔQ = 150 cal


Lab 31 Pre Lab Questions

300 calories of heat is added to 100 ml of water at 20 ºC.

  • What is the mass of water?

    100g

  • What is the ΔT of the water?

  • What would the final temperature be?

    23o C

ΔQ = m · c · ΔT300 cal = 100g · 1 · ΔT

ΔT = 3o C


m = ?

T2

∆Q

T1

1

cal

g

Lab 31 Pre Lab Questions

  • What mass of water is in a calorimeter if it takes 1500 calories to heat it from 10 ºC to 65 ºC?

DQ = m c DT

1500 cal = m · 1 · (65 ºC - 10 ºC)

m = 27.27 g


Heat of fusion

cal

g

http://www.skthew.com/upload/image/time_ice_melting.jpg

Heat of Fusion

heat needed to melt 1 g of a solid.

Heat of Fusion =

For ice,

Heat of Fusion = 80


Lab 32 terms
Lab 32 Terms

Law of Conservation of Energy =

Energy is neither created nor destroyed…Only Transferred.

Heat Lost = Heat Gained


http://www2.uni-siegen.de/~pci/versuche/pics/exo4.jpg

http://www.photospin.com/content/photos/thumb/PS012046.jpg

  • Exothermic =

    a reaction that releases heat…Feels HOT!

    Temp Rises

  • Endothermic =

    a reaction that absorbs heat…Feels COOL!

    Temp Decreases


Lab 32 heat of crystallization
Lab 32: Heat of Crystallization

Heat is released

Heat


Heat of crystallization

65

35.1

cal

cal

g

g

Heat of Crystallization

Heat of Crystallization

Heat released as 1 g of a substance crystallizes.

For wax =

For napthalene =


Lab 33 specific heat capacity
Lab 33: Specific Heat Capacity

  • Symbol: c

  • heat needed to raise 1 g of substance 1oC

  • Units:

  • cwater = 1

  • cAl = 0.21

  • cCu = 0.09

  • cFe = 0.11


Review which is holding more heat

2 L H2O at 50º C

2 L H2O at 90º C

Review:Which is holding more heat?


Examples which is holding more heat

10 L H2O at 50º C

2 L H2O at 50º C

Examples:Which is holding more heat?


Examples which is holding more heat1

2 kg gold at 50º C

2 kg H2O at 50º C

Examples:Which is holding more heat?


Lab 34
Lab 34

  • “Error of the Air”

    assuming room temp is about 20 ºC by starting our experiments at 10 ºC and finishing at 30 ºC we can limit the error from heat gained from the room or lost to the room.

finish

30oC

20oC

10oC

start


Heat of combustion

33,000

7,000

Joules

cal

g

g

Heat of Combustion

Heat of Combustion

Heat released as 1 g of a substance burns.

For candle =

For nut =

Remember, 1 cal = 4.18 J



Cooling curve
Cooling Curve

Heat of fusion

Heat of crystallization

Heat of combustion

DQ = mcDT

g

g  l

DQ = mcDT

temperature

l

l s

DQ = mcDT

s

time


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