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Heat and Phase Change. What affects heat transfer?. The material: conductor or insulator? Area. Ex: a bigger window vs. a smaller one. Thickness. Ex: heat escapes a thinner styrofoam cup than a thicker one. Heat and State of Matter.

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Presentation Transcript
what affects heat transfer
What affects heat transfer?
  • The material: conductor or insulator?
  • Area. Ex: a bigger window vs. a smaller one.
  • Thickness. Ex: heat escapes a thinner styrofoam cup than a thicker one.
heat and state of matter
Heat and State of Matter
  • Does the absorption or release of heat always cause a temp change? NO
  • Ex: para-dichlorobenzene (chemical in mothballs) has a melting point of 54 oC.
  • If you were to heat this to 80 oC, then insert it into room temperature water, you’d see the temp steadily decrease until you hit 54 oC.
  • At this point it is becoming a solid.
slide4

It will stay at 54 oC until the last of the liquid becomes a solid. Then it will decrease again.

  • Was heat being transferred during that time? YES
slide6

Phase transitions occur at a constant temperature.

  • Heat transfers with no change in temp are referred to as latent heat.
slide7

For melting and freezing:

Q= mΔHFusion

  • For vaporization and condensation:

Q= mΔHVaporization

  • ΔHFusion = specific heat of fusion per gram
  • ΔHVaporization= specific heat of vaporization per gram
slide8

Solid Water: Cp=2.00 J/g°C

  • Liquid Water: Cp= 4.18 J/g°C
  • Gaseous Water: Cp= 2.01 J/g°C
examples
Examples
  • Elise places 48.2 grams of ice in her beverage. What quantity of energy would be absorbed by the ice (and released by the beverage) during the melting process? The heat of fusion of water is 333 J/g.
slide10

What is the minimum amount of liquid water at 26.5 degrees that would be required to completely melt 50.0 grams of ice? The specific heat capacity of liquid water is 4.18 J/g°Cand the specific heat of fusion of ice is 333 J/g.

slide11

Given info about ice:

m = 50.0 gΔHfusion = 333 J/g

Given info about water:

C = 4.18 J/g°CTinitial = 26.5°C Tfinal = 0.0°CΔT = -26.5°C (Tfinal - Tinitial )

slide12

The energy gained by the ice is equal to the energy lost from the water.

Qice = -Qliquidwater

Qice = mΔHfusion = (50.0 g)(333 J/g)

Qice = 16650 J

slide13

16650 J = -Qliquid water

16650 J = -mliquidwaterCliquid waterΔTliquid water

16650 J = -mliquidwater (4.18 J/g°C)(-26.5°C)

16650 J = -mliquidwater(-110.77 J/g)

mliquid water = 150.311 g

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