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# Thermal Equilibrium - PowerPoint PPT Presentation

Heat Examples Heat up Water from 21C to 95C m = 1 kg c = 4180 Q = mc D T Q = (1)(4180)(95 – 21) Q = 309,320 Joules. Thermal Equilibrium. Two objects at different temperatures will eventually reach a common temp T f ? Heat Gained = Heat Given mc D T = mc D T

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## PowerPoint Slideshow about ' Thermal Equilibrium' - ulema

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### Heat Examples Heat up Water from 21C to 95Cm = 1 kg c = 4180Q = mcDTQ = (1)(4180)(95 – 21)Q = 309,320 Joules

• Two objects at different temperatures will eventually reach a common temp Tf?

• Heat Gained = Heat Given

• mcDT = mcDT

• Example Soup (m = 1 kg, c = 4180, T=95)

Mixed with a bowl (m=.8kg, c=800, T=20)

(1)(4180)(95-T) = (.8)(800)(T-20)

397100 – 4180T = 640T – 12800

409900 = 4820T T = 85 0C

Latent Heat = the heat required to change the state of the object (liquid to gas, solid to liquid)

• Lf = 3.35 E 5 J/kg (to melt ice, freeze water)

• Lv = 2.26 E 6 J/kg (to make steam, condense water)

• Q = mL

• Water at densest point at 4 0C

• If ice is below 0, use mcDT and the c for ice, to warm ice.

• At 0 the ice melts, use mLf

• Between 0 and 100, it is water, use mcDT and the c for water, to warm water.

• At 100 the water converts to steam, use mLv.

• Above 100, use mcDT and the c for steam.

• All objects expand with increase in temperature, contract with decrease in temperature.

• DL= L0aDT L0 is original length at original temperature, a is the expansion coefficient, and DT is the change in temperature. You are solving for the change in length.