Heat Examples Heat up Water from 21C to 95C m = 1 kg c = 4180 Q = mcDT Q = (1)(4180)(95

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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
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Thermal Equilibrium

• 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, T95)
• Mixed with a bowl (m.8kg, c800, T20)
• (1)(4180)(95-T) (.8)(800)(T-20)
• 397100 4180T 640T 12800
• 409900 4820T T 85 0C

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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

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Problem Solving Hints

• 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.

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Linear Expansion

• 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.