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
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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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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

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


Thermal equilibrium
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, 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
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


Problem solving hints
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.


Linear expansion
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.


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