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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 95Cm = 1 kg c = 4180Q = mcDTQ = (1)(4180)(95 – 21)Q = 309,320 Joules
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) • 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 • 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 • 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.
