Heat Examples Heat up Water from 21C to 95C m = 1 kg c = 4180 ...

Heat Examples Heat up Water from 21C to 95C m = 1 kg c = 4180 Q = mcT Q = (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 mcT = mcT 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 mcT and the c for

ice, to warm ice. At 0 the ice melts, use mLf Between 0 and 100, it is water, use mcT and the c for water, to warm water. At 100 the water converts to steam, use mLv. Above 100, use mcT and the c for steam. Linear Expansion All objects expand with increase in temperature, contract with decrease in

temperature. L= L0T L0 is original length at original temperature, is the expansion coefficient, and T is the change in temperature. You are solving for the change in length.