1. Heat Examples Heat up Water from 21C to 95C m = 1 kg c = 4180 Q = mc  T Q = (1)(4180)(95 – 21) Q = 309,320 Joules

2. Thermal Equilibrium Two objects at different temperatures will eventually reach a common temp T f ? Heat Gained = Heat Given mc  T = mc  T 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 0 C

3. Latent Heat = the heat required to change the state of the object (liquid to gas, solid to liquid) L f = 3.35 E 5 J/kg (to melt ice, freeze water) L v = 2.26 E 6 J/kg (to make steam, condense water) Q = mL Water at densest point at 4 0 C

4. Problem Solving Hints If ice is below 0, use mc  T and the c for ice, to warm ice. At 0 the ice melts, use mL f Between 0 and 100, it is water, use mc  T and the c for water, to warm water. At 100 the water converts to steam, use mL v. Above 100, use mc  T and the c for steam.

5. Linear Expansion All objects expand with increase in temperature, contract with decrease in temperature.  L= L 0  T L 0 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.