1. Thermal Equilibrium

2. Heat always travels from hot to cold. So if something hot is put near something cold, the heat tries to move to the cold object. This makes the hot object colder and the cold object hotter. When will it stop?

3. Thermal Equilibrium If a hot object is placed next to a cold object, heat passes between the two until they reach the same temperature. Two objects at the same temperature are in Thermal Equilibrium. Therm- means heat. Equi- means equal. Librium- means scales or balance. Thermal Equilibrium means an equal balance of heat.

4. Thermal Equilibrium If I have a 100g of water at 20°C and add 100g of water at 40°C, what will happen? The 20° water will get warmer and the 40° water will get cooler. What will the final temperature be? What if I add another 100g of 40° water?

5. Specific Heat Capacity What happens if I put 100g of 40°C iron into 100g of 20°C water? What will the equilibrium temperature be? Water heats up more slowly that iron. Can you still just average their temperatures?

6. Specific Heat Capacity Water has a higher specific heat capacity (it holds more heat). c pw = 4.18J/g°C c pi = 0.449J/g°C Heat (energy) travels from the iron from into the water. We can find the amount of energy that is transferred by multiplying the mass of the iron by its specific heat by the change in temperature. Q i = mass*c p *∆T

7. Specific Heat Capacity We could find it using the mass and specific heat of water, but the sign will be opposite. Q i = – Q w mass i *c pi * ∆T = – mass w *c pw * ∆T mass i *c pi *(T F – T i ) = – mass w *cp w *(T F – T w ) 100g*(0.449J/g°C)*(T F – 40°C) = – 100g*(4.18J/g°C)*(T F – 20°C) 44.9J/°C*(T F – 40°C) = – 418J/°C*(T F – 20°C) T F *44.9J/°C – 40°C (44.9J/°C) = T F * – 418J/°C + 418J/°C(20°C) T F *44.9J/°C – 1796J = T F * – 418J/°C + 8360J T F *44.9J/°C + T F *418J/°C = 8360J + 1796J T F (462.9J/°C) = 10156J T F = 21.94°C

8. How ‘bout a cold one? If I place 100g of ice (-5°C) in 100g of water (20°C), what will the equilibrium temperature be? It starts out easy—the ice warms to 0°C and the water cools to 15°C. But don’t forget it takes energy to change solid ice into liquid water (a lot of energy in fact). The latent heat of fusion of water is 333.55J/g. (It takes 333.55J to change 1g of ice into 1g of water.) The water can only transfer heat to the ice while it is hotter (until it reaches 0°C). (4.18J/g*K)*(15K)*(100g) = 6270J That’s only enough energy to melt 18.8g of ice.