1. Partner Talk…
It is a warm day and you go to the park. Is everything going to feel the same? Why/Why not?

2. Unit 4 lecture 5:
PS 3-1 Create a computation model to calculate the change in the energy of one component in a system when the change in energy of the other component(s) and energy flows in and out of the system are known.

3. Specific Heat (Heat Capacity)
Specific heat (c): the amount of heat to increase 1g of substance by 10C
The larger the heat capacity the better at resisting changes in temperature! (the slower it will heat up

4. Specific Heat of some common substances:
Iron (0.46 J/g0C)
Plastic (~1 J/g0C)
Paper (1.4 J/g0C)
Water (4.18 J/g0C)
Which of the above substances would heat up the fastest? WHY? 
Which of the above substances would heat up the slowest (most resistant to heat change)? WHY?

5. Specific heat demonstration- cups and candles
1. What do you expect to happen to the water filled cup when over the candle?
2. What happened?
3. What do you expect to happen when the water is taken out of the cup and it is over the candle?
4. Explain the different results using the term specific heat.

6. Heat Energy (Q) can be found using the equation… q=mcT
Q = energy change (J)
m= mass of substance being heated (grams)
C = specific heat capacity of substance (J/g0C)
T= change in temperature (0C)

7. Mass = 7.40 g Q = x Q= m Cp ∆T Q= (7.40) (4.184) (17.0) Q= 526J
Example 1– Calculate the amount of heat energy (in joules) needed to raise the temperature of 7.40 g of water from 29.0°C to 46.0°C.
Specific Heat of Water = 4.184
Mass = 7.40 g
Q = x
Temperature Change = 46.0°C – 29.0°C = 17.0°C
Q= m Cp ∆T
Q= (7.40) (4.184) (17.0)
Q= 526J 22

8. Example 2: Calculate the amount of heat energy (in joules) needed to raise the temperature of 50.0 g of Iron from °C to 5.0°C (heat capacity of iron is 0.46 j/g0C).