1. Specific Heat IB1 Chemistry

2. Specific Heat Adding Energy to a material Causes the
Temperature to go up.
Taking energy away from a substance causes the temp. to
Go down!

5. Specific Heat
The amount of energy required to raise the temperature of a material (substance).
It takes different amts of energy to make the same temp change in different substances.
We call the amt required: Specific Heat!

6. Specific Heat of water
The Cp is high because H2O mols. form strong bonds w/each other.
It takes a lot of energy to break the bonds so that the the molecules can then start to move around faster (HEAT UP).

7. Example: Specific Heat of Water
Cp = 4,184 Joules of energy to raise the temperature of 1kg 1°C.
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Why Cp?
Cp Stands for “Heat Capacity”

8. Calculating Specific Heat
The Greek letter Δ means “change in”

9. EXAMPLE : p162 Mass = 45kg Q = 203,000J Δt = 40°-28° Δt = 12° Cp = ?
Q =m x Cp x Δt
Q/(m x Δt) = Cp
Cp = 376 J/(kg °C)

10. SPECIFIC HEAT
Determine the energy (in kJ) required to raise the temperature of g of water from 20.0 oC to 85.0 oC?
m = g DT = Tf -Ti = oC = 65.0 oC
q = m x s x DT s (H2O) = J/ g - oC
q = (100.0 g) x (4.184 J/g-oC) x (65.0oC)
q = J (1 kJ / 1000J) = 27.2 kJ
Determine the specific heat of an unknown metal that required 2.56 kcal of heat to raise the temperature of g from 15.0 oC to oC?
S = cal /g -oC

11. LAW OF CONSERVATION OF ENERGY
The law of conservation of energy (the first law of thermodynamics), when related to heat transfer between two objects, can be stated as:
The heat lost by the hot object = the heat gained by the cold object
-qhot = qcold
-mh x sh x DTh = mc x sc x DTc
where DT = Tfinal - Tinitial

12. LAW OF CONSERVATION OF ENERGY
Assuming no heat is lost, what mass of cold water at 0.00oC is needed to cool g of water at 97.6oC to 12.0 oC?
-mh x sh x DTh = mc x sc x DTc
- (100.0g) (1 cal/goC) ( oC) = m (1 cal/goC) ( oC)
8560 cal = m (12.0 cal/g)
m = cal / (12.0 cal/g)
m = 713 g
Calculate the specific heat of an unknown metal if a g piece at 100.0oC is dropped into mL of water at 17.8 oC. The final temperature of the mixture was 39.4oC.
s (metal) = cal/g oC

13. PRACTICE PROBLEM #7
1. Iron metal has a specific heat of J/goC. How much heat is transferred to a 5.00 g piece of iron, initially at 20.0 oC, when it is placed in a beaker of boiling water at 1 atm?
2. How many calories of energy are given off to lower the temperature of g of iron from oC to 35.0 oC?
3. If 3.47 kJ were absorbed by 75.0 g H2O at 20.0 oC, what would be the final temperature of the water?
4. A 100. g sample of water at 25.3 oC was placed in a calorimeter g of lead shots (at 100 oC) was added to the calorimeter and the final temperature of the mixture was 34.4 oC. What is the specific heat of lead?
5. A 17.9 g sample of unknown metal was heated to oC. It was then added to g of water in an insulted cup. The water temperature rose from oC to 23.98oC. What is the specific heat of the metal in J/goC?
180. J
1.23 x 103 cal
31.1 oC
1.28 J/g oC
0.792 J/goC

14. GROUP STUDY PROBLEM #7
_____1. A g metal bar requires kJ to change its temperature from 22.0oC to 100.0oC. What is the specific heat of the metal in J/goC?
_____2. How many joules are required to lower the temperature of g of iron from 75.0 oC to 25.0 oC?
_____ 3. If 40.0 kJ were absorbed by g H2O at 10.0 oC, what would be the final temperature of the water?
_____ 4. A 250 g of water at oC is mixed with mL of water at 5.0 oC. Calculate the final temperature of the mixture.
_____5. A 400 g piece of gold at 500.0oC is dropped into 15.0 L of water at 22.0oC. The specific heat of gold is J/goC or cal/goC. Calculate the final temperature of the mixture assuming no heat is lost to the surroundings.