Transcript Specific Heat IB1 Chemistry

Specific Heat IB1 Chemistry
10
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!
15
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!
16
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).
Example:
Specific Heat of Water
• Cp = 4,184 Joules of energy
to raise the temperature of
1kg 1°C.
17
• video clip
Why Cp?
Cp Stands for “Heat Capacity”
18
Calculating Specific Heat
The Greek letter Δ means “change in”
19
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)
SPECIFIC HEAT
• Determine the energy (in kJ) required to raise
the temperature of 100.0 g of water from 20.0
oC to 85.0 oC?
m = 100.0 g DT = Tf -Ti = 85.0 - 20.0 oC = 65.0 oC
q = m x s x DT
s (H2O) = 4.184 J/ g - oC
q = (100.0 g) x (4.184 J/g-oC) x (65.0oC)
q = 27196 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 150.00 g from 15.0 oC to
200.0 oC?
S = 0.0923 cal /g -oC
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
LAW OF CONSERVATION OF ENERGY
• Assuming no heat is lost, what mass of cold water
at 0.00oC is needed to cool 100.0 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) (12.0-97.6oC) = m (1 cal/goC) (12.0 - 0.0 oC)
8560 cal = m (12.0 cal/g)
m = 8560 cal / (12.0 cal/g)
m = 713 g
• Calculate the specific heat of an unknown metal if a 92.00 g piece at
100.0oC is dropped into 175.0 mL of water at 17.8 oC. The final
temperature of the mixture was 39.4oC.
s (metal) = 0.678 cal/g oC
PRACTICE PROBLEM #7
1. Iron metal has a specific heat of 0.449 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?
180. J
2. How many calories of energy are given off to lower the temperature of 100.0 g of
iron from 150.0 oC to 35.0 oC?
1.23 x 103 cal
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?
31.1 oC
4. A 100. g sample of water at 25.3 oC was placed in a calorimeter. 45.0 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?
1.28 J/g oC
5. A 17.9 g sample of unknown metal was heated to 48.31 oC. It was then added to
28.05 g of water in an insulted cup. The water temperature rose from 21.04 oC
to 23.98oC. What is the specific heat of the metal in J/goC?
0.792 J/goC
GROUP STUDY PROBLEM #7
• _____1. A 250.0 g metal bar requires 5.866 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
100.0 g of iron from 75.0 oC to 25.0 oC?
• _____ 3. If 40.0 kJ were absorbed by 500.0 g H2O at 10.0 oC, what would
be the final temperature of the water?
• _____ 4. A 250 g of water at 376.3 oC is mixed with 350.0 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 0.131 J/goC or 0.0312 cal/goC.
Calculate the final temperature of the mixture assuming no heat is lost
to the surroundings.