Specific Heat IB1 Chemistry

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!

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!

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. video clip Why Cp? Cp Stands for “Heat Capacity”

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

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? 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? 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. 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? 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? 180. J 1.23 x 103 cal 31.1 oC 1.28 J/g oC 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.