reaction energy

Reaction Energy

Specific Heat

The specific heat of any substance is the amount of heat required to raise the temperature of one gram of that substance by one degree Celsius.

Because different substances have different compositions, each substance has its own specific heat.

Exothermic and Endothermic

Exothermic: Heat flows out of the system (to the surroundings). The value of ‘q’ is negative.

Endothermic: Heat flows into the system (from the surroundings). The value of ‘q’ is positive.

Specific Heat

q = m Cp ΔT

q = heat (J)

Cp = specific heat (J/(g.°C)

m = mass (g)

ΔT = change in temperature = Tf – Ti (°C)

Measuring Heat Heat changes that occur during

chemical and physical processes can be measured accurately and precisely using a calorimeter.

A calorimeter is an insulated device used for measuring the amount of heat absorbed or released during a chemical or physical process.

A coffee-cup calorimeter made of two Styrofoam cups.

Phase Changes

Solid

Liquid

Gas

Melting Vaporization

Condensation

Freezing

Liquid

Sublimation

Melting Vaporization

Deposition

Condensation

Solid

Freezing

Gas

Latent Heat

q = m Hv

HHvv = latent heat of vaporization (J/g) = latent heat of vaporization (J/g)

HHff = latent heat of fusion (J/g) = latent heat of fusion (J/g)

q = m Hf

Energy and Phase Change Heat of vaporization (Hv) is the energy

required to change one gram of a substance from liquid to gas.

Heat of fusion (Hf) is the energy required to change one gram of a substance from solid to liquid.

Specific Heat Problem1) The temperature of a sample of iron

with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 114 J heat. What is the specific heat of iron?

q = ∆TCironm10.0-114 = (25.0 – 50.4)

Ciron = 0.449 J/g°C

2) A piece of metal absorbs 256 J of heat when its temperature increases by 182°C. If the specific heat of the metal is 0.301 J/g°C, determine the mass of the metal.

m = 4.67 g

Specific Heat Problem

3) If 335 g water at 65.5°C loses 9750 J of heat, what is the final temperature of the water?

Tf = 58.5 °C

Specific Heat Problem

4) As 335 g of aluminum at 65.5°C gains heat, its final temperature is 300.°C. Determine the energy gained by the aluminum.

q = 70500 J

Specific Heat Problem

5) How much heat does it take to melt 12.0 g of ice at 0 C?

q = Hfm12.0 (334)

q = 4010 J

Latent Heat Problem

6) How much heat must be removed to condense 5.00 g of steam at 100 C?

q = Hvm5.00 (2260)

q = 11300 J

Latent Heat Problem

7) If 335 J of heat are added to melting 5.00 g of gold, what is the latent heat of fusion for gold in J/g?

Lf = 67 J/g

Latent Heat Problem

8) The latent heat of fusion for platinum is 119 J/g. Platinum absorbs 735 J of heat. What is the mass of platinum?

m = 6.18 g

Latent Heat Problem

HEATING AND COOLING CURVES

The heating curve has 5 distinct regions.

The horizontal lines are where phase changes occur.

melting

vaporization

During any phase change, temperature is constant.

melting

vaporization

Use q = m Cp ΔT for all diagonal lines. Use q = m Hf for melting and q = m Hv for boiling.

melting

vaporization

Heating Curve for WaterHeating Curve for WaterSection A:Section A:

Ice is being heatedIce is being heated to the melting to the melting

point.point.

Section B:Section B:The iceThe ice

is melting.is melting.

Section C:Section C:Water is Water is

being heatedbeing heated to the boiling to the boiling

point.point.Section E:Section E:Steam isSteam is

being heated.being heated.

Section D:Section D:Water isWater isBoiling.Boiling.

Heating Curve for WaterHeating Curve for Water

Section ASection Aq = m Cq = m Cpp ΔT(solid)

Section BSection Bq = mHq = mHff

Section CSection Cq = m Cq = m Cpp ΔT(liquid)

Section ESection Eq = m Cq = m Cpp ΔT(gas)

Section DSection Dq = mHq = mHvv

Solving Problems

For water:

HeatIce

Below 0 C

+

Melt Ice

At 0 C

+Heat

Water

0 C -100 C

+

Boil Water

At100 C

+Heat

Steam

Above100 C

Specific Heat and Latent Heat9) How much heat does it take to heat

12 g of ice at – 6 C to 25 C water? Round to a whole number.

You begin at ice below 0 °C. Note: The final temperature for this process cannot exceed 0 °C.

q = ∆TCicem12(2.05)(0 - – 6)q = 148 J

How much heat does it take to heat 12 g of ice at – 6 C to 25 C water? Round to a whole number.

Since the temperature needs to rise to 25 °C, you must melt the ice next.

q = Hfm12 (334)

q = 4008 J

Specific Heat and Latent Heat, cont.

How much heat does it take to heat 12 g of ice at – 6 C to 25 C water? Round to a whole number.

You now have water at 0 °C. The final temperature of the water should be 25 °C.

q = ∆TCwaterm12 (4.18) (25 - 0)

q = 1254 J

Specific Heat and Latent Heat, cont.

How much heat does it take to heat 12 g of ice at – 6 C to 25 C water? Round to a whole number.

Finally, add all of the q values together.

q = 148 + 4008 + 1254

q = 5410 J

Specific Heat and Latent Heat, cont.

10) How much heat does it take to heat 35 g of ice at 0 C to steam at 150 C? Round to a whole number.

You begin with ice at 0 °C, so you should melt it first.

q = Hfm35(334)

q = 11690 J

Specific Heat and Latent Heat

How much heat does it take to heat 35 g of ice at 0 C to steam at 150 C? Round to a whole number.

You now have water at 0 °C. The final temperature of the steam is to be 150 °C. You must take the water to 100 °C before you can even convert it to steam.

q = ∆TCwaterm35 (4.18)(100 - 0)

q = 14630 J

Specific Heat and Latent Heat, cont.

How much heat does it take to heat 35 g of ice at 0 C to steam at 150 C? Round to a whole number.

Now you have water at 100 °C. Convert this to steam.

q = Hvm35 (2260)

q = 79100 J

Specific Heat and Latent Heat, cont.

How much heat does it take to heat 35 g of ice at 0 C to steam at 150 C? Round to a whole number.

You must now take the steam at 100 °C to 150 °C.

q = ∆TCicem35 (2.02)(150 - 100)

q = 3535 J

Specific Heat and Latent Heat, cont.

How much heat does it take to heat 35 g of ice at 0 C to steam at 150 C? Round to a whole number.

Finally, add all of the Q values together.

q = 11690 + 14630 + 79100 +

q = 108955 J

3535

Specific Heat and Latent Heat, cont.

11) How much heat does it take to convert 16.0 g of ice to water at 0 C?

(5340 J)

Specific Heat and Latent Heat

12) How much heat does it take to heat 21.0 g of water at 12.0 C to water at 75.0 C?

(5530 J)

Specific Heat and Latent Heat

13) How much heat does it take to heat 14.0 g of water at 12.0 C to steam at 122.0 C?

(37400 J)

Specific Heat and Latent Heat

Calorimetry For calorimetry problems, use the

equation: – m Cp ΔT = m Cp ΔT

which is based on the law of conservation of energy.

Heat lost equals heat gained.

14) 125 g of water at 25.6°C is placed in a foam-cup calorimeter. A 50.0 g sample of the unknown metal is heated to a temperature of 115.0°C and placed into the water. Both water and metal have attain a final temperature of 29.3°C. Determine the specific heat of the metal.

Calorimetry Problem

125 g of water at 25.6°C is placed in a foam-cup calorimeter. A 50.0 g sample of the unknown metal is heated to a temperature of 115.0°C and placed into the water. Both water and metal have attain a final temperature of 29.3°C. Determine the specific heat of the metal.

Find the heat gained by the water.

q = ∆TCwaterm125 (4.18) (29.3 – 25.6)q = 1930 J

Since heat lost equals heat gained, determine the specific heat of the metal.

q = ∆TCmetalm50.0-1930 = (29.3 – 115.0)Cmetal = 0.450 J/g°C

125 g of water at 25.6°C is placed in a foam-cup calorimeter. A 50.0 g sample of the unknown metal is heated to a temperature of 115.0°C and placed into the water. Both water and metal have attain a final temperature of 29.3°C. Determine the specific heat of the metal.

15) You put 352 g of water into a foam-cup calorimeter and find that its initial temperature is 22.0°C. What mass of 134°C lead can be placed in the water so that the equilibrium temperature is 26.5°C?

m = 477 g

Calorimetry Problem

16) You put water into a foam-cup calorimeter and find that its initial temperature is 25.0°C. What is the mass of the water if 14.0 grams of 125°C nickel can be placed in the water so that the equilibrium temperature is 27.5°C?

m = 58.0 g

Calorimetry Problem