1. Specific Heat

2. Temperature is the average kinetic energy of the particles in a substance. The SI unit for temperature is the Kelvin. a. K = C (10 C = 283 K) b. C = K – 273 (10 K = -263 C)
Thermal Energy – the total of both the kinetic and potential energy of all the particles in a substance.

3. Specific Heat Specific heat
is the amount of heat energy required to raise the temperature of 1 g of a substance by 1 °C
is different for different substances
in the SI system has units of J/g C
in the metric system has units of cal/g C

4. Units for Energy
The SI unit for energy is the Joule. English units for energy include the unit calories.
1000 calories (little c) are in one Calorie (Big C), which is used to measure the energy in foods (that the human body can make use of).
4.186 J = 1 cal cal = 1 Calorie

5. Have you ever been super hot at the beach, only to run down to the water and realize it is so cold? How can this be?
Why does the land heat up faster and cool down quicker than the water?
Sand =
664 J / kg C
Water = J / kg C

6. water metal Why do materials have different specific heats?
Water molecules form strong bonds with each other;(hydrogen bonding), therefore it takes more heat energy to break the molecules apart.
Metals have weaker bonds (metallic – electron sea model) and do not need as much energy to break the metal atoms apart.

7. Learning Check
A. If the same amount of heat is added to two different substances, the one with a large specific heat….
1) has a smaller increase in temperature
2) has a greater increase in temperature
B. When ocean water cools, the surrounding air
1) cools 2) warms 3) stays the same
C. Sand in the desert is hot in the day and cool
at night. Sand must have a
1) high specific heat ) low specific heat

8. Solution
A. For the same amount of heat added, a substance with a large specific heat
1) has a smaller increase in temperature
B. When ocean water cools, the surrounding air
2) warms
C. Sand in the desert is hot in the day and cool at night. Sand must have a
2) low specific heat

9. Examples of Specific Heats

10. Specific Heat Equation
Q = m x C(p) x T
Q = Change in Thermal Energy (Joules or cal)
M = mass of substance (Grams of Kilograms)
C = specific heat of substance (p) = what phase the material is in
T = change in temperature (Tf – Ti) (Celsius or Kelvin)

11.  Calculating Heat with Temperature Increase
How many joules are absorbed by 45.2 g of aluminum if its temperature rises from C to 76.8 C? (The specific heat of Aluminum is J/goC
Given: Mass = 45.2 g
Specific heat (C) for aluminum = J/g C
Initial temperature = 12.5 C
Final temperature = 76.8 C
Need: Heat (q) in joules (J)

12.  Calculating Heat Loss
A 225-g sample of hot tea cools from 74.6 C to 22.4 C. How much heat, in kilojoules, is lost, assuming that tea has a specific heat of J/goC?
Given: Mass = 225 g Specific Heat (C) for tea = J/g C
Initial temperature = 74.6 C Final temperature = 22.4 C
Need heat (q) in kilojoules (kJ)

13.  Calculating Mass Using Specific Heat
Ethanol has a specific heat of 2.46 J/g C. When 655 J are added to a sample of ethanol, its temperature rises from 18.2 C to 32.8 C. What is the mass of the ethanol sample?
Given: Heat (q) = 655 J
Specific Heat (C) for ethanol = 2.46 J/g C
Initial temperature = 18.2 C Final temperature = 32.8 C
Need: Mass of ethanol sample

14. Need to do a heating curve of Water Test

15. But what happens when we go through a phase change?
Heat is transferred to the particles as they heat up. It’s what causes the molecules to speed up. This is what we have been calculating.
q = (mass)(c)(∆T)
But what happens when we go through a phase change?
Can we use the same equation?

16. Remember the Heating/Cooling Curve for Water?
What did we learn from this activity?
As we go through a phase change, the temperature doesn’t change!
Q = m x T x C(p)
If we did used the same equation, then the heat added would always equal 0, because the temperature doesn’t change, and that can’t be true!

17. So what do we do? Find a new equation!
Heat of Fusion – It’s the amount of heat absorbed in order to convert all of a solid into its liquid.
Heat of Vaporization– It’s the amount of heat absorbed in order to convert all of a liquid into its gas form.

18. Heat of Fusion Calculation – Energy required to convert all a solid to liquid at a given temperature.
Q = (heat of fusion) x (mass)
Heat of Vaporization Calculation – Energy required to convert all of a liquid to gas at a given temperature.
Q = (heat of vaporization) x (mass) * (Heat of fusion and vaporization values will come from a chart)

19. Example: What is the heat required to melt 25. 0 grams of ice
Example: What is the heat required to melt 25.0 grams of ice? Is this Heat of Fusion or Vaporization?
Q = (heat of fusion) x (mass) (Heat of fusion of water is 334 J/g)
Q = (334 J/g) x (25.0 g) = 8350 Joules

20. Example: What is the heat required to evaporate 25. 0 grams of water
Example: What is the heat required to evaporate 25.0 grams of water? Is this Heat of Fusion or Vaporization?
Q = (heat of vaporization) x (mass) (Heat of vaporization of water is 2260 J/g)
Q = (2260 J/g) x (25.0 g) = Joules

21. So… if I want the total heat required to melt ice and turn it to steam what would I need?

22. 1) To melt the ice I need to multiply the heat of fusion with the mass q =(heat of fusion)x(mass)
2) Then, there is moving the temperature from 0 C to 100C… for this there is a change in temperature so we can use… q = (mass)x(∆T)x(c)
3) But wait, that just takes us to 100 C, what about vaporizing the molecules? Well, then we need q = (heat of vaporization)x(mass)

23. Practice Problem: What quantity of heat is required to melt 500
Practice Problem: What quantity of heat is required to melt g of ice and then heat the water to steam at 100 oC?
Heat of fusion of ice = 333 J/g
Specific heat of water = 4.2 J/g•K
Heat of vaporization = 2260 J/g

24. How much heat is required to melt 500
How much heat is required to melt 500. g of ice and heat the water to steam at 100 oC?
1. To melt ice
q = (500. g)(333 J/g) = x 105 J
2. To raise water from 0 oC to 100 oC
q = (500. g)(4.2 J/g•K)( )K = 2.1 x 105 J
3. To evaporate water at 100 oC
q = (500. g)(2260 J/g) = x 106 J
4. Total heat energy = 1.51 x 106 J = 1510 kJ