1. What’s the MATTER: Specific Heat of Matter

2. Matter, Specific Heat of Matter At the conclusion of our time together, you should be able to: 1. Define specific heat 2. Use specific heat to determine energy changes

3. Thermochemistry  Some Definitions: Calorimeter – a device to measure the energy absorbed or released as heat in a chemical or physical change Temperature – measure of the average kinetic energy of the particles in a sample of matter Joule – the SI unit of heat

4. Thermochemistry  Some Definitions: Heat – energy transferred between samples of matter Specific Heat – the amount of energy required to raise the temperature of one gram of a substance by one Celsius degree or one Kelvin 1 Calorie/4.184 Joules – will do the above with water

5. Thermochemistry  Some Definitions: Enthalpy (Heat) of Fusion – amount of energy gained or lost by a system as heat during melting or freezing Enthalpy (Heat) of Vaporization – amount of energy gained or lost by a system as heat during boiling or condensation

6. Specific Heat Calculations:  q = c p x m x t: q = energy lost or gained c p = specific heat of the substance at a specific pressure m = mass of the sample t = change in temperature (final – initial)

7. Practice #1  q = c p x m x t: q = 59.912 J c p = x m = 36.359 g t = 152.0 o C 59.912 J = (x)(36.359 g)(152.0 o C) = 0.01084 J/g o C

8. Practice #2  q = c p x m x t: q = -800. J c p = 0.4210 J/g o C m = 73.174 g t = (x – 102.0 o C) -800. J = 0.4210 J / g o C (73.174 g)(x – 102.0 o C) -800. = 30.81x – 3142 2342 = 30.81x = 76.0 o C

9. Matter, Specific Heat of Matter Let’s see if you can : 1. Define specific heat 2. Use specific heat to determine energy changes

10. Define Specific Heat Specific Heat – the amount of energy required to raise the temperature of one gram of a substance by one Celsius degree or one Kelvin

11. Practice #3  q = c p x m x t: q = -185.4 J c p = 0.440 J/g o C m = x g t = -1475 o C -185.4 J = (0.440 J/g o C )(x)(-1475 o C) -185.4 J = -649 Jg = 0.29 g

12. Practice #4  q = c p x m x t: q = x J c p = 0.0335 cal/g o C (4.184 J/cal) m = 152.00 g t = -51.5 o C x = (0.140164 J/g o C )(152.00 g)(-51.5 o C) = -1.10 x 10 3 J