What’s the MATTER:Specific Heat of Matter
Be Afraid…
Be Very Afraid…
Matter, Specific Heat of Matter
At the conclusion of our time together, you should be able to:
1. Define specific heat2. Use specific heat to determine energy
changes
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
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
Calorie – will do the above
15 Helpful Hints On The Lab Report from
Mr. T’s Vast Lab Experience!!!
Hint #9. Teamwork is essential. It allows you to blame someone else.
Thermochemistry
Some Definitions: Enthalpy of Fusion – amount of energy
lost by a system as heat during condensation or deposition
Enthalpy of Vaporization – amount of energy gained by a system as heat during boiling or sublimation
Calvin’s Definition of a Liquid:
Specific Heat Calculations:
q = cp x m x t:
q = energy lost or gained cp = specific heat of the substance
at a specific pressure m = mass of the sample t = change in temperature
(final – initial)
Practice #1
q = cp x m x t:
q = 59.912 J cp = x
m = 36.359 g t = 152.0 oC
= 0.01084 J/gK
Practice #2
q = cp x m x t:
q = -800. J cp = 0.4210 J/g oC
m = 73.174 g t = (x – 102.0 oC)
-800. J = 0.4210 J/goC (73.174 g)(x – 102.0 oC)
-800. = 30.81x – 31422342 = 30.81x
= 76.02 oC
Matter, Specific Heat of Matter
Let’s see if you can :
1. Define specific heat2. Use specific heat to determine energy
changes
Maxine’s Thoughts on Intelligence!!
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
Practice #3
q = cp x m x t:
q = -185.4 J cp = 0.440 J/g oC
m = x g t = -1475 oC
-185.4 J = (0.440 J/goC )(x)(-1475 oC)-185.4 J = -649 Jg
= 0.29 g
Practice #4
q = cp x m x t:
q = x J cp = 0.0335 cal/goC (4.184 J/cal)
m = 152.00 g t = -51.5oC
x = (0.140164 J/goC )(152.00 g)(-51.5 oC)
= -1.10 x 102 J
Maxine asks: What do you get with a woman who has PMS and a GPS??
A b---- who will find you!!!