What’s the MATTER:Specific Heat of Matter

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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

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