Lab 12: Heat, Energy, and TemperatureThis is it!!

Today we are going to measure the specific heat of an unknown metal.

Important terms:Temperature, T: a measure of a body’s hotness or coldness. An measure of the average kinetic energy of the molecules

Heat, Q: a from of energy used to do work. Specifically, heat is the energy that must be transferred

Specific Heat, c: the amount of heat required to raise the the temperature of an object by one degree Celsius.

Tm

Qc

Q = change in heatT = change in temperaturem = mass of the object

How much heat is required to raise 5 kg of water for 20 degrees C to 80 degrees C?

JCCCkg

JkgTmcQ 1257000)2080)(4190)(5(

TmcQ This equation is not valid when a substance undergoes a “change of phase”(ice water, water vapor)

When heat energy is added to change to phase of a substance this energy is used to break the bonds of the molecules. During this process, the temperature does not change.

Consider a 1 kg block of ice at –50 C. Slowly adding heat causes the ice to warm. Let’ssay we warm the ice until it reaches 0 C.

J

CkgCJkgTmcQ

11100

)50)(/2220)(1(

Energy

Tem

pera

ture

-50 C

0 Cice

Once the temperature reaches ) C, the temperature will not change as more heatenergy is added. All the will happen is that the ice will change phase.

The heat required to melt the ice is called the latent heat of fusion, f.

JkgJkgmfQ 333000)/333000)(1( So, the 333000 J is the energy needed to change the ice completely into water.

Energy

Tem

pera

ture

-50 C

0 Cice

ice + water mix

Once we have water at 0 C, we will continue to add heat energy to heat the water to 100 C

J

CkgCJkgTmcQ

419000

)100)(/4190)(1(

Energy

Tem

pera

ture

-50 C

0 Cice

ice + water mix

100 C

water

water + steam

Q= m V

Today we are going to do two experiments.

Power = Energy/time = E/t so then E = Power x time = P t

mc

tPTTTmctP i

Here Ti is the initial temperature. You are essentially going to measure the temperature rise of the water as a function of the heating time.

Tem

pera

ture

Time

the slope in this regionis equal to:

)(slopem

Pc

mc

Pslope

The next experiment will allow you to measure the specific heat of a metal.

You will take a hot piece of metal and drop it into a cool cup of water,from the temperature increase you can work out the specific heat.

Consider a piece of metal, an aluminum cup containing some mass of water.

The metal starts out with an initial temperature equal to, metaliT while the

cup and water start out at the same temperature,cupi

wateri TT

BEFORE AFTERmetaliT

cupi

wateri TT

When the metal is dropped into thewater, all three objects equalize tothe same temperature, Tf

Tf

Since this is a closed system, the water and cup gain the energy lost by the meta.So that means:l

)()()( cupif

cupcupif

waterwaterf

metali

metalmetal

cupwatermetal

TTcmTTcmTTcm

QQQ

Solving for the specific heat of the metal gives us:

)(

)()(

fmetali

metal

cupif

cupcupif

waterwatermetal

TTm

TTcmTTcmc