Lab 12: Heat, Energy, and Temperature
This 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.
Q = change in heat
T = change in temperature
m = mass of the object
Q
c
mT
How much heat is required to raise 5 kg of water for 20 degrees C to 80 degrees C?
J
Q  mcT  (5kg)( 4190
)(80C  20C )  1257000 J
kg C
Q  mcT
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’s
say we warm the ice until it reaches 0 C.
Temperature
Q  mcT  (1kg)( 2220 J / kgC)(50C )
 11100 J
0C
-50 C
Energy
Once the temperature reaches ) C, the temperature will not change as more heat
energy 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.
Q  mf  (1kg)(333000 J / kg)  333000 J
Temperature
So, the 333000 J is the energy needed to change the ice completely into water.
ice + water mix
0C
-50 C
Energy
Once we have water at 0 C, we will continue to add heat energy to heat the water
to 100 C
Q  mcT  (1kg)(4190 J / kgC)(100C )
 419000 J
Q= m V
water + steam
Temperature
100 C
ice + water mix
0C
-50 C
Energy
Today we are going to do two experiments.
Power = Energy/time = E/t so then E = Power x time = P t
P t
P  t  mcT  T  Ti 
mc
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.
Temperature
the slope in this region
is equal to:
P
P
slope 
c
mc
m(slope )
Time
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.
Ti metal while the
Ti water  Ti cup
The metal starts out with an initial temperature equal to,
cup and water start out at the same temperature,
BEFORE
Ti metal
When the metal is dropped into the
water, all three objects equalize to
the same temperature, Tf
AFTER
Tf
Ti water  Ti cup
Since this is a closed system, the water and cup gain the energy lost by the meta.
So that means:l
Q metal  Q water  Q cup
m metalc metal (Ti metal  T f )  m water c water (T f  Ti )  m cup c cup (T f  Ti cup )
Solving for the specific heat of the metal gives us:
c
metal

m
water water
c
(T f  Ti )  m c
cup cup
mmetal (Ti metal  T f )
(T f  Ti
cup
)