Determining the specific heat of three metals
Method of mixtures:
When different parts of an isolated system are at different temperatures, heat will
flow from the part of higher temperature to the part of lower temperature.
Therefore, from conservation of energy:
Heat lost by hot object = Heat gained by cold object.
Aim:
This experiment is conducted in order to determine the specific heats of three
different metals using the method of mixtures.
Method of this experiment :
(copper ,ead ,Aluminum).
1. Measure the mass of the three different supplied cubic metals.
2. Heat the supplied cubic masses of metal in boiling water and record the
temperature.
3. Place some cold water into a separate container of known mass, so that this
is enough to just cover the cubic masses. Record the mass and temperature
of the water.
4. Transfer the hot metal quickly into a container of cold water.
5. Wait until the temperature does not fluctuate and then record the final
temperature.
6. Calculate the specific heat of the metal.
Data collection and calculation :
Zinc :
Mass
Zinc(m1)
of
±0.001kg
Mass
of empty
cup(m2)
±0.001kg
Mass of cup
with water(m3)
±0.001kg
Temperature of
cold water(T1)
±0.1□
Temperature
of metal with
object(T2)
±0.1□
Trial 1
Trial 2
Trial 3
0.054
0.054
0.054
0.002
0.002
0.002
0.040
0.041
0.046
20.0
19.0
18.0
86.0
89.0
79.0
Temperature
26.0
25.0
in
equilibriu
m (T3the
) ±0.1□
From
Conservation of energy, we can know:
23.0
Heat lost by Copper = Heat gained by cold water So
the Zinc : cm(T2-T3) = Cold water: cm(T3-T1) Trial 1:
–1
c(water)=4200 J · kg–l ·
M(water)=M3-M2=0.040-0.002=0.038kg
(±0.001kg)
C(zinc)×0.054×(86.0-26.0)=4200×0.038×(26.0-20.0)
C(zinc)=296 J · kg–l · –1
Trial 2:
c(water)=4200 J · kg–l ·
–1
m(water)=0.041-0.002=0.039 kg (±0.001kg)
C(zinc)×0.054×(89-25)=4200×0.039×(25-19)
C(zinc)=283 J · kg–l ·
–1
Trial 3:
M(water)=m3-m2=0.046-0.002=0.044 kg (±0.001kg)
m(water)=0.046-0.002=0.044 kg (±0.001kg)
C(zinc)=271J · kg–l · –1
C(zinc)×0.054×(79-23)=4200×0.039×(23-18)
The average specific heat of
zinc =(270.833+283.375+295.556)/3=283.255 J · kg–l ·
–1
Experiment result –accurate value
✕lOO% = percentage error of the
accurate va lue accurateSo the percentage of error is„|(283.255380)/380|×100%=25.46%
Aluminum :
Trial 1
Trial 2
Trial 3
Mass of Al(m1)
0.020
0.020
0.020
±0.001kg
Mass of empty
cup(m2)
±0.001kg
0.002
0.002
0.002
Mass of cup
with water(m3)
±0.001kg
Temperature of
cold water(T1)
±0.1□
Temperature
of metal with
object(T2)
±0.1□
Temperatur
e
in
equilibriu
0.046
0.042
0.043
18.0
21.0
17.0
84.0
87.0
74.0
23.0
30.0
24.0
m
the Aluminum : cm(T3-T2) = Cold water: cm(T3-T1)
(T3) ±0.1□
Trial 1:
c(water)=4200 J · kg–l · –1
M(water)=M3-M2=0.046-0.002=0.044kg (±0.001kg)
C(AI)=924 J · kg–l · –1
C(AI)×0.020×(84.0-23.0)=4200×(0.046-0.002)×(23.0-18.0)
Trial 2:
c(water)=4200 J · kg–l ·–1
m(water)=
0.042-0.002=0.040kg
(±0.001kg)
C(AI)×0.020×(87.0-30.0)=4200×0.040×(30.0-21.0)
–1
C(AI)=1326 J · kg–l ·
Trial 3:
M(water)=m3-m2=0.041
kg
(±0.001kg)
C(Al)×0.020×(74.0-24.0)=4200×0.041×(24.0-17.0)
C(Al)=1205 J · kg–l ·
–1
The average specific heat of
Aluminum =(924+1326+1205)/3=1152 J · kg–l ·
Experiment result –accurate value
–1
✕lOO% = percentage error of the accurate va lue
accurate
So the percentage of error is„
|(1152-910)/910|×100%=26.59%
Copper:
Mass of
Copper(m
1)
Mass±0.001kg
of empty
cup(m2)
±0.001kg
Mass of cup
with water(m3)
±0.001kg
Temperature of
cold water(T1)
±0.1□
Trial 1
Trial 2
Trial 3
0.062
0.062
0.062
0.002
0.002
0.002
0.041
0.040
0.039
19.0
18.0
18.0
Temperature
89.0
86.0
of metal with
object(T2)
Temperatur
±0.1□
25.0
27.0
e
in
equilibriu
m
the Copper : cm(T3-T2) = Cold water: cm(T3-T1)
(T3) ±0.1□
Trial 1:
c(water)=4200 J · kg–l · –1
M(water)=M3-M2=0.039kg
C(copper)×0.062×(89.0-25.0
C(AI)= 248 J · kg–l · –1
85.0
25.0
(±0.001kg)
)=4200×0.039×6
Trial 2:
c(water)=4200 J · kg–l ·
–1m(water)=0.038 kg (±0.001kg)
C(Copper)×0.062×(89.0-25.0)=4200×0.038×(25.0-18.0)
–1
C(Copper)=282 J · kg–l ·
Trial 3:
M(water)=m3-m2=
0.037kg
(±0.001kg)
C(Copper)×0.062×(85.0-18.0)=4200×0.037×(25.0-18.0)
C(Copper)=262 J · kg–l ·
–1
The average specific heat of
Copper =(248+282+262)/3=264 J · kg–l ·
Experiment result –accurate value
–1
✕lOO% = percentage error of the accurate va lue
accurate
So the percentage of error is„
|(264-385)/385|×100%=31.43%
Evaluation:
Limitation:
In the experiment of the three numbers of specific heat are both
higher than real specific heat of those three. Why like this? I think is :
1. We can’t move the metal from boil water to cold water
immediately. When the metal on the air, it will lose heat because
of the temperature of metal is much higher than air. So the
temperature of the metal will decrease.
2. The thermometer is cold, when put into the hot water, the heat of
the hot water will transfer to the thermometer, so the temperature
will decrease.
3. Sometimes not parallel to see the temperature, it will cause the
value incorrect.
4. when we use thermometer sometimes it touched the bottom of
the beaker. So the temperature of metal will be hotter.
Error:
We got the Parallax error during measure the temperature of using
thermometer. Because if we not parallel with the top of the mercury in the
thermometer, it will makes the result to be a little bit different. But not
affect a lot.
Improvement:
1. When you put the hot metal into the container of the cold water, you
must be careful not to let the metal touch the container and make the
container be occluded immediately so that the heat can be transfer into
the cold water.
2 The big problem is the heat losing on the way that the metal is
transferred into the container of the cold water. But I think if we can do
this experiment in vacuum the heat lost will be decrease and the data
we collected will more and more close to the data in theory.