# Experiment 4 Latent Heat of Fusion of Ice

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"Experiment 4 Latent Heat of Fusion of Ice"

```					Experiment 4: Latent Heat of
Fusion of Ice

PHYS 4C

July 3, 2008

Rui Lu

John J.A. Lin
Introduction and Objective:
Latent heat of fusion is involved during substance’s phase transition of fusion or
melting. At substance’s melting point, latent heat of fusion is needed to melt 1 kg
of such substance without changing in temperature. For calculating such heat, we
follow the formula Q=mLf. The purpose of this experiment is to compare the
experimented latent heat with the standard value of latent heat. By comparison,
we can calculate out our percentage error. For this experiment, possible ideal
condition may be an environment where ice does not melt into extra water. Our
hypothesis states that we can possibly get the standard latent heat of fusion of ice
under such ideal condition. Since our condition is not as perfect as the ideal one
and we hypothesizes that there should be some small percentage error between the
experimental value and standard value. Such comparison will lead us prove the
standard value of latent heat of fusion of ice. Our assumption proves correct by
our experiment.

Set-up:
- Measure mass of inner cup with the stirrer.
- Mix hot water from hot plate and cold water in the inner cup.
- Fill cup to one third full.
- Measure mass of the warm water.
- Cover calorimeter. Insert a thermometer into the rubber stopper.
- Dry some ice cube with a towel. Put ice into the inner cup and cover
Raw Data:

mw = mtot − mc
mw = 160.6 g − 59.1g
mw = 101.5 g

Note: calculation of latent heat of ice is present in Data Analysis section

Mass of the cup with
Temperature of warm water
stirrer and warm water 160.6 g = 0.1606 kg                                    36.5 °C
(T1)
(mtot)
Mass of the cup with                                Temperature of ice
stirrer                59.1 g = 0.0591 kg                                        -0.5 °C
(Ti）
(mc)
Mass of the warm
Temperature of final mixture
water                  101.5 g = 0.1015 kg                                       12.0 °C
(T2）
(mw）
Latent Heat of ice
Mass of the ice
31.9 g = 0.0319 kg       (experiment, assume ice is at 3.17×105 J/kg
(mi）
0°C）
Latent Heat of ice
Latent Heat of ice                                  (experiment, experimental
3.33×105 J/kg            temperature of ice at -       3.16×105 J/kg
(standard)
0.5°C）

Data Analysis:

A. Calculations
1. Latent Heat of Ice
a) Assume ice is at 0°C
Qwater = cw mw | (T2 − T1 ) |
J
Qwater = (4186          )(0.1015kg ) | (12.0°C − 36.5°C ) |
kg ⋅°C
Qwater   = 10409.5355 J

Qcup = cc mc | (T2 − T1 ) |
J
Qcup = (900          )(0.0591kg ) | (12.0°C − 36.5°C ) |
kg ⋅°C
Qcup   = 1303.155 J
Qice = cw mi | (T2 − Ti ) |
J
Qice = (4186        )(0.0319kg ) | [12.0°C − 0.0°C )] |
kg ⋅°C
Qice = 1602.4008 J

Qwater + Qcup = Q fusion + Qice
Q fusion = Qwater + Qcup − Qice
mi L f = cw mw | (T2 − T1 ) | + cc mc | (T2 − T1 ) | −cw mi | (T2 − Ti ) |
cw mw | (T2 − T1 ) | + cc mc | (T2 − T1 ) | −cw mi | (T2 − Ti ) |
Lf =
mi
10409.5355 J + 1303.155 J − 1602.4008 J
Lf =
0.0319kg
J
L f = 3.17 × 105
kg

b) Use experimental temperature of ice at -0.5°C
Qice ( −0.5°C →0.0°C ) = ci mi | (0.0°C − Ti ) |
J
Qice ( −0.5°C →0.0°C ) = (2090          )(0.0319kg ) | [0.0°C − (−0.5°C )] |
kg ⋅°C
Qice ( −0.5°C →0.0°C )   = 33.3355 J

Qice (0.0°C →12.0°C ) = cw mi | (T2 − 0.0°C ) |
J
Qice (0.0°C →12.0°C ) = (4186           )(0.0319kg ) | [12.0°C − 0.0°C ] |
kg ⋅°C
Qice (0.0°C →12.0°C )    = 1602.4008 J

Qwater + Qcup = Q fusion + Qice ( −0.5°C →0.0°C ) + Qice (0.0°C →12.0°C )
Q fusion = Qwater + Qcup − Qice ( −0.5°C →0.0°C ) − Qice (0.0°C →12.0°C )
mi L f = cw mw | (T2 − T1 ) | + cc mc | (T2 − T1 ) | −ci mi | (T2 − Ti ) | −cw mi | (T2 − Ti ) |
cw mw | (T2 − T1 ) | + cc mc | (T2 − T1 ) | −ci mi | (T2 − Ti ) | −cw mi | (T2 − Ti ) |
Lf =
mi
10409.5355 J + 1303.155 J − 33.3355 J − 1602.4008 J
Lf =
0.0319kg
J
L f = 3.16 × 105
kg
Compare the two results for Lf, 3.17×105 J/kg and 3.16×105 J/kg are
about the same. Because the temperature of ice of -0.5 °C is very close
to 0.0 °C, and then Qice(-0.5°C 0.0°C) only contributes a very small part of
the whole heat exchange. It does not affect Lf so much.

2.   Percent Difference
a) Assume ice is at 0°C
(valueexp erimental ) − (valueaccepted )
%difference =|                                          |
(Valueaccepted )
J                 J
(3.17 ×105      ) − (3.33 × 105 )
kg                kg
%difference =|                                      |
5 J
3.33 × 10
kg
%difference =| −0.0480 |
%difference = 4.80%

b) Use experimental temperature of ice at -0.5°C
(valueexp erimental ) − (valueaccepted )
%difference =|                                          |
(Valueaccepted )
J                J
(3.16 × 105     ) − (3.33 ×105 )
kg               kg
%difference =|                                     |
5  J
3.33 × 10
kg
%difference =| −0.0511|
%difference = 5.11%

B. Questions
1. What will happen to your percentage error if you put too much warm
water in the inner cup?
The percentage error would become bigger. By putting too much warm water
in the inner cup, the temperature of warm water (T1) would be increased by
certain amount. Since we still add the same amount of ice, and the ice would
absorb same amount of heat. Thus, the final temperature of the whole
mixture (T2) would be increased by the as same amount as T1 has increased.
It means that |(T2 – T1)| remains same. Because we still add the same ice, and
the temperature of ice (Ti) is same. With T2 getting higher, |(T2 – Ti)|
becomes bigger. According to the following equation:
c m | (T − T ) | +cc mc | (T2 − T1 ) | −cw mi | (T2 − Ti ) |
Lf = w w 2 1
mi
(cw mw + cc mc ) | (T2 − T1 ) | −cw mi | (T2 − Ti ) |
Lf =
mi
When |(T2 – T1)| remains the same, and |(T2 – Ti)| becomes bigger, then Lf
would become smaller. Compare to our experimental Lf, the new Lf is
smaller. Since our experimental Lf is already smaller than the accepted Lf,
then the new Lf is also smaller than then accepted Lf.
Given the following equation:
(valueexp erimental ) − (valueaccepted )
%difference =|                                           |
(Valueaccepted )
Because new Lf is less than accepted Lf, then the smaller valueexperimental
means |valueexperimental – valueaccepted| becomes bigger. The resulting
percentage difference becomes bigger as well. Therefore, if putting too much
warm water in the inner cup, the percentage error becomes larger.

2. Why must be the ice cubes dried off?
Without drying the ice cubes, the ice cubes may contain the extra water from
melting. When mixing the ice cubes with the warm water in calorimeter, this
extra water does not go through the phase transition of fusion because they
are already in the state of liquid. By drying off the ice cubes, we can avoid
the extra water’s effect on heat exchange. It helps make the calculation of
latent heat of fusion of ice more accurate.

Moreover, we want our experiment’s condition to be as ideal as possible.
Hence we want to achieve a state that solid ice is just itself. By drying off the
ice cubes, we can reduce the amount of extra water melted on ice’s surface.
Afterwards, we can measure the mass of pure solid ice cubes (mi) more
accurately without the mass of extra water. Moreover, by drying off, we can
reduce the possible heat exchange between the solid ice and the surrounding.
During the drying process, we are sort of “isolating” ice cubes in the towel
so the heat exchange can be at minimum. It helps maintain the temperature
of ice (Ti) after taking out of the bucket. Therefore, drying off the ice cubes
helps reduce the error in measuring mass of ice cubes (mi) and temperature
of ice (Ti).

3. When should you measure the mass of ice? Why?
We should measure the mass of ice after we dry the ice and before we put
them into the inner cup of the calorimeter. Ideally, we want to measure the
solid ice’s mass by itself without the extra water melted on it. Moreover, all
ice would be melt after mixing with warm water in the inner cup. It then is
very difficult to measure the mass of ice cubes accurately. Thus, it would be
more accurate to measure the mass of ice after drying off and before putting
into the calorimeter.

4.   What are the major sources of error for these experiments? How can
these errors be reduced?
Since the condition is not ideal, and then drying off cannot eliminate all extra
water melted on ice cubes. It cannot fully avoid heat exchange between ice
cubes and surrounding either. Thus, the mass and temperature of ice cubes
could not stay constant. Also, some extra melted water on ice’s surface can
have an effect on heat exchange. There would be possible mass loss,
temperature rise, and extra water emerging. These errors can be reduced if
we had an isolated ice storage area where the temperature gradient between
the surrounding and ice itself is very small. If measurements are taken in
place under this condition, the change of mass and temperature of ice will be
reduced. Thus, the related errors can be reduced as well.

C. Summary Charts

Experiment, assume ice is at 0°C
Latent Heat of Ice                         3.17×105 J/kg
%difference                                4.80%

Experiment, assume ice is at -0.5°C
Latent Heat of Ice                         3.16×105 J/kg
%difference                                5.11%

Possible Change of Condition               Possible Effects
Too much warm water in inner cup           % error increases
Not drying off ice cubes                   1. extra water melted from ice does not
go through fusion & calculation of
latent heat of fusion of ice becomes
less accurate
2. mass and temperature of ice are less
constant & measurements of mi and
Ti become less accurate

Error Analysis:

A. Possible maximum error for all masses (m)
The mass we measure for ice cubes (mi) is 31.9 g. For measuring this mass, we can
possibly read it to about 32 g.
∆mi mi − max error − mi − exp ermental
=
mi          mi − exp ermental
∆mi   32.0 g − 31.9 g
=
mi       31.9 g
∆mi
= 0.00313 = 0.313%
mi

Note: continued at next page
The mass we measure for the cup with stirrer (mc) is 59.1 g. For measuring this mass,
we can possibly read it to about 59 g.
∆mc mc − max error − mc − exp ermental
=
mc          mc − exp ermental
∆mc   59.0 g − 59.1g
=
mc       59.1g
∆mc
= −0.00169 = −0.169%
mc

The mass we measure for the cup with stirrer and warm water (mtot) is 160.6 g. For
measuring this mass, we can possibly read it 160.5 g.
∆mtot mtot − max error − mtot − exp ermental
=
mtot          mtot − exp ermental
∆mtot   160.5 g − 160.6 g
=
mtot        160.6 g
∆mtot
= −0.0006227 = −0.06227%
mtot

As we already know,
mc = 59.1g ± 0.10 g (0.169%)
mtot = 160.6 g ± 0.10 g (0.06227%)
Then,
mw = mtot − mc
mw = 101.5 g ± 0.20 g

∆mw mw − max error − mw − exp ermental   ∆mw mw − max error − mw − exp ermental
=                                        =
mw          mw − exp ermental            mw          mw − exp ermental
∆mw 101.7 g − 101.5 g                    ∆mw 101.3g − 101.5 g
=                                        =
mw      101.5 g                          mw      101.5 g
∆mw                                      ∆mw
= 0.001970 = 0.1970%                     = −0.001970 = −0.1970%
mw                                       mw

Note: continued at next page
B. Possible maximum error for all temperatures (T)
The temperature we measure for ice cube (Ti) is -0.5 °C. For measuring this
temperature, we can possibly read it to either about -0.7 °C or -0.3°C.
∆Ti Ti − max error − Ti − exp ermental ∆Ti Ti − max error − Ti − exp ermental
=                                      =
Ti           Ti − exp ermental         Ti           Ti − exp ermental
∆Ti   − 0.7°C − (−0.5°C)            ∆Ti   − 0.3°C − (−0.5°C)
=                                   =
Ti         −0.5°C                   Ti         −0.5°C
∆Ti                                 ∆Ti
= 0.4 = 40%                         = −0.4 = −40%
Ti                                  Ti

The temperature we measure for warm water (T1) is 36.5 °C. For measuring this
temperature, we can possibly read it to either about 36.3 °C or 36.7°C.
∆T1 T1− max error − T1− exp ermental ∆T1 T1− max error − T1− exp ermental
=                                    =
T1          T1− exp ermental         T1          T1− exp ermental
∆T1   36.3°C − 36.5°C               ∆T1   36.7°C − 36.5°C
=                                   =
T1        36.5°C                    T1        36.5°C
∆T1                                 ∆T1
= −0.00548 = −0.548%                = 0.00548 = 0.548%
T1                                  T1

The temperature we measure for final mixture (T2) is 12.0 °C. For measuring this
temperature, we can possibly read it to either about 11.8 °C or 12.2°C.
∆T2 T2 − max error − T2 − exp ermental ∆T2 T2 − max error − T2 − exp ermental
=                                     =
T2          T2 − exp ermental          T2          T2 − exp ermental
∆T2 11.8°C − 12.0°C                 ∆T2 12.2°C − 12.0°C
=                                   =
T2      12.0°C                      T2      12.0°C
∆T2                                 ∆T2
= −0.0167 = −1.67%                  = 0.0167 = 1.67%
T2                                  T2

C. Possible maximum error for latent heat of fusion of ice (Lf)
We already know the following:
Ti = −0.5°C ± 0.2°C (40%)
T1 = 36.5°C ± 0.2°C (0.548%)
T2 = 12.0°C ± 0.2°C (1.67%)
Then,
| (T2 − T1 ) |= 24.5°C ± 0.4°C (1.63%)
| (0.0°C − Ti ) |= 0.5°C ± 0.2°C (40%)
| (T2 − 0.0°C ) |= 12.0°C ± 0.2°C (1.67%)

Note: continued at next page
a) Possible maximum error for heats (Qwater, Qcup, Qice)
Qwater = cw mw | (T2 − T1 ) |
ln Qwater = ln cw + ln mw + ln | (T2 − T1 ) |
∆Qwater      ∆mw ∆ | (T2 − T1 ) |
= 0+     +
Qwater        mw   | (T2 − T1 ) |
∆Qwater
= 0 + 0.1970% + 1.63%
Qwater
∆Qwater
= 1.83%
Qwater

Qcup = cc mc | (T2 − T1 ) |
ln Qcup = ln cc + ln mc + ln | (T2 − T1 ) |
∆Qcup             ∆mc ∆ | (T2 − T1 ) |
= 0+         +
Qcup              mc   | (T2 − T1 ) |
∆Qcup
= 0 + 0.169% + 1.63%
Qcup
∆Qcup
= 1.80%
Qcup

Qice ( −0.5°C →0.0°C ) = ci mi | (0.0°C − Ti ) |
ln Qice ( −0.5°C →0.0°C ) = ln ci + ln mi + ln | (0.0°C − Ti ) |
∆Qice ( −0.5°C →0.0°C )          ∆mi ∆ | (0.0°C − Ti ) |
= 0+      +
Qice ( −0.5°C →0.0°C )          mi   | (0.0°C − Ti ) |
∆Qice ( −0.5°C →0.0°C )
= 0 + 0.313% + 40%
Qice ( −0.5°C →0.0°C )
∆Qice ( −0.5°C →0.0°C )
= 40.31%
Qice ( −0.5°C →0.0°C )

Note: continued at next page
Qice (0.0°C →12.0°C ) = cw mi | (T2 − 0.0°C ) |
ln Qice (0.0°C →12.0°C ) = ln cw + ln mi + ln | (T2 − 0.0°C ) |
∆Qice (0.0°C →12.0°C )                ∆mi ∆ | (T2 − 0.0°C ) |
= 0+         +
Qice (0.0°C →12.0°C )             mi   | (T2 − 0.0°C ) |
∆Qice (0.0°C →12.0°C )
= 0 + 0.313% + 1.67%
Qice (0.0°C →12.0°C )
∆Qice (0.0°C →12.0°C )
= 1.98%
Qice (0.0°C →12.0°C )

b) Possible maximum error for heat (Qfusion)
Now, we know the following:
Qwater = 10409.5355 J ± 190 J (1.83%)
Qcup = 1303.155 J ± 23.5 J (1.80%)
Qice ( −0.5°C →0.0°C ) = 33.3355 J ± 13.4 J (40.31%)
Qice (0.0°C →12.0°C ) = 1602.4008 J ± 31.7 J (1.98%)

Then,
Qwater + Qcup = Q fusion + Qice ( −0.5°C →0.0°C ) + Qice (0.0°C →12.0°C )
Q fusion = Qwater + Qcup − Qice ( −0.5°C →0.0°C ) − Qice (0.0°C →12.0°C )
Q fusion = 10076.9542 J ± 258.6 J (2.57%)

c) Possible maximum error for Lf
mi L f = Q fusion
ln mi + ln L f = ln Q fusion
∆mi ∆L f ∆Q fusion
+    =
mi   Lf   Q fusion
∆L f           ∆Q fusion        ∆mi
=                ±
Lf         Q fusion          mi
∆L f
= 2.57% ± (0.313%)
Lf
∆L f
(          ) max error = 2.88%
Lf

Note: continued at next page
Due to the limitations of our human eyes, we cannot read the scales that
accurately. Thus, 2.88% of possible maximum error would be built on our
experimental value for latent heat of fusion of ice. The possible Lf with maximum
errors are the following:

L f max error = L f + ∆L f                    L f max error = L f − ∆L f
L f max error = L f + L f (2.88%)             L f max error = L f − L f (2.88%)
L f max error = L f (102.88%)                 L f max error = L f (97.12%)
J                                             J
L f max error = (3.16 × 105      )(102.88%)   L f max error = (3.16 × 105      )(97.12%)
kg                                            kg
J                                             J
L f max error   = 3.25 × 105                  L f max error   = 3.07 × 105
kg                                            kg

Conclusion:
Latent heat of fusion of ice is involved during the phase transition of fusion or
melting of ice. This heat can be calculated out through Q = mLf. The standard
latent heat of fusion of ice is 3.33×105 J/kg. Assume ice at 0°C, our experimental
latent heat of fusion of ice is 3.17×105 J/kg with percentage difference of 4.80%.
By using our experiment temperature of ice at -0.5°C, our experimental latent heat
of fusion of ice is 3.16×105 J/kg with percentage difference of 5.11%. Through
this experiment, one trend we discovered is that percentage error would increase
if putting too much warm water in inner cup of calorimeter. Moreover, other
percentage errors may be resulted if ice cubes are not dried off. One important
reason is that the extra water melted from ice does not go through phase transition
of fusion. By drying off, we are able to calculate latent heat of fusion of ice more
accurately and keep the measurements of mass and temperature of ice more
accurately too. More ideally, if we are able to have an environment with very
small temperature gradient between ice cubes and surrounding, we are able to
minimize the amount of extra water melted from ice. If we are able to make our
experiment condition closer to the ideal condition, then the resulting experimental
Lf should be more accurate and closer to the standard Lf.

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