Heat
Instructor:
Dr. Lab3
Latent Mohebi
Jiajie Huo(Jacky)
Member:
Ruihong Xiao
TEJ Shiwakoti
Anthony Hernandez
Dana Saiez
Rneiang Ye
WAJIH RAHMAN
2. introduction.
In this experiment, we are generally focusing on Measuring the
latent heat of fusion for water.
1. Q=mc^T, where Q is the heat, m is the mass, C is the specific
heat capacity, and ^T is the change in temperature.
2. In this experiment, the final equation that we would use is
miCi(0-Ti)+miCw(Tf-0)+miL=mwCw(Tw-Tf)+mcCc(Tw-Tf), while
mw refers to the Mass of the water, Cw refers to the specific
heat of the water, mi refers to the mass of ice, Tf refers to the
final temperature while the water reach the heat equilibrium,
Tw refers to the temperature of water in normal atmosphere
before the heating. Tf refers to the temperature to which the
water had been heated.
3. Procedure.(additional: all temperature are in degree.)
a. we found the mass of the calorimeter, so, mc=55.8g.
b. We heated the water in a beaker to some degree above room
temperature, , and the room temperature we measure is 22C.
c. We transferred some of the heated water into the calorimeter
and weigh the total mass of the container and the water, and the
result is Tw=33C, mc+w=205.1g, so the mass of the water that had
been transferred mw=149.3g.
e. We recorded the temperature of the ice, so Ti=-1C.
f. We added some ice to the calorimeter until the water reaches
an equilibrium temperature and do not move any more, so Tf=11C.
g. We weigh the calorimeter and mc+w+i=255.7g, so the mass of
ice that had been added is mi=50.6g
h. we calculated the latent heat of fusion of water and the
above repeat procedures for three more times.
3. Date sheets, spread sheets.
Measured Date
Mw(g
Tw(C
)
Mi(g) )
Ti(C)
Sample Trail
Calculation
Tf(C)
Lf
Water
1
149.3
50.6
33
-1
11 246
Water
2
137
41.9
37
-1
Water
3
181.9
50.8
38
-1
Water
4
133.8
43
35
-1
12 320
36
12
3
31
10
1
Lfo (Lf-Lfo)/Lfo
33
3
26%
33
3
4.00%
33
3
9%
33
3
6.70%
Sample calculation:
mw=149.3g, Cw=4186J/Kg*K, mc=55.8g,Cc=910J/Kg*K,
mi=50.6g,Ti=-1C, Tw=33C, Tf=11C,
and miCi(0-Ti)+miCw(Tf-0)+miLf=mwCw(Tw-Tf)+mcCc(Tw-Tf),
So, Lf= [mwCw(Tw-Tf)+mcCc(Tw-Tf)- miCi(0-Ti)-miCw(Tf-0)]/mi
=246J/Kg*K.
As the standard heat of fusion for water is 333,
So Error=(Lfo-Lf)/Lfo=(333-246)/333=26%.
6. Discussion and conclusion.
In this experiment, we repeat four times and the final result are
246, 320,363,311, the error would be 26%,4%,9% and 6.7%
respectively. During the whole experiment, there are many
behaviors that would make the errors larger. First, the container
can not completely rid the energy transfer from the outside, so
there would be some energy lost during the processes. Second, at
the first trail, we forget to place the container into the calorimeter
after we place the ice into the water, so the energy transfer between
the water and the outside atmosphere is much amplified. However,
after that, we corrected the mistake and the error is much smaller.
Third, after placing the ice into the water, we could not determine
the time when its precisely in energy equilibrium, so the error is
larger, above all, there are two trails that the final error are 4% and
6.7%, and it is within the acceptable range.