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MELTING RATE OF ICE
ATROOM TEMPERATURE
NUR SAFIKAH BT MASURINOOR SARAH AKMAL BT ABU BAKAR
NADHRAH BT MURAD
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1. To find the melting rate of ice at room
temperature
2. To study the physics concept of lever
system
OBJECTIVES
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Retort Stand
Paper Ring Clip
Force SensorWeight
Potentiometer
Comint 2013Beaker, Long And Small Rods, Long Holder, Ice,
Plastic Cup.
APPARATUS
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The harder the force,
the lower the resistance
that resulting in
increase of outputvoltage
can detect location of
exerted pressure by
applying a voltage
across the
potentiometer
THEORY
the longer the position
of contact, the higher
the voltage
Force sensor
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system lever is used
and attached to the
potentiometer
first system lever isapplied
the further the distance
between the fulcrumand the load, the more
effort needed to move
the object
SYSTEM LEVER
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melting point of pure
ice is 0 or 32
particle begin to meltas it gain energy
from the surrounding
the temperaturedoes not change
until the melting
process is complete
MELTING POINT OF ICE
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PROCEDURES
1. The apparatus were set up as figure below.2. The USB Port of COMINT 2013 was connected to the computer.
3. The long and small rod was entered into one-end of the hole at long holder.
4. The Java programme was started up.
5. Take starting reading of the empty plastic cup.
6. The Java programme was then ran.
7. Calibration process just started by putting some weight to comparemeasurement known value of mass and voltage.
8. Next step was the actual experiment of melting rate of ice at room temperature.
9. An amount of ice cube is put in a filter tunnel straight up above to the plastic
cup.
10. The melting of ice were let to flow through by the hole to the plastic cup.11. The reading of voltage is taken after each of 5 minutes.
12. Step 3 until 11 were then repeated with changing the hole of long holder to its
center and another one-end of it.
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DATA COLLECTIONSDATA 1A
DISTANCE BETWEEN LOAD AND FULCRUM ( 47.5 0.1 ) CM
Mass( x 1 ) g Voltage( y 0.001) V0 0.16020 0.19040 0.21860 0.23280 0.243
100 0.254120 0.266140 0.275
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DATA 2A
DISTANCE BETWEEN LOAD AND FULCRUM ( 43.5 0.1 ) CM
Mass( x 1 ) g Voltage( y 0.001) V0 0. 02120 0.15240 0.22160 0.23680 0.247
100 0.261120 0.273140 0.286
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DATA 3A
DISTANCE BETWEEN LOAD AND FULCRUM ( 39.5 0.1 ) CM
Mass( x 1 ) g Voltage( y 0.001) V0 0.028
20 0.09140 0.12860 0.16080 0.179
100 0.202120 0.215140 0.225
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DATA COLLECTION FOR MELTING ICE
DATA 1B
Time ( x 1 ) min Voltage( y 0.001) V0 -0.0045 0.045
10 0.05915 0.06820 0.08125
0.094
30 0.10535 0.11940 0.12945 0.14150 0.15355 0.16760 0.17565 0.18670 0.20175 0.20680 0.210
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DATA 2BTime ( x 1 ) min Voltage( y 0.001) V
0 0.0015 0.046
10 0.05515 0.06520 0.07625 0.09130 0.12135 0.13540 0.15745 0.17750 0.19255 0.21160 0.22165 0.23670 0.24375 0.25380
0.262
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DATA 3B
Time ( x 1 ) min Voltage( y 0.001) V0 0.0045 0.01010 0.01615 0.01820 0.03625 0.04430 0.06935 0.09140 0.10845 0.13050 0.15155 0.16760 0.18565 0.19770 0.20575 0.21680 0.230
DATA ANALYSIS ( AFTER CALIBRATION )
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DATA ANALYSIS ( AFTER CALIBRATION )DATA 1C
Mass( x 1 ) g Voltage( y 0.001) V0 -0.015
20 0.01540 0.04360 0.05780 0.068100 0.079120 0.091140 0.100
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DATA 2C
Mass( x 1 ) g Voltage( y 0.001) V0 -0.083
20 0.04840 0.11760 0.13280 0.143100
0.157
120 0.169140 0.182
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DATA 3C
Mass( x 1 ) g Voltage( y 0.001) V0 -0.032
20 0.03140 0.06860 0.10080 0.119100 0.142120 0.155140 0.165
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MELTING RATE OF ICE ( CALIBRATION )
DATA 1D
Time ( x 1 ) min Mass( y 0.01 ) g0 0.105 5.60
10 7.3015 8.5020 10.1025 11.7030 13.1035 14.8040 16.1045 17.6050 19.10
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DATA 2D
Time ( x 1 ) min Mass( y 0.01 ) g0 0.535 30.53
10 36.5315 43.2020 50.5325 60.5330 80.5335 89.8740 104.5345 117.8750 127.86
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DATA 3D
Time ( x 1 ) min
Mass( y 0.01 ) g
0 2.925 7.5910 12.1515 13.6920 27.5425 34.0030 53.0035 69.8540 82.9245 99.8550 116.00
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DISCUSSION
3 different parts which is depends on distance
fulcrum and load.
The distance between load and force at the first
part is ( 47.50.1)cm, second part is
( 43.50.1)cm and the last part is( 39.50.1)cm.
the change of voltage per mass for graph 1 isthe highest and graph 3 is the lowest
(follow the theory)
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GRAPH 1M = ( 0.008 7.663E-5) V/G (9.873% )
Y = (0.00040.006412) V (1539%)
GRAPH 2
M =( 0.0015 0.0003673) V/G (28.82% )
Y = (0.00020.03073) V (1.84E-4%)
GRAPH 3
M =( 0.0013 0.0001536) V/G (11.252% )Y = (0.00020.01285) V (7713%)
GRAPH MASS VERSUS TIME FOR EACH PARTS TO
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GRAPH MASS VERSUS TIME FOR EACH PARTS TO
FIND THE MELTING RATE.
THE GRADIENT BY USING MATHEMATICIAN
METHOD. FROM CALCULATION
>>WHEN T= (101)MIN
PART 1 : 0.5384 PART 2 : 2.2596 PART 3 : 1.263
>>WHEN T = (301)MIN
PART 1 : 0.6544 PART 2 : 2.4316 PART 3 : 2.695
>>WHEN T = (501)MIN
PART 1 : 0.7704 PART 2 : 2.6036 PART 3 : 4.127
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THEORY THE GRAPH FOR MELTING RATE
SHOULD BE CURVE AS SHOWN BELOW
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GRAPH PART 1
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GRAPH PART 2
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GRAPH PART 3
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PROBLEMS
the designing of the system lever might be
imperfect set up.
the sensitivity of the supplied force sensoris quite low (measure the mass only
between 100g-10kg).
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PRECAUTION
Use another force sensor that more
sensitivity
Designing the experiment moresystematic
Take more result to get the bestaverage
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CONCLUSION
The melting rate of ice at room temperature(27) are not constant. It is dependson time.
When t = (101)min
Part 1 : ( 0.5384 0.0001 ) g/min
Part 2 : ( 2.2596 0.0001 ) g/min
Part 3 : ( 1.2630 0.0001 ) g/min When t = (301)min
Part 1 : ( 0.6544 0.0001 ) g/min
Part 2 : ( 2.4316 0.0001 ) g/min
Part 3 : ( 2.6950 0.0001 ) g/min
When t = (501)minPart 1 : ( 0.7704 0.0001 ) g/min
Part 2 : ( 2.6036 0.0001 ) g/min
Part 3 : ( 4.1270 0.0001 ) g/min
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The further the distance between fulcrum and load, themore output voltage will the detected by force sensor.
Graph 1m =( 0.008 7.663e-5) V/g (9.873% )
y = (0.00040.006412) V (1539%)
Graph 2
m =( 0.0015 0.0003673) V/g (28.82% )
y = (0.00020.03073) V (1.84e-4%)
Graph 3
m =( 0.0013 0.0001536) V/g (11.252% )y = (0.00020.01285) V (7713%)
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