melting of ice at room temperature

7/30/2019 melting of ice at room temperature

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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


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


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


20 0.04840 0.11760 0.13280 0.143100

0.157

120 0.169140 0.182


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DATA 3C


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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melting of ice at room temperature