In this lab , I chose a variable and investigated the effect of alter it on the specific latent genus Oestrus of fusion of field glass .
Materials:
* Electric Immersion alterer
* Funnel
* Glass beaker
* An ammeter
* A voltmeter
* Power turn in
* Stopwatch
* parry stand & clamp
* Ice
* Mass balance
Hypothesis :
Simply, If m repositions it wont change the latent heat of fusion in glassful.
Used Equations:
zero in Joules = Voltage x Current x cartridge clip (s)
Latent heat = Energy (joules) / mass (kg)
Procedure :
1- check up the apparatus as shown in the diagram but without the ice .
2-weigh you beaker.
3-Crush the ice and put it in the funnel trying to relative majority it closely round the heater .
4-Turn on the power supply and start the stop watch at the same time.
5-Record the potential V and the current I .
6-Leave the heater on for 3 minutes.
7-Switch off , remove the beaker and weigh it .
8-Calculate the specific latent heat of fusion of ice .
9-Change a factor and repeat the examine.
10-Explain the descent between your chosen variable and the specific latent heat of fusion of ice and comment on your results .
Note: The experiment was d single at room temperature and not 0 degrees so of course , the result will be different than the one known, which is 300000 j/kg , but as the main place of this experiment is to see if latent heat changes with the change of time , so it wont affect the experiment .
Results :
crop A: Time = 3 minutes = 180 seconds
Voltage(v)| Current(A)| freight of empty beaker (kg)| heaviness of beaker after melting of ice (kg)| weighting difference| Heat Energy (J)| Latent Heat (J/kg)|
12V| 2.17A| 0.398 kg| 0.404 kg| 0.006 kg| 4687.2 J| 781200 j/kg|
Set B: Time = 2minutes = 120 seconds
Voltage(v)| Current(A)| Weight of empty beaker(kg)| Weight of beaker after melting of ice(kg)| Weight difference| Heat Energy...If you indigence to get a full essay, order it on our website: Orderessay
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