Physics w08 BIG PROBLEM!

[arabiannight975]

Ok so in the physics w08 qp ( http://www.xtremepapers.net/CIE/Cambrid ... _qp_03.pdf )

Qestion 5 c Using a 40 W heater, 16.3 g of ice is melted in 2.0 minutes. The heater is then switched
off. In a further 2.0 minutes, 2.1 g of ice is melted.
Calculate the value of the specific latent heat of fusion of ice from these results.

This means that the heater has melted 16.3g of Ice and the other 2.1 is irrelevant as it says that its been melted after the heater has been switiched off

BUT IN THE MARKING SCHEME ( http://www.xtremepapers.net/CIE/Cambrid ... 8_ms_3.pdf )

(mass of ice melted by heater = 16.3 – 2.1) = 14.2 g C1
ml in any form, words, symbols or numbers C1
Wt or Pt in any form, words, symbols or numbers accept VIt C1
338 J/g OR 338 000 J/kg c.a.o A1

I DONT understand how the heater melted 14.2g?? its in the question that it melted 16.3 and the other was meleted after it was switiched off.




Now the other question same paper Q6 b) Check it from the paper

n = sinr/sini OR n = sini/sinr in any form C1
sinr/sin30 = 1.49 OR sinr = 1.49 × sin30 C1
48.0° – 48.2° A1

this is the answer. I dont understand that the formula for refractive index is SinI/Sinr
its stated in the answer above too. THEN HOW CAN IT GET Sinr=1.49 x Sin30
It should be sinr=Sin30/1.49


PLZZZ HELp

 

[PlanetMaster]

Ok so in the physics w08 qp ( http://www.xtremepapers.net/CIE/Cambrid ... _qp_03.pdf )

Qestion 5 c Using a 40 W heater, 16.3 g of ice is melted in 2.0 minutes. The heater is then switched
off. In a further 2.0 minutes, 2.1 g of ice is melted.
Calculate the value of the specific latent heat of fusion of ice from these results.

This means that the heater has melted 16.3g of Ice and the other 2.1 is irrelevant as it says that its been melted after the heater has been switiched off

BUT IN THE MARKING SCHEME ( http://www.xtremepapers.net/CIE/Cambrid ... 8_ms_3.pdf )

(mass of ice melted by heater = 16.3 – 2.1) = 14.2 g C1
ml in any form, words, symbols or numbers C1
Wt or Pt in any form, words, symbols or numbers accept VIt C1
338 J/g OR 338 000 J/kg c.a.o A1

I DONT understand how the heater melted 14.2g?? its in the question that it melted 16.3 and the other was meleted after it was switiched off.

After the heater was turned off, that 2.1g of ice did melt because of heater.
If there wasn't any heater, would this 2.1g had melted?
Still didn't got? OK
Imagine yourself driving a car. You accelerated for 5 secs and turned of the engines.
You consumed (lets say) 0.1 liter of fuel. During your acceleration your car covered (lets say) 30m.
After you turned your engines off, you covered 10m more. Where did that 10m come from?
Obviously your 0.1 liter fuel although it wasn't being consumed by car during that 10m.

Now the other question same paper Q6 b) Check it from the paper

n = sinr/sini OR n = sini/sinr in any form C1
sinr/sin30 = 1.49 OR sinr = 1.49 × sin30 C1
48.0° – 48.2° A1

this is the answer. I dont understand that the formula for refractive index is SinI/Sinr
its stated in the answer above too. THEN HOW CAN IT GET Sinr=1.49 x Sin30
It should be sinr=Sin30/1.49

From what CIE's marking schemes is saying is that:
n = sinr/sini OR n = sini/sinr
i.e n = 1/n (Amazing!!! :mrgreen: )
I'm pretty sure its something else.
Always remember that light moves away from normal when moving from denser to rarer medium and vice versa.

 

[arabiannight975]

If there wasn't any heater, would this 2.1g had melted?
Still didn't got? OK
Imagine yourself driving a car. You accelerated for 5 secs and turned of the engines.
You consumed (lets say) 0.1 liter of fuel. During your acceleration your car covered (lets say) 30m.
After you turned your engines off, you covered 10m more. Where did that 10m come from?
Obviously your 0.1 liter fuel although it wasn't being consumed by car during that 10m.

From what CIE's marking schemes is saying is that:
n = sinr/sini OR n = sini/sinr
i.e n = 1/n (Amazing!!! :mrgreen: )
I'm pretty sure its something else.
Always remember that light moves away from normal when moving from denser to rarer medium and vice versa.

So this means that in the first question the if the other 2.1 g was melted becuz of the hearter too then the total ice melted by the hearted was 16.3+2.1=18.4
but it says its 16.3-2.1=14.1

I dont UNDERSTAND THIS AT ALL..


and about the second question...
so you mean the marking scheme is wrong?

 

[PlanetMaster]

So this means that in the first question the if the other 2.1 g was melted becuz of the hearter too then the total ice melted by the hearted was 16.3+2.1=18.4
but it says its 16.3-2.1=14.1

I dont UNDERSTAND THIS AT ALL..

Nope!
The latent heat of fusion is the amount of energy required to convert a given mass of substance from solid to liquid (or vice-versa) without a change in the temperature of the surroundings – all absorbed energy goes into the phase change.
After the heater was switched off, 2.1g melted because of temperature going back to room temperature.
Clear?

and about the second question...
so you mean the marking scheme is wrong?

I never said that its wrong.
I meant we aren't getting what the marking scheme is trying to say.

 

[shoaib672002]

 

[shoaib672002]

 

[shoaib672002]

 

[Xylferion]

Ok so in the physics w08 qp ( http://www.xtremepapers.net/CIE/Cambrid ... _qp_03.pdf )

Qestion 5 c Using a 40 W heater, 16.3 g of ice is melted in 2.0 minutes. The heater is then switched
off. In a further 2.0 minutes, 2.1 g of ice is melted.
Calculate the value of the specific latent heat of fusion of ice from these results.

This means that the heater has melted 16.3g of Ice and the other 2.1 is irrelevant as it says that its been melted after the heater has been switiched off

BUT IN THE MARKING SCHEME ( http://www.xtremepapers.net/CIE/Cambrid ... 8_ms_3.pdf )

(mass of ice melted by heater = 16.3 – 2.1) = 14.2 g C1
ml in any form, words, symbols or numbers C1
Wt or Pt in any form, words, symbols or numbers accept VIt C1
338 J/g OR 338 000 J/kg c.a.o A1

I DONT understand how the heater melted 14.2g?? its in the question that it melted 16.3 and the other was meleted after it was switiched off.




Now the other question same paper Q6 b) Check it from the paper

n = sinr/sini OR n = sini/sinr in any form C1
sinr/sin30 = 1.49 OR sinr = 1.49 × sin30 C1
48.0° – 48.2° A1

this is the answer. I dont understand that the formula for refractive index is SinI/Sinr
its stated in the answer above too. THEN HOW CAN IT GET Sinr=1.49 x Sin30
It should be sinr=Sin30/1.49


PLZZZ HELp

The ray in 6 b is traveling from a denser medium to a rarer medium. When traveling from denser to rarer we use.

n = Sir "r" / Sin " i "

If we were going from rarer to denser it would be Sin "i" / Sin "r".