AQA Physics PHYA5 - Thursday 18th June 2015 [Exam Discussion Thread]

Oh sorry my bad

Yeah u find the number if moles then divide by avagadro constant to get number of particles or in this case number of molecules

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oh ok thanks, why cant you use the normal formula with N?

Reply 1101

sykik

4

Hi ... I don't get why you do 600/0.35 instead of 600 * 0.35 ??

Original post by imyimy

oh ok thanks, why cant you use the normal formula with N?

I dont know tbh i dont want to give u wrong advice

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Reply 1103

a.a.k

2

Original post by imyimy

oh ok thanks, why cant you use the normal formula with N?

You got the right answer it is +21
So you CAN use pV=NkT
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(edited 8 years ago)

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Reply 1104

sykik

4

Hi I am really stuck on this question... Any help would be appreciated

Original post by sykik

Hi ... I don't get why you do 600/0.35 instead of 600 * 0.35 ??

It should be 600*0.35
Whats the answer btw.

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Reply 1106

Cosmocos

2

Original post by sykik

Hi ... I don't get why you do 600/0.35 instead of 600 * 0.35 ??

Hmm, I think it's because of this:

The Power output is 6000MW, but this is only 35% of the overall energy supplied by the fuel rods, so dividing by 0.35 gives you 100% of the energy supplied by the fuel rods,, I guess from then you can work out the decrease in mass (based on part a? idk..)

1.14x10^-2 kg

Original post by Cosmocos

Hmm, I think it's because of this:

The Power output is 6000MW, but this is only 35% of the overall energy supplied by the fuel rods, so dividing by 0.35 gives you 100% of the energy supplied by the fuel rods,, I guess from then you can work out the decrease in mass (based on part a? idk..)

doesn't 600/0.35 give u the input power?

Reply 1109

a.a.k

2

Original post by sykik

1.14x10^-2 kg

Close i got 1.41x10^-3 kg

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Reply 1110

a.a.k

2

Actually its not close enough

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Reply 1111

Cosmocos

2

Original post by sykik

doesn't 600/0.35 give u the input power?

Isn't the power input what is supplied by the fuel rods? lol I;m not sure tbh

Reply 1112

sykik

4

[QUOTE="a.a.k;56386187"]Close i got 1.41x10^-3 kg

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Here is MS Answer

This is what i got

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1433077192837.jpg34.4KB

Reply 1114

Cosmocos

2

Original post by sykik

Hi I am really stuck on this question... Any help would be appreciated

After probably doing it the longest way possible. I got a mass of 80623 kg...lol

Reply 1115

a.a.k

2

600MW is only elextrical energy which is only 35% other 75% energy is wasted whixh you need to take in account so you need to fins 100% energy by 600x100/35

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