https://i.gyazo.com/539d05581fa48e2d96d2c1ec695d408a.png

VERY VERY CONFUSED ON THIS TWO MARKER.

ATTEMPTED TO USE E=PT BUT TURNS OUT YOU DON'T NEED TO DO THAT.

VERY VERY CONFUSED ON THIS TWO MARKER.

ATTEMPTED TO USE E=PT BUT TURNS OUT YOU DON'T NEED TO DO THAT.

Hi Behemouth,

Personally I did use E=Pt

P = 4500W

t = 60s

Therefor energy needed is 270,000J.

Since burning 1m3 of the gas gives 39x10^6J, the volume needed to gain 270,000J is:

V = 270,000/(39x10^6)

V = 6.92x10^-3 as stated in the question.

Sorry, the answer's a bit all over the place with typing.

Regards,

G

Personally I did use E=Pt

P = 4500W

t = 60s

Therefor energy needed is 270,000J.

Since burning 1m3 of the gas gives 39x10^6J, the volume needed to gain 270,000J is:

V = 270,000/(39x10^6)

V = 6.92x10^-3 as stated in the question.

Sorry, the answer's a bit all over the place with typing.

Regards,

G

Original post by GeronimoFrankie

Hi Behemouth,

Personally I did use E=Pt

P = 4500W

t = 60s

Therefor energy needed is 270,000J.

Since burning 1m3 of the gas gives 39x10^6J, the volume needed to gain 270,000J is:

V = 270,000/(39x10^6)

V = 6.92x10^-3 as stated in the question.

Sorry, the answer's a bit all over the place with typing.

Regards,

G

Personally I did use E=Pt

P = 4500W

t = 60s

Therefor energy needed is 270,000J.

Since burning 1m3 of the gas gives 39x10^6J, the volume needed to gain 270,000J is:

V = 270,000/(39x10^6)

V = 6.92x10^-3 as stated in the question.

Sorry, the answer's a bit all over the place with typing.

Regards,

G

Thanks so much

Original post by GeronimoFrankie

Hi Behemouth,

Personally I did use E=Pt

P = 4500W

t = 60s

Therefor energy needed is 270,000J.

Since burning 1m3 of the gas gives 39x10^6J, the volume needed to gain 270,000J is:

V = 270,000/(39x10^6)

V = 6.92x10^-3 as stated in the question.

Sorry, the answer's a bit all over the place with typing.

Regards,

G

Personally I did use E=Pt

P = 4500W

t = 60s

Therefor energy needed is 270,000J.

Since burning 1m3 of the gas gives 39x10^6J, the volume needed to gain 270,000J is:

V = 270,000/(39x10^6)

V = 6.92x10^-3 as stated in the question.

Sorry, the answer's a bit all over the place with typing.

Regards,

G

sorry again but little bit confused why t=60, I know that 1 min = 60 sec but I'm still not sure aaaaa. Sorry for this.

I recommend paying attention to units... and trying to get your head around algebra. It will really help when you can't think of any equations to use.

Regards,

G

Regards,

G

Original post by GeronimoFrankieHi Behemouth,

Personally I did use E=Pt

P = 4500W

t = 60s

Therefor energy needed is 270,000J.

Since burning 1m3 of the gas gives 39x10^6J, the volume needed to gain 270,000J is:

V = 270,000/(39x10^6)

V = 6.92x10^-3 as stated in the question.

Sorry, the answer's a bit all over the place with typing.

Regards,

G

Personally I did use E=Pt

P = 4500W

t = 60s

Therefor energy needed is 270,000J.

Since burning 1m3 of the gas gives 39x10^6J, the volume needed to gain 270,000J is:

V = 270,000/(39x10^6)

V = 6.92x10^-3 as stated in the question.

Sorry, the answer's a bit all over the place with typing.

Regards,

G

I got confused since I thought you had to use E=39X10^6 and P=4500 and USE E=PT for the time and got sooo lost

Original post by GeronimoFrankie

I recommend paying attention to units... and trying to get your head around algebra. It will really help when you can't think of any equations to use.

Regards,

G

Regards,

G

Yes, t=60s because it says to "show that the volume of gas that must be burned each MINUTE" in the question, so need to convert down to base SI units.

Regards,

G

Original post by Behemouth

I got confused since I thought you had to use E=39X10^6 and P=4500 and USE E=PT for the time and got sooo lost

The thing with this is the units... E=39x10^-3 has the units J/m-3 (Joule per metre cubed)... as it's the energy obtained from 1m3 of the gas... so you can't use it in the E=Pt equation. Always pay attention to units! It will make the difference between 1-2 marks on 4-5 questions and it could swing what grade you get.

Regards,

G

Original post by GeronimoFrankie

The thing with this is the units... E=39x10^-3 has the units J/m-3 (Joule per metre cubed)... as it's the energy obtained from 1m3 of the gas... so you can't use it in the E=Pt equation. Always pay attention to units! It will make the difference between 1-2 marks on 4-5 questions and it could swing what grade you get.

Regards,

G

Regards,

G

Got confused, didn't understand what the energy obtained from a cubic metre was meaning but it's just joules/m^3. thats why I got confused. Thanks

Original post by GeronimoFrankie

The thing with this is the units... E=39x10^-3 has the units J/m-3 (Joule per metre cubed)... as it's the energy obtained from 1m3 of the gas... so you can't use it in the E=Pt equation. Always pay attention to units! It will make the difference between 1-2 marks on 4-5 questions and it could swing what grade you get.

Regards,

G

Regards,

G

wait do you just convert 1 minute to 60 seconds since seconds is a SI unit and power is joules/seconds.. I understand thanks. question, why do you keep saying G..

Original post by GeronimoFrankie

The thing with this is the units... E=39x10^-3 has the units J/m-3 (Joule per metre cubed)... as it's the energy obtained from 1m3 of the gas... so you can't use it in the E=Pt equation. Always pay attention to units! It will make the difference between 1-2 marks on 4-5 questions and it could swing what grade you get.

Regards,

G

Regards,

G

Thanks for the help, I really understood it. Was struggling for quite a while.

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