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# Edexcel Phy 2 watch

1. look on page one mike A1, please tell me how this is rong, (if it is :P)
2. the very botton of page one has my workign for that ques btw
3. for the specific heat capacity I got 5.9 J per second someone tell me its right
4. i dont think that right, did u realise for the grad there was only a change of 40 kelvin from 60 to 20, and take this into account on your working
5. i think i got 100%, that was the easiest phy2 paper i have ever done!
phy1 was far harder.
6. page one very botton see my working,please somebody tell me if or where i have gone wrong!
7. for the rise in temperature, you don't convert it to kelvin, a rise is a rise

Q=m* c* rise in temp

Q=0.064*4200*37.5=10080J

rate=10080/(30*60)=5.6
8. taylor, think about it. since the first R was 10 M ohms, and u were given the V2, u should subtract this V2 from 9V(pd of supply), get d current, this current 'since series' will be equal. Then u get R from there, understand?
9. Easier to argue it in terms of ratios.

3 volts across resistor means it has 1/3 of the volts => it has 1/3 of the resistance. You can use the formula or just think for 10 secs and see that the other resistor is 10Mohms, so a 1/3 of the resistance would mean that this resistor must be 5 Mohms.

10 + 5 = 15, 5/15 = 1/3
10. yea the kelvin didnt matter but u had to change it to seconda and realise the change in temp was 40 kelvin or 40 celcius either didnt matter, i made my initial gradient 0.1 degreeC per second, this led to a power of 29.9 J per second
11. (Original post by grotto)
for the rise in temperature, you don't convert it to kelvin, a rise is a rise

Q=m* c* rise in temp

Q=0.064*4200*37.5=10080J

rate=10080/(30*60)=5.6
True, however i got the rate of rise to be 0.14Ks^-1, so did all the clever people in my college. You draw the tangent and you get like 60-20/[60x[5-0]] = 1.33degrees/s. Notice the x-axis was in minutes.]

this is the same as kelvins [write a lil note and change it to kelvin, if you don't do this, for the next question you will lose 1 method mark for putting degrees instead of kelvin, even though the final answer is the same.
12. (Original post by grotto)

Q=m* c* rise in temp

Q=0.064*4200*37.5=10080J

rate=10080/(30*60)=5.6
On that question did we have to do Q=mct for the whole of the graph! I thought it wanted it just for the begining so I did

0.064*4200*0.016666 = 4.48J per second
13. no manyea but if u work it out u get the current into the combo and the p.d across the combo use this to find the total R of the resistor AND the voltmeter, and u wil lsee the resistance of the combination is 5Mohm, but this for a resisor and a voltmeter in parallel, apply then the resistors in parallel rule and u can see the R of the resistor is infact 10Mohm this about it liek that
14. I did for whole graph
15. (Original post by grotto)
for the rise in temperature, you don't convert it to kelvin, a rise is a rise

Q=m* c* rise in temp

Q=0.064*4200*37.5=10080J

rate=10080/(30*60)=5.6
I'm in big trouble, I did something crazy. I did this....
rate of Q= mass*4200
does this make sense?
8.
a) T1 - initial temperature
T2 - final temperature
b) initial temp = 213 degree C
c) make T2 = 0 degree C
QUOTE]

should have been T1 - temp at hot source
and T2 - Temp at cold sink
17. i dont think so because in the next question which was clearly related they said "find the initial temperature" and it was clear to useto formula, i simply wrote what they defined it to be
18. tktaylor, u know I was just about to do that. But if u think about it. They kinda gave us the value of d voltage. Normally we neglect it bcos. it's a shunt, right?
19. (Original post by tktaylor6)
i dont think so because in the next question which was clearly related they said "find the initial temperature" and it was clear to useto formula, i simply wrote what they defined it to be
they said find the initial temperature that dosen't mean that T1 is the initial temperature.
"should have been T1 - temp at hot source
and T2 - Temp at cold sink"
Is what i wrote and what the textbook defines them as.
20. they gave us the voltage i assume to find the current flowing into our little combination, i get your ratios argument but i think u will find 5Mohm is the resistance of the entire combination, and then u had to use the 1/R thingy with the resistance of the 2nd voltmeter

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