The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

AQA Physics PHYA5 - 28th June 2016 [Exam Discussion Thread]

Scroll to see replies

How did people find the ISAs on Capacitance and SHM?
Reply 61
Avatar for cowie
cowie
7
Original post by henry.philips
How did people find the ISAs on Capacitance and SHM?


Not fantastic, got 41 on SHM one, hbu?


Posted from TSR Mobile
Is unit 5 synoptic at all?
anyone doing the unit 1 exam on Tuesday?
Original post by Exams987
anyone doing the unit 1 exam on Tuesday?


i am!!!!
Original post by kother2015
i am!!!!


i have made a thread here

http://www.thestudentroom.co.uk/showthread.php?t=4087081&p=64958147#post64958147
Original post by Exams987
i have made a thread here

[url="http://www.thestudentroom.co.uk/showthread.php?t=4087081&p=64958147#post64958147
http://www.thestudentroom.co.uk/showthread.php?t=4087081&p=64958147#post64958147[/QUOTE]


thanx

r u re sitting unit 2 as well?
out of interest what ums did you get last year ? and what u predicted?
in
Reply 68
Avatar for Nerrad
Nerrad
7
Following. Doing Turning Points here, let's hope they don't accelerate a potato through a fluorescent tube this time.
Reply 69
Avatar for Nerrad
Nerrad
7
For those who are taking Turning Points, I'm making flash cards as I go along my revisions and past papers. Hope you'll find it useful.
http://www.cram.com/flashcards/phy5a-7323575
Can anyone help with this q?
Three ice cubes each of mass 25 g are dropped into a glass containing 330 g of water at a temperature of 22 °C. The ice is at 0 °C and melts so that the temperature of the water decreases.Calculate the final temperature of the water when all of the ice has melted. Assume that no heat is lost to the glass or the surroundings.specific latent heat of fusion of ice = 3.3 × 10^5 J kg^1 specific heat capacity of water = 4.2 × 10^3 J kg^1 K^1
Thanks:smile:
I keep answering questions with the wrong number of sig figs...

Is it true that you can only lose 1 mark in the paper to wrong number of significant figures?? or am i getting confused
Reply 72
Avatar for scrlk
scrlk
7
Original post by Lili1998
Can anyone help with this q?
Three ice cubes each of mass 25 g are dropped into a glass containing 330 g of water at a temperature of 22 °C. The ice is at 0 °C and melts so that the temperature of the water decreases.Calculate the final temperature of the water when all of the ice has melted. Assume that no heat is lost to the glass or the surroundings.specific latent heat of fusion of ice = 3.3 × 10^5 J kg^1 specific heat capacity of water = 4.2 × 10^3 J kg^1 K^1
Thanks:smile:


QIce = (0.025 × 3) × 330000 = 24750 J
QIce = 0.075 × 4200 × ΔT = 315ΔT
QWater = 0.330 × 4200 × ( 22 - ΔT) = 1386 × ( 22 - ΔT)

QIce = QWater therefore

24750 + 315ΔT = 1386 × ( 22 - ΔT)

Rearrange to find ΔT, giving a ΔT of 3.37 ºC

Hope this helps.
Original post by Lili1998
Can anyone help with this q?
Three ice cubes each of mass 25 g are dropped into a glass containing 330 g of water at a temperature of 22 °C. The ice is at 0 °C and melts so that the temperature of the water decreases.Calculate the final temperature of the water when all of the ice has melted. Assume that no heat is lost to the glass or the surroundings.specific latent heat of fusion of ice = 3.3 × 10^5 J kg^1 specific heat capacity of water = 4.2 × 10^3 J kg^1 K^1
Thanks:smile:


Hey I had a go but i'm still in the process of revising this so don't take my word for it hahah..

Tf (final temp)

heat energy supplied by water = energy to melt ice + heat energy gained by the melted ice

0.330 x 4200 x (22-Tf) = 0.025 x 3.3x10^5 + 0.025 x 4200 x Tf

which rearranges to..

Tf = 14.9ºC

not 100% sure but the answer seems fairly reasonable ahah
is there a mark scheme?
Original post by Calchioni
Hey I had a go but i'm still in the process of revising this so don't take my word for it hahah..

Tf (final temp)

heat energy supplied by water = energy to melt ice + heat energy gained by the melted ice

0.330 x 4200 x (22-Tf) = 0.025 x 3.3x10^5 + 0.025 x 4200 x Tf

which rearranges to..

Tf = 14.9ºC

not 100% sure but the answer seems fairly reasonable ahah
is there a mark scheme?



WELLLL i can already see where i've gone wrong, I didn't multiple by the number of ice cubes...
This is a perfect example of where i screw up my exams -.-
Original post by scrlk
QIce = (0.025 × 3) × 330000 = 24750 J
QIce = 0.075 × 4200 × ΔT = 315ΔT
QWater = 0.330 × 4200 × ( 22 - ΔT) = 1386 × ( 22 - ΔT)

QIce = QWater therefore

24750 + 315ΔT = 1386 × ( 22 - ΔT)

Rearrange to find ΔT, giving a ΔT of 3.37 ºC

Hope this helps.


Thanks... that makes sense :smile:
Reply 76
Avatar for Yo12345
Yo12345
2
How's everyone feeling about the exam?


Posted from TSR Mobile
What is the differrence between a cell and a high voltage unit?

Because they are both in a thermionic emission circuit (Turning points module), so what do they both do?

Any input appreciated :smile:
HELP:An isotope of technetium Tc which is in a metastable state, decays emitting only γ rays. When the isotope is placed 20 cm from a γ ray detector the count rate is 25 counts per second. The background count rate is 120 counts per minute. Calculate the count rate, in counts per second, when the detector is placed 30 cm from the isotope.
I keep getting the answer 9.1, but the mark scheme says the answer is 12.2, how do they do this?
Original post by Alby1234
HELP:An isotope of technetium Tc which is in a metastable state, decays emitting only γ rays. When the isotope is placed 20 cm from a γ ray detector the count rate is 25 counts per second. The background count rate is 120 counts per minute. Calculate the count rate, in counts per second, when the detector is placed 30 cm from the isotope.
I keep getting the answer 9.1, but the mark scheme says the answer is 12.2, how do they do this?


Which textbook is this?

I would have thought it was 25/1.5^2 = 11.1

So I don't get that answer either for the corrected count rate

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you