igcse physics ppr 6 DOUBT

[moon]

please can u help in Jun.1999 ppr 6 Q.3 (a) & (b ii)

[holychalice]

can u put up d paper ?? or d question ??

[nid404]

please can u help in Jun.1999 ppr 6 Q.3 (a) & (b ii)

yes please...i don'y have da paper

[moon]

ok, I will try.

[moon]

here is the question

[holychalice]

ok
1. a - Avg = (26.8 + 26.8 + 27 + 26.9 + 27 + 27) / 6 = 26.9 ^OC
      - 0.1 ^OC
         The largest temperature difference is very low ... Less than even 5% or even 1% of d avg temperature .. so we can take it as constant
    b - 0 cm3   as the volume remained 14 aall d tym
         3.1 x 2 = 6.2 cm3                                 This is because d volumes given are h/cm    but we have a 2cm3 of tube so double.
         13.9 x 2 = 27.8 cm3 
     Avg u can do it = 27.8/5 = 5.6 cm3

I guess this is d way ...

[moon]

Avg = (26.8 + 26.8 + 27 + 26.9 + 27 + 27) / 6 = 26.9 OC

why did u divide it by 6???

 Avg u can do it = 27.8/5 = 5.6 cm3

and here why did u divide it by 5???

[holychalice]

c d table noe :
dere six values for room temperature :
for avg temperature .. u hav 6 observations ..
at 0hr , 1hr, 2hr, 3hr, 4hr and 5hr ...

nd divided by 5 ... cuz we r checking d rate of water lost in 5 hrs ... dats d experiment ... so its kept for 5 hrs ... ... so for 5 hrs .. 27.8 divided by 5 ...nd u hav 5 hrs of observation for d water loss so divided by 5

got it ?
read this again ... ive modified it

[moon]

Why don't we take the initial time

[Ghost Of Highbury]

Why don't we take the initial time

ok..its simple ratio

27.8cm³ of water lost in 5 hours

how much lost in 1 hour?

5 hours - 27.8

1 hours - 27.8/5 *1 = 5.56cm³/h

[moon]

I've got it. Thank u both 4 helping me.

[Ghost Of Highbury]

I've got it. Thank u both 4 helping me.

ur welcome..