Third day..(sigh....long weekend no more....)

one thing wants to point out. the 4B SI will be the same SI for both our and Prof. Vahe's class.

doubt there will be many peope there...the SI named Mi Lan(米蘭)

i've take few classes with her before. she's an expert physicist...(even though it's not her major..)

anyways....since i dont really sure the time... so...if anybody know. please reply to this topic....

thx.
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1st quiz! (review)

1. calculate changing T.

1st need to know the equations.


Area(cylinder) : pi * r^2 * h (pi times r square times height)


Q= m * c * delta T


delta V = Vo * beta * delta T


Density = mass / volume (unit: g/cm^3......not really cm^3, but since Cm^3 = mL...in the other way it is mass / cm^3..but need to convert)


Watt = Joule / second (pretty much the same thing....)

now we have all the equation.....then let's do the 1st question.


1.calculate the final temperture....after 1 minute
(given: r=0.5m, h=1m, Vwater = 0.75 m^3 , Watt = 200)


Q= m * c * delta T

200 (J/second) * 60 (s) = 0.75 (m^3) *1000 (kg/m^3) * 4200 (J/kgC) * delta T (C or K)


recall: the reason why it's 1000 (kg/m^3)....since it's original 1 (g/cm^3) <~ density of water

then, 1 (g/cm^3) * [100*100*100](cm^3/m^3) * [1/1000](kg/g)

remember to write out all unit....and match the unit for both side...it's a most powerful way to identify the equation is right or wrong.

then we can get the delta T = (200*60) / [0.75 * 1000 * 4200] = 0.0038K (or C)

for the 2nd part, just plug the delta T in to the equation delta V = Vo * beta * delta T


3rd, what temperture will the water overflow.

i.e water(total valume) = tank(total volume) (well...actually this is the maximum limit)

so here we go, 0.75 + (0.75*beta*[Tf - To] ) = pi*r^2*h + (pi*r^2*h*beta*[Tf - To])

plug in number.....and solve for Tf.

4th, i dont quite remember the question....calculate the energy i think?

so by part 3 we can get the Tf. then we will be able to find the delta T....

then use that delta T.....plug back to Q = m*c*delta T.....

but remember the Q here is in SECOND....so just conver that to whatever the time unit the question ask.

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lecture time.

i am not a big fan of typing..... so i will just go brifly for the important stuff.

oh....btw here's the link of the notes....

1st note
http://docs.google.com/Doc?id=d3w6sww_297pvmhcn7q

2nd note
http://docs.google.com/View?docid=d3w6sww_350d2fch2dp





























Temperature remains CONSTANT until change of phase is complete .

remember this....very important....


Lf = Heat of Fusion

Lf => 80(cal/gram) = 3.35 x 10^5 (J/kg)

Lv= Heat of Evaoiration

Lv=> 540(cal/gram) = 2.26 x 10^6 (J/kg)


and remember!!!!!! Qgain = Qlost


pretty much every Assignment#3 problem is talking about this!!!!


EX. finding Heat(energy) for 30kg ice(initialy -20C) to melt.
specific heat of ice to be 2100 J/(kg*K).

so. at first we need to know that when ice is melting....the ice cube(s) and the ice melted (water) will remind at 0C until the ice melt completely. and the energy that requires to melt all of the ice is the Latean Heat (Lf = Heat of Fusion Lf => 80(cal/gram) = 3.35 x 10^5 (J/kg) )

so... this question will simply become......

Q = Mice * Cice * (Changing T) + Mice * Lf

= 30kg * 2100 J/(kg*K) * [0 - (-20)]K or C + 30kg * 3.35 * 10^5 J/(kg*K)

notice that we can factor out Mice.....

=30kg *[ 2100 J/(kg*K) * [0 - (-20)]K or C + 3.35 * 10^5 J/(kg*K)]

see that the units are perftctly cancel out....and only J left...and that's the energy unit we want!

=11310000 = 1.131 x 10^7 J


and that's pretty much it.....about the Lf , and the Le is the same...except it's for BOILING.

oh....btw....this same equation will only help calculate the Energy require to MELT!!!!!

i.e....does not include the energy after 0C....what do I mean? well...let's see.. same question and
let's change to...let's say 10C.

then the equation will be.....

Q = Mice * Cice * (Changing T (0 - initial )) + Mice * Lf + Mice * Cwater * (Changing T (final - 0))

= 30kg * 2100 J/(kg*K) * [0 - (-20)]K or C + 30kg * 3.35 * 10^5 J/(kg*K) + 30kg * 4190J/(kg*K) * (10 - 0) K or C


and that's it.....good luck and have fun with the Assignment #3.......

oh....and see u guys at physic's study room on monday


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