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A glass of neglible heat capacity contains 100g of water at 20 degrees celsius
a) how many mlecules of water are inside the glass?
b) what mass of ice at -20 degrees celsius should be added to water to cool it down to 10 degrees celsius?

2007-12-25 06:37:48 · 7 answers · asked by Anonymous in Science & Mathematics Physics

7 answers

What have we got 2 Hydrogen and one Oxygen makes up a water molecule
2 Hydrogen - 2 protons and 2 electrons mass 2x 1.00794 AU
1 oxygen= 8 protons 8 neutrons and 8 electrons mass 15.9994AU

Water molecule weighs 15.9994 + 2.015884=18.015284AU
1 AU= 1.660538782 E−27 kg
a) number of molecules = total mass/ (mass of a single molecule)
number of molecules =.100/ (18.015284 x 1.660538782 E−27 )=3.34 E+24 water molecules
b) If mass of ice is m2 theb
Q1=m2 Lf
Q2= m2Cp(0-10)

Q3= m1 Cp( 20 -10) where

Lf =latent heat of fusion = 334 (J./(g K)) fro water
Cp= specific heat =4.18 (J./(g K))

Q1+Q2=Q3
m1( Lf+ Cp)= m1Cp(10)
m1= m1Cp(10)/( Lf+ Cp)

2007-12-25 08:40:48 · answer #1 · answered by Edward 7 · 0 0

H=2
O=16
H2O=4+16=20, 20 g=one mole
100/20=5moles, 5 x Avogadro's number=molecules in your glass @ STP

sorry don't remember the rest of it

2007-12-25 06:45:28 · answer #2 · answered by klby 6 · 1 1

Hey I'm far too tired to do that right now, but just to let you know Jimmy there who tried to do the question said H=2.
H=1 actually so the rel. molecular mass of water is 18, not 20.

2007-12-25 06:49:49 · answer #3 · answered by dk_falcon 2 · 0 1

37 and a block full of ice

2007-12-25 06:41:16 · answer #4 · answered by elizabeth; 2 · 0 1

a) No. of molecules in 100 g of water:
1 kmol of water = 1.0080(2) + 15.9994(1)
1 kmol of water = 17.0074 kg of water

No. of molecules in 100 g of water = Avogadro's Number time Number of kmol of water

No. of molecules in 100 g of water = 6.022 x 10^26 molecules/kmol x 0.100kg(1kmol water/17.0074 kg water)

No. of molecules in 100 g of water = 3.54 x 10^24 molecules of water ANS

b) heat change of water + heat change of ice = 0
(MwCwΔT)water + (MiCiΔT)ice = 0
4.1813J/gK = Cw, specific heat of water
2.114 J/gK = Ci, specific heat of ice

(100g)(4.1813 J/gK)(283K - 293K) + Mi(2.114 J/gK)(283K - 253K) = 0
-4181.3 + Mi(63.42) = 0
Mi = 4181.3/63.42
Mi = 65.9 g of ice ANS

Hope I help.

teddy boy

2007-12-25 16:59:37 · answer #5 · answered by teddy boy 6 · 0 0

37 mibacycles and a whole bunch of ice

2007-12-25 06:40:15 · answer #6 · answered by Anonymous · 0 1

not sure, but if you are done with the ice, can I stick the rest in my vodka?

2007-12-25 06:43:41 · answer #7 · answered by Mac 3 · 0 1

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