A cylindrical vessel is filled with 0.005 m^3 of pure steam at 227 degree C at a

Question

A cylindrical vessel is filled with 0.005 m^3 of pure steam at 227 degree C at a pressure of 200 kPa. (a) How many moles of water are in the container? (b) How many molecules of water are in the container? (c) How many grams of water are in the container? The steam is cooled to 100 degree C then condensed into liquid water, and the container collapses to a tiny size. What is the percent change of the volume? (e) The liquid is then cooled to the freezing point. Throughout this whole process (steam rightarrow hot water rightarrow cold water) it releases heat into a separate 2 kg block of ice held at 0 degree C. How many grams of the ice melt by the time the condensed steam and ice block reach equilibrium at 0 degree C?

Explanation / Answer

a) we use PV=nRT in this case to get number of moles where

P=200KPa =1.97atm, T=227+273K=500K V=.005m^3=5L, R=.08206L.atm/mol/K

using them we get n=PV/RT=.2405369892068mole

b) We use avogardo number in this case

we know that 1 mole=6.022*10^23 molecules

then for .2405369892068mole we have = .2405369892068*6.022*10^23= 1.448*10^23 water molecules

c) now we get the volume of water=5l =.005m^3 in the container

that means we use a conversation from m^3 to gm

1m^3 = 10^6gm

.005m^3= 5000gm

the container has 5000gm of water

d) using V1/T1=V2/T2 we get

5/500=V2/273

or V2= 2.73l

then% change in volume= 5-2.73/5*100=45.4%

e) heat released from stream to hot water =mst= 5000*1(227-100)=635000J

heat realesed from hot to cold water=mst=5000*1*(100-0) =500000J

Total heat realesed =5000*[127+100]=1135000J

1g ice melting required 333J heat

2000g ice melting required=333*2000=666000J

therefore 666000/1135000=.58 or 2000*.58= 1.16Kg ice will be melted.

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