IIT JEE , AIEEE Forum - Physics , Thermal Physics

urmilamip

1 kg of ice at 0 C is mixed with 1 kg of steam at 100 C.What will be the composition of the system when thermal equilibrium is reached?Latent heat of fusion of ice =3.36 x 10^5 J/Kg and latent of vapourization of water = 2.26 x 10^6.

Will surely give 5 salutes to well explained answers......

pannaguma

final temp of mix will be same.
let it be T.
now heat energy gained by ice = heat energy lost by steam

But
3.36*10^5*1 + 4200*100 < 2.26*10^6*1
where 4200 is specific heat of water per degree per Kg
and T increase is 100.

thus not all steam is coverted to water at 100'C.
3.36*10^5*1 + 4200*100 = 2.26*10^6*XKg
equating X=0.334Kg
thus 1.34 Kg is water at 100'C
while rest is steam.

ramyani

heat reqd to melt all ice= 3.36 x 10^5 J

heat reqd to condense all steam = 2.26 x 10^6 J

thus, all ice will be melted and the water that will be formed will reach the

temperature of 100 degree, but only that amount of steam will be condensed so

that the amount of heat reqd for the above mentioned process is supplied. let m

be the mass of steam.

3.36*10^5*1 + 4200*100 = 2.26*10^6*m

m= .335 kg, which turns to water.

so composition oe the system is 1.335 kg of water and 665g steam

chetan_kp

ya both the answers are perfect

krishna.gopal

Perfects amswers. It will be 1.335 kg water and remaining stean at 100 degree celcius

rohit.mehra

First of all ice at 0 C will be converted into water at 0 C.for this heat required is:

    Q=mL=1*3.36*10^5=3.36*10^5

the 1kg of water at 0 C will be converted into water at 100 C.For this heat required is

   Q=1*4200*100=420000

so total heat required is
   Q=336,000+420,000=756,000

Now this heat should be supplied by steam.When steam gets converted into water it releases heat.The amount of heat released when 1 kg of steam is converted is given as

   Q=1*2.26*10^6=2,260,000

Now this heat is greater than the total heat required.So the whole of steam is not to be converted.Let m be the mass of steam gets converted
so      m*2.26*10^6=756000
                          m=.335g
so this mass of steam finally gets converted into water.The remaining steam i.e 1-.335=.665 kg is still present in the system.

so the total water at the end of the reaction is 1kg(from ice)+.335 kg(from steam)=1.335 kg
also there is .665 kg of steam

nitish0991

ur explanation/answer is the best rohit..

Ask & Discuss Questions with Community & Experts