Q. 1kg of ice at 0 degree is mixed with 1 kg of steam at 100 degree .What will be composition of the system when thermal equilibrium is reached?latent heat of fusion of ice =3.36*10^5 J kg^-1 and latent heat of vapourisation of water =2.26*10^6 J kg^-1?

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Q. 1kg of ice at 0 degree is mixed with 1 kg of steam at 100 degree .What will be composition of the system when thermal equilibrium is reached?latent heat of fusion of ice =3.36*10^5 J kg^-1 and latent heat of vapourisation of water =2.26*10^6 J kg^-1?

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Q. 1kg of ice at 0 degree is mixed with 1 kg of steam at 100 degree .W...

The final mixture shall be in liquid phase

Latent heat of fusion of ice(L)=3.36 x 100000 J/kg

we have 1 kg of ice, energy required to convert it into water =mL=3.36 x 100000 J(+)

now Latent heat of vapourization of water = 2.26 x 1000000J/kg

so energy released to get to liquid form =22.6 x 100000J(-ve)

now both are in water phase

The amount of energy required to increase the temperature of water from 0 to 100.

Q = mC ∆T

Specific heat = 4186 J/(kg * degreeC)

c = specific heat of water = 4186 J / kg C

Total energy required(for 1kg ice) = 3.36 * 10^5 + 418,600 = 7.546 * 10^5 J

however the energy is significantly less than what we are getting from steam which means that only fraction of steam is condensed

mL=7.546 * 10^5 J

m∗2.26∗10^6=7.546∗10^5

m=.334kg

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Q. 1kg of ice at 0 degree is mixed with 1 kg of steam at 100 degree .What will be composition of the system when thermal equilibrium is reached?latent heat of fusion of ice =3.36*10^5 J kg^-1 and latent heat of vapourisation of water =2.26*10^6 J kg^-1? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Q. 1kg of ice at 0 degree is mixed with 1 kg of steam at 100 degree .What will be composition of the system when thermal equilibrium is reached?latent heat of fusion of ice =3.36*10^5 J kg^-1 and latent heat of vapourisation of water =2.26*10^6 J kg^-1? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Q. 1kg of ice at 0 degree is mixed with 1 kg of steam at 100 degree .What will be composition of the system when thermal equilibrium is reached?latent heat of fusion of ice =3.36*10^5 J kg^-1 and latent heat of vapourisation of water =2.26*10^6 J kg^-1?.

Solutions for Q. 1kg of ice at 0 degree is mixed with 1 kg of steam at 100 degree .What will be composition of the system when thermal equilibrium is reached?latent heat of fusion of ice =3.36*10^5 J kg^-1 and latent heat of vapourisation of water =2.26*10^6 J kg^-1? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of Q. 1kg of ice at 0 degree is mixed with 1 kg of steam at 100 degree .What will be composition of the system when thermal equilibrium is reached?latent heat of fusion of ice =3.36*10^5 J kg^-1 and latent heat of vapourisation of water =2.26*10^6 J kg^-1? defined & explained in the simplest way possible. Besides giving the explanation of Q. 1kg of ice at 0 degree is mixed with 1 kg of steam at 100 degree .What will be composition of the system when thermal equilibrium is reached?latent heat of fusion of ice =3.36*10^5 J kg^-1 and latent heat of vapourisation of water =2.26*10^6 J kg^-1?, a detailed solution for Q. 1kg of ice at 0 degree is mixed with 1 kg of steam at 100 degree .What will be composition of the system when thermal equilibrium is reached?latent heat of fusion of ice =3.36*10^5 J kg^-1 and latent heat of vapourisation of water =2.26*10^6 J kg^-1? has been provided alongside types of Q. 1kg of ice at 0 degree is mixed with 1 kg of steam at 100 degree .What will be composition of the system when thermal equilibrium is reached?latent heat of fusion of ice =3.36*10^5 J kg^-1 and latent heat of vapourisation of water =2.26*10^6 J kg^-1? theory, EduRev gives you an ample number of questions to practice Q. 1kg of ice at 0 degree is mixed with 1 kg of steam at 100 degree .What will be composition of the system when thermal equilibrium is reached?latent heat of fusion of ice =3.36*10^5 J kg^-1 and latent heat of vapourisation of water =2.26*10^6 J kg^-1? tests, examples and also practice JEE tests.

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