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1gram of ice is mixed with 1gram of steam. At themeal equllibrium ,the temperature of mixture is

1gram of ice is mixed with 1gram of steam. At themeal equllibrium ,the temperature of mixture is
 

Grade:12th pass

1 Answers

Arun
25758 Points
3 years ago

1 gm of ice requires 334 joules of heat to change it into water as we know specific heat of water is 4.2J/gk

1 gm of water is converted into steam by giving 2230J/g of heat energy specific heat of steam is 1.996kj/kg k.

For attaining equilibrium temperature= ice to water latent heat + water latent heat to change to 0degree to 100degree change in temperature /latent heat of vaporization

Mass of steam =334+ 420/2230=0.338g.

Q=m×c×t

80 degree Celsius final temperature

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