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

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