1 gram of ice is mixed with 1 gram of steam. At thermal equilibrium, the temperature of the mixture is

Options

A. `100^(@)C`
B. `55^(@)C`
C. `0^(@)C`
D. `50^(@)C`

Answer

A

Explanation

Here,
Mass of ice , `m_(ice)=1g`
Mass of steam , `m_(steam)=1g`
Latent heat of fusion of ice , `L_(ice)=80calg^(-1)`
Latent heat of steam , `L_(steam)=540calg^(-1)`
Specific heat of water , `s_(water)=1calg^(-1).^(@)C^(-1)` Heat required to convert ice at `0^(@)C` to water at `100^(@)C` is
`Q_(1)=m_(ice)L_(ice)+m_(ice)s_(water)DeltaT`
`=(1g)(80" cal " g^(-1))+(1g)(1" cal "g^(-1).^(@)C^(-1))(100^(@)C-0^(@)C)`
`=80cal+100cal=180`cal
Heat released by steam at `100^(@)C` to condense into water at `100^(@)C` is
`Q_(2)=m_(steam)L_(steam)`
`=(1g)(540 " cal "g^(-1))=540cal`
As `Q_(1)ltQ_(2)`, whole of the steam will not condense, so the temperature of the mixture is `100^(@)C`.