1 g of ice at `0^@C` is mixed with 1 g of steam at `100^@C`. After thermal equilibrium is achieved, the temperature of the mixture is

Options

A. `0^(@)C`
B. `100^(@)C`
C. `55^(@)C`
D. `80^(@)C`

Answer

B

Explanation

1 gm of ice `0^(@)C`
1 gm of steam at `100^(@)C`
problem on insufficient heat
`Q_(s)-mL =1xx540`
`Q_(1)=mL+ mcDelta T`
`=1xx80+1xx1xx100=180`

`Q_(s) gt Q_(1) implies` entire steam will not condense temperature `100^(@)C`