`5g` ice at `0^(@)C` is mixed with `5g` of steam at `100^(@)C`. What is the final temperature?

Explanation

Given, mass of the steam, m=1g
Temperature `DeltaT = T_(2)-T_(1)` = 100-0=`100^(@)`C
Heat required to melt ice at
`0^(y@)C=Q_(1)=m_(i)c_(w)DeltaT = 5xx1xx100`=500cal
This heat rejected can melt the ice completely but cannot raise the temperature of water from 0 to `100^(@)`C as there is a deficiency of heat = `Q_(1) + Q_(2)-Q_(3)`=(900-540)cal = 360 cal
`therefore` Resulting temperature
`=100^(@)`C-(Heat deficient)/(Thermal capacity of system i.e., 6 g water)
`=100^(@)C-(360 cal)/(6gxx1calg^(-1^(@))C^(-1))=100^(@)C -(360^(@)C)/6
=100^(@)C-60^(@)C=40^(@)`C