lg steam at `100^(@)` C is passed in to an insulating vessel having lg of ice at `0^(@)` C. Find the equlibrium temperature of the mixture, neglecting heat capacity of tlie vel'l·el.

Explanation

Heat available on steam ( changes from steam to water) = mL = 1 `xx` 540 = 540 cal
Heat gained by ice to change into water at `0^(@)`C and then rise its temperature to `100^(@)` C
`m_("ice") L_(ice) + M_("water") S_(w) Delta T `
`= 1 xx 80 + 1 xx 1 xx (100 - 0) = 180 ` cal.
The above calculations show that some part of steam will condense to change the ice into water at `100^(@0` C. Let .x. is the mass of steam condensed, then
x `xx ` 540 = 180 or
x = `(180)/(540) = (1)/(3)` g
Final contents of the mixture : ice = 0 g
water = 1 + `(1)/(3) = (4)/(3) ` g,
steam = 1 - `(1)/(3) = (2)/(3) ` g .
Final temperature of the mixture = `100^(@)` C