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

Answer

D

Explanation

d) According to principle of calorimetry, total heat given by a hotter body is equal to the total heat received by colder body i.e., heat lost by hotter body = heat gained by colder body
Heat required to melt 1g of ice at `0^(@)`C into 1g of water at `0^(@)`C is 80 cal.
Heat required to convert 1g of water at `0^(@)`C into 1g of water at `100^(@)`C = `1 xx 1xx100 = 100 cal`
Heat required to condense 1g of steam = `1xx540` cal = 540 cal
Clearly, whole of steam is not condensed, So, temperature of the mixture is `100^(@)`C.