When 20 grams of Ice at 0°C is added to 30 grams of water at 0 °C,

20 gram of water gains the heat of 20(t-0)s which is equal to the heat lost by the 30 gms of water

i,e, 30 s (30-t).

Here,

 t is the resultant temperature

s is the specific heat of the water.

So 20s t = 30s (30 - t)

 or 20s t = 30 x 30 - 30s t

or, ( 30 + 20 )t = 30 x 30

or,  t =30 x 30/50 = 18° C

The temperature of mixture is 18° C.