water of mass 75g at 100 C is added to ice of mass 20g at -15 C. whatis the resulting temprature

Heat lost by the hot body = heat gain by the cold body

specific heat of water s = 1cal/g/^oc

let the resulting temperature is T_o

Heat lost by the water = m s ΔT

= 75 x 1 x (100 - T_o)

Heat gained by the ice

1.) from -15 c to 0 c = m x sx ΔT

= 20 x .5 x (0+15) = 150 cal.

2.) in converting into water at 0^o c
   = m x L = 20 x 80

1600 cal.

3.) in raising  the temp. of water formed from 0 to   T_o

m  s (T_o - 0 )

= 20 x 1 x T_o

= 20T_o

from first

75 (100 - T_o) = 150 + 1600 + 20 T_o

 95T_o = 5750

T_o =5750/95

=60.5^o c