To answer this problem, we assume that upon the addition, the ice just melts without the temperature change and just uses the latent heat. The water only uses the sensible heat. Hence the equation is 8.5 g * (80cal/g)*4.1858 (J/g)= 255 g 8(4.18 J/gK)*(ΔT). 80cal/g is the latent heat of vaporization of ice. ΔT is equal to 2.67 Kelvin.
