An 8.5 g ice cube is placed into 255g of water. calculate the temperature change of water when the ice melts

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.

Ithink the answer is a or d.

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