I take 2.0 kg of ice and dump it into 1.0 kg of water and, when equilibrium is reached, I have 3.0 kg of ice at 0$\degree$C. The water was originally at 0$\degree$C. The specific heat of water = 1.00 kcal/kg·$\degree$C, the specific heat of ice = 0.50 kcal/kg·$\degree$C, and the latent heat of fusion of water is 80 kcal/kg. The original temperature of the ice was:
A) "one or two degrees below 0$\degree$C."
B) "-80$\degree$C."
C) "-120$\degree$C."
D) "The whole experiment is impossible."

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The Answer of I take 2.0 kg of ice and dump it into 1.0 kg of water and, when equilibrium is reached, I have 3.0 kg of ice at 0$\degree$C. The water was originally at 0$\degree$C. The specific heat of water = 1.00 kcal/kg·$\degree$C, the specific heat of ice = 0.50 kcal/kg·$\degree$C, and the latent heat of fusion of water is 80 kcal/kg. The original temperature of the ice was:
A) "one or two degrees below 0$\degree$C."
B) "-80$\degree$C."
C) "-120$\degree$C."
D) "The whole experiment is impossible."

I Take 20 Kg of Ice and Dump It into 1