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°C.The water was originally at 0°C.The specific heat of water = 1.00 kcal/kg·°C,the specific heat of ice = 0.50 kcal/kg·°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°C.
B) −80°C.
C) −120°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°C.The water was originally at 0°C.The specific heat of water = 1.00 kcal/kg·°C,the specific heat of ice = 0.50 kcal/kg·°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°C.
B) −80°C.
C) −120°C.
D) The whole experiment is impossible.

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