I take 2.0 kg of ice and dump it into 2.3 kg of water and,when equilibrium is reached,I have 4.3 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 = 0.500 kcal/kg⋅°C,and the latent heat of fusion of water is 80.0 kcal/kg.The original temperature of the ice was
A)one or two degrees below 0°C.
B)-46.0°C.
C)-184°C.
D)184°C.
E)The whole experiment is impossible.

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The Answer of I take 2.0 kg of ice and dump it into 2.3 kg of water and,when equilibrium is reached,I have 4.3 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 = 0.500 kcal/kg⋅°C,and the latent heat of fusion of water is 80.0 kcal/kg.The original temperature of the ice was
A)one or two degrees below 0°C.
B)-46.0°C.
C)-184°C.
D)184°C.
E)The whole experiment is impossible.

I Take 20 Kg of Ice and Dump It into 2