Calculate Mass of Ice Cube from Calorimetry Experiment

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To calculate the mass of the ice cube in the calorimetry experiment, the heat lost by the aluminum calorimeter and water must equal the heat gained by the melting ice. The equation used is m(al)C(al)(T(al) - Tf) + m(w)C(w)(Tw - Tf) = m(ice)C(ice)(Tf - Ti) + m(ice)Lfusion. The latent heat of fusion for the ice must be included on the right side of the equation, as the ice melts and transitions from solid to liquid. The user is unsure about the correct application of the latent heat and seeks clarification on their equation setup. Properly incorporating the latent heat of fusion is crucial for obtaining the correct mass of the ice cube.
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A cube of ice is taken from the freezer at -8.5C and placed in a 95g aluminium calorimeter filled with 310g of water st room temperature of 20C. The final situation is observed to be all water at 17C. What was the mass of the ice cube?

C (al)= 900, C (water)= 4186, C (ice)= 2100

m(al)= 0.095kg, m(water) = 0.310kg, m (ice)= ??
T (ice)= -8.5 C,T(final)= 17C, Twater= 20C, I assume the temperature of the aluminium container is the same of the water at 20C so T(al)= 20C,

heat lost= heat gained

I got --> m(al)C(al){ Tal- Tf} + m(w) C(w) { Tw- Tf) = m(ice) C(ice) { Tf- Ti}.
And then of course I would derive the formula to find the missing m(ice). However I am not getting the right answer. Can someone show tell me what I neeed to change in the formula?
 
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Did you take into account the latent heat of the solid ←→ liquid phase change?
 
Oh, I forgot that, umm.. so I would add m(w) Lfusion on the left side and m(ice) Lfusion on the right ??
 
Not both sides, just one. It melts, and stays melted.
 
oh right I was thinking I needed it for the water, but it doesn't change state, SO + m(ice) Lfusion on the right side? Is the rest of my equation right?
 
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