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For a 8.88 gram sample of ice at -43.6°C , what is the minimum amount of heat (in kJ) needed to convert the sample to liquid water at 0.00°C?

Physical data for water:
specific heat (ice) = 2.10 J/g°C
specific heat (water) = 4.184 J/g°C
heat of fusion = 0.335 kJ/g
heat of vaporization = 2.258 kJ/g

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2 answers
  1. q1 = heat to warm ice from -43.6 C to zero C.
    q1 = mass ice x specific heat ice x (Tfinal - Tinital) = ?

    q2 = heat to melt the ice
    q2 = mass ice x heat fusion ice = ?

    q total = q1 + q2 = ?. Note that q1 answer is in J and q2 is in kJ. Since the problem asks for the answer in kJ, you should calculatae q1 and convert to kJ before adding to q2. Post your work if you get stuck.

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    DrBob222
  2. 1) If the specific heat capacity of ice is 0.5 cal/gC°, how much heat would have to be added to 200 g of ice, initially at a temperature of -10°C, to raise the ice to the melting point?

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