First you will have to compute the final temperature and determine if all the ice melts. (It does, in this case)
Use the formula they gave you for the entropy lost by the cooling liquid water.
The entropy gained by the melting ice is [80 cal/g]*3000 g/273K
There will be additional entropy gain
ÄS = m c ln(Tf/Ti),
as the melted ice incgeases to the final temperature Tf.
Ti = 273 K m(ice) = 3000 g
Add up all the entropy gains and losses for the final answer.
After 6.00 kg of water at 77.3 oC is mixed in a perfect thermos with 3.00 kg of ice at 0.0 oC, the mixture is allowed to reach equilibrium. When heat is added to or removed from a solid or liquid of mass m and specific heat capacity c, the change in entropy can be shown to be ΔS = mc ln(Tf/Ti), where Ti and Tf are the initial and final Kelvin temperatures. Using this expression and the change in entropy for melting, find the change in entropy that occurs.
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