Entropy change for melting ice

by chriswilson
Tags: entropy, melting
 P: 3 1. The problem statement, all variables and given/known data 1. The problem statement, all variables and given/known data Calculate the entropy change when 1 mole of ice at 268 K is melted to form water at 323 K. The heat capacity of ice is 3.8 J K-1 kg-1 and that of water is 75 J K-1 kg-1. The enthalpy of fusion of ice at 273 K is 6.02 kJ mol-1. I know the entropy change by the melting of the ice is given by delta(S)=delta(Q)/T and that this is worked out by the enthalpy of fusion. My question is how do I calculate the entropy change caused by the change in temperature since it is not at a constant temperature does this mean the first equation cannot be used? Also this isn't a homework question it is an exam question from a previous year and my exam is tomorrow. Not sure whether this should be in here or in other sciences catagory
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P: 11,689
 Quote by chriswilson 1. The problem statement, all variables and given/known data 1. The problem statement, all variables and given/known data Calculate the entropy change when 1 mole of ice at 268 K is melted to form water at 323 K. The heat capacity of ice is 3.8 J K-1 kg-1 and that of water is 75 J K-1 kg-1. The enthalpy of fusion of ice at 273 K is 6.02 kJ mol-1.
I think your heat capacity units should be J K-1 mol-1, and the value for ice should be 38, not 3.8.
 I know the entropy change by the melting of the ice is given by delta(S)=delta(Q)/T and that this is worked out by the enthalpy of fusion. My question is how do I calculate the entropy change caused by the change in temperature since it is not at a constant temperature does this mean the first equation cannot be used?
For a given molar quantity M of substance with heat capacity constant C, the total heat held by the substance at absolute temperature T is

Q = M*C*T

Differentiating:

dQ = M*C*dT

So your equation for the change in entropy becomes an integral over the temperature change.

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