One end of a copper rod is immersed in boiling water at 100 degreeC and the othe

Question

One end of a copper rod is immersed in boiling water at 100 degreeC and the other end in an ice-water mixture at 0 degree C. Thesides of the rod are insulated. After steady-state conditions havebeen achieved in the rod, 0.150 kg of ice melts in a certain timeinterval. For the time interval mentioned in theintroduce, find the entropy change of the boiling water? For the time interval mentioned in theintroduce, find the entropy change of the ice-water mixture? For the time interval mentioned in theintroduce, find the entropy change of the copper rod? And
For the time interval mentioned in theintroduce, find the total entropy change of the entire system?
Please help me, thanks in advance.
For the time interval mentioned in theintroduce, find the entropy change of the boiling water? For the time interval mentioned in theintroduce, find the entropy change of the ice-water mixture? For the time interval mentioned in theintroduce, find the entropy change of the copper rod? And
For the time interval mentioned in theintroduce, find the total entropy change of the entire system?
Please help me, thanks in advance.
For the time interval mentioned in theintroduce, find the entropy change of the ice-water mixture? For the time interval mentioned in theintroduce, find the entropy change of the copper rod? And
For the time interval mentioned in theintroduce, find the total entropy change of the entire system?
Please help me, thanks in advance.
For the time interval mentioned in theintroduce, find the entropy change of the copper rod? And
For the time interval mentioned in theintroduce, find the total entropy change of the entire system?
Please help me, thanks in advance.
For the time interval mentioned in theintroduce, find the total entropy change of the entire system?
Please help me, thanks in advance.

Explanation / Answer

The heat exchanged is the amount needed to melt the ice, whichis .    Q = mL = 150 grams * 333J/g =   49950 Joules . This is the amount lost by the boiling water and gained by theice mixture, so... .     entropy change of boiling water =Q /T =    -49950 J / 373 K =   -133.914 J/K .    entropy change of ice-water = Q /T   = 49950 / 273   =    182.967J/K .    entropychange of copper is zero, because copper does not gain orlose heat (it only transfers it) . Total change in entropy of the system =  182.967 - 133.914   =    49.053J/K

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