A physics student immerses one end of a copper rod in boiling water at 100 C and
ID: 2020491 • Letter: A
Question
A physics student immerses one end of a copper rod in boiling water at 100 C and the other end in an ice-water mixture at 0 C . The sides of the rod are insulated. After steady-state conditions have been achieved in the rod, an amount of ice with a mass of 0.260 kg melts in a certain time interval. For this time interval find
The entropy change of the boiling water
S=
the entropy change of the ice water mixture,
S=
The entropy change of the copper rod.
S=
And the total entropy change of the entire system.
Ssystem =
Explanation / Answer
Mass of ice melt = m = 0.260kg boiling temparature of the water =T' = 100oC or T ' = 273.15+ 100 = 373.15 K Ice -water mixture temparature , T = 0+ 273.15 = 273.15 K (a)The entropy change of the boiling waterS=Q/T' = - mLf /T'= - (0.260kg)(334x103 J/kg)/(373.15K) = - 232.72 J /K (b)the entropy change of the ice water mixture,
S=Q/T = mLf /T = (0.260kg)(334x103 J/kg)/(273.15K) = + 317.92 J/K (c)The entropy change of the copper rod.
S=0 Explanation: from the time equilibrium has been reached, there is no net heat exchange between the rod and its surroundings, so the entropy change of the copper rod is zero. (d)And the total entropy change of the entire system. Ssystem = - 232.72 J /K + + 317.92 J/K = 85.2 J /k
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