A physics student immerses one end of a copper rod in boiling water at 100°C and
ID: 1590691 • 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, 0.155 kg of ice melts in a certain time interval. For this time interval find the following.
(a) the entropy change of the boiling water
_______ J/K
(b) the entropy change of the ice-water mixture
________ J/K
(c) the entropy change of the copper rod
_____ J/K
(d) the total entropy change of the entire system
______ J/K
Explanation / Answer
(a)
DS = Q/T
= - mLf/T
= - (0.155 kg)(334 x 10^3 J/kg)/(373.15 K)
= -138.74 J/K.
(b)
DS = Q/T
= mLf/T
= (0.155 kg)(334 x 10^3 J/kg)/(273.15 K)
= 189.53 J/K.
(c)
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)
189.53 J/K - 138.74 J/K= 50.79 J/K.
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