A physics student immerses one end of a copper rod in boiling water at 100°C and
ID: 1590780 • 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 in J/K
(b) the entropy change of the ice-water mixture in J/K
(c) the entropy change of the copper rod in J/K
(d) the total entropy change of the entire system in J/K
Explanation / Answer
Heat required to melt 0.155kg of ice
Q = 0.155x 335x1000 J = 51925 J
a )
entrophy change of boiling water
S 1= -Q/T1
T1 = 373 k
S1 = -139.21 J/k
b)
S2 = Q/T2
T2 = 273 K
S2 = 190.2 J/k
c)
entrophy change of copper rod is zero as it is in steadt state
d)
total entrophy change
S = S1+S2+S3
S = 50.99 J/K
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