One end of a copper rod is immersed in boiling water at 100 degrees C and the ot

Question

One end of a copper rod is immersed in boiling water at 100 degrees C and the other end in an ice-water mixture at 0 degrees C. The sides of the rod are insulated. After steady-state conditions have been achieved in the rod, 0.165 kg of ice melts in a certain time interval.

A) For the time interval mentioned in the introduce, find the entropy change of the boiling water?

B) For the time interval mentioned in the introduce, find the entropy change of the ice-water mixture?

C) For the time interval mentioned in the introduce, find the entropy change of the copper rod?

D) For the time interval mentioned in the introduce, find the total entropy change of the entire system?

Explanation / Answer

heat gained by ice = heat lost by water = mL = 0.165 * 333000 = 54945 Joules

A) entropy change for water = heat lost / temp in kelvin =

= -54945 / 373 = -147.3 J/K

B) entropy change for ice = heat gained / temp =

= 54945 / 273 = 201.3 J/K

C) entropy change for copper = 0 (because copper doesnt change)

D) total entropy change = -147.3 + 201.3 + 0 = 54.0 J/K

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.