Two metal rods, one silver and the other copper, are both attached to a steam ch
ID: 1568318 • Letter: T
Question
Two metal rods, one silver and the other copper, are both attached to a steam chamber as shown in the figure, with a temperature of 100 degree C, at one end, and an ice water bath, with a temperature of 0degree C, at the other. The rods are 1, 7 cm long and have a square cross-section, 4, 5 cm on a side. When steady state has been reached, how much heat (kJ) flows through the two rods in 3 min? The thermal conductivity of silver is 417 W/(m.K), and that of copper is 395 WI(m K). No heat is exchanged between the rods and the surroundings, except at their ends. Answer in two decimal places.Explanation / Answer
The length of silver and copper rod is ls = lb = 7.0 cm
ls = lb = 7.0 cm*(1m/100cm) = 0.07 m
Both the rods have square cross section with s = 5.0 cm = 0.05 m on side. Thus cross sectional area is
A = (0.05 m)2 = 25.0*10-4 m2
Thermal conductivity of silver is , ks = 417 W/m.K and for copper it is kc = 395 W/m.K
Hot end of the rods is at TH = 100oC = (100+273)K = 373 K
cold end of the rods is at TL = 0oC = (0+273)K = 273 K
Pconduction = ks*A(TH – TL)/Ls
Pconduction = (417 W/m.K)*( 25.0*10-4 m2)*(373 K – 273 K) / (0.07 m)
Pconduction = 1489.29 W
Pconduction = 1489.29 J/s
But Pconduction = Q/t
Thus in 3 minute = 180 s, the heat Qs flowed through the silver rod is given by,
Qs = Pconduction *t
Qs = (1489.29 J/s)*(180s)
Qs = 268.07*103 J
Qs = 268.07 kJ
Similarly for copper,
Rate of conduction of heat (heat current) for copper is given by,
Pconduction = kc*A(TH – TL)/Lc
Pconduction = (395 W/m.K)*( 25.0*10-4 m2)*(373 K – 273 K) / (0.07 m)
Pconduction = 1410.71 W
Pconduction = 1410.71 J/s
But Pconduction = Q/t
Thus in 3 minute = 180 s, the heat Qc flowed through the copper rod is given by,
Qc = Pconduction *t
Qs = (1410.71 J/s)*(180s)
Qs = 253.93*103 J
Qs = 253.93 kJ
Thus total heat that flows through the rods is given by,
Q = Qs + Qc
Q = 268.07 kJ + 253.93 kJ
Q = 522 kJ
Thus the answer is 522 kJ
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.