Two metal rods, one silver and the other copper, are both immersed at one end in
ID: 2000850 • Letter: T
Question
Two metal rods, one silver and the other copper, are both immersed at one end in a steam chamber 16) at a temperature of 100°C. The other end of each one is in an ice water bath at 0°C. The rods are 5.0 cm long and have a square cross-section that is 2.0 cm on a side. No heat is exchanged between the rods and the surroundings, except at the ends. How much total heat flows through the two rods each minute? The thermal conductivity of silver is 417 W/m · K, and that of copper is 395 W/m · K. Answer should be 39 kJ
Explanation / Answer
Solution: The length of silver and copper rod is ls = lb = 5.0 cm
ls = lb = 5.0cm*(1m/100cm) = 0.05 m
Both the rods have square cross section with s = 2.0 cm = 0.02 m on side. Thus cross sectional area is
A = (0.02 m)2 = 4.0*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
No heat is exchanged with the surrounding except at the ends of the rod and we assume that the steady state heat current flows through the rods.
Rate of conduction of heat (heat current) for silver is given by,
Pconduction = ks*A(TH – TL)/Ls
Pconduction = (417 W/m.K)*( 4.0*10-4 m2)*(373 K – 273 K) / (0.05 m)
Pconduction = 333.6 W
Pconduction = 333.6 J/s
But Pconduction = Q/t
Thus in 1 minute = 60 s, the heat Qs flowed through the silver rod is given by,
Qs = Pconduction *t
Qs = (333.6 J/s)*(60s)
Qs = 20.02*103 J
Qs = 20.02 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)*( 4.0*10-4 m2)*(373 K – 273 K) / (0.05 m)
Pconduction = 316.0 W
Pconduction = 316 J/s
But Pconduction = Q/t
Thus in 1 minute = 60 s, the heat Qc flowed through the copper rod is given by,
Qc = Pconduction *t
Qs = (316 J/s)*(60s)
Qs = 18.96*103 J
Qs = 18.96 kJ
Thus total heat that flows through the rods is given by,
Q = Qs + Qc
Q = 20.02 kJ + 18.96 kJ
Q = 38.98 kJ
Thus the answer is 38.98 kJ
(or 39 kJ if rounded)
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