Two rods, one made of brass and the other made of copper, are joined end to end.

Question

Two rods, one made of brass and the other made of copper, are joined end to end. The length of the brass section is 0.300 m and the length of the copper section is 0.700 m . Each segment has cross-sectional area 0.00600 m2 . The free end of the brass segment is in boiling water and the free end of the copper segment is in an ice-water mixture, in both cases under normal atmospheric pressure. The sides of the rods are insulated so there is no heat loss to the surroundings.

1. What is the temperature of the point where the brass and copper segments are joined?

2. What mass of ice is melted in 6.00 min by the heat conducted by the composite rod?

Explanation / Answer

Given: length of brass rod Lb = 0.3 m; length of copper rod Lc = 0.7 m; cross sectional area of both A = 0.006 m2

Now the thermal conductivity of brass kb = 109 W/m-K; thermal conductivity of copper kc = 385 W/m-K

Heat conduction is just like electric current (rate of heat transfer) under the applied voltage (temperature difference), where the role of resistance is played by R = L/kA

Hence this is a series configuration so the net resistance is:

R = R1+ R2  = Lb/(kb*A) + Lc/(kc*A) = 0.3/(109*0.006) + 0.7/(385*0.006) = 0.76 K/W

Now the temperature difference is T = 100 K (boling point of water - melting point of water)

So heat flow rate dQ/dt = T/R = 100/0.76 = 131.58 W

2. Hence heat supplied in 6 min Q = 131.58*6*60 = 47368.8 J

So the amount of ice melted with this heat is given by: m = Q/L ;

where the latent heat capacity of water 'L' = 334000 J/Kg

Hence m = 47368.8/334000 = 0.142 Kg

1. Let the temperature of the joint be T we have:

dQ/dt = (T - 0)/R2 or T = (Lc/kc*A)*dQ/dt = 0.7*131.58/(385*0.006) = 39.870C

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