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 0.280m and the length of the copper section is 0.720m . Each segment has cross-sectional area 5.90

Answer 1

for Copper rate of heta transfer is

P_Cu=k_CuA(T_x-T_Cu)/L_Cu, where T_x is temperature at intersection

and for Brass

P_B=k_BA(T_B-T_x)/L_B

P_Cu=P_B (condition for steady state transfer)

ie.k_CuA(T_x-T_Cu)/L_Cu=k_BA(T_B-T_x)/L_B

solving th ebaove equation we get

T_x=(k_CuT_CuL_B+k_BT_BL_Cu)/(k_CuL_B+k_BL_Cu)

putting all values

T_x=(400x273x0.28+109x373x0.72)/(400x0.28+109x0.72)=314.20~314K=41.0^oC

P=A(T_B-T_Cu)/)(L_B/K_B+L_Cu/K_Cu)

P=5.9x10^-3(373-273)/(0.28/109+0.72/400)=135.32J/Sec

since P=Q/t

Q=Pt=135.32x5.x60=40596J (heat transfer in 5.3min)

also

Q=mL (where L for ice=334J/g)

m=Q/L=40596/334=121.55g~122g of ice is melted into water by conduction of composite system