Question #207517

1) A copper rod and an aluminium rod of the same length and cross-sectional area are attached end to end. The copper end is placed in a furnace maintained at a constant temperature of 272°C. The aluminium end is placed in an ice bath held at a constant temperature of 0.00°C.

Calculate the temperature at the point where the two rods are joined. 


 


1
Expert's answer
2021-06-16T14:26:50-0400

Answer:-

thermal conductivity of copper and aluminium are respectively 380 and 200 W/m/◦ C.

Temperature of hot end

-from the information

ΔQΔt=KAlTL=P\frac{\Delta Q}{\Delta t}=\frac{KA_lT}{L}=P

in steady state

Qculoss=QalGainedQ_{cu}loss=Q_{al}Gained

Pal=PcuP_{al}=-P_{cu}

KALAALΔTALLAL=KcuAcuΔTcuLcu\frac{KA_LA_{AL}\Delta T_{AL}}{L_{AL}}=\frac{-K_{cu}A_{cu}\Delta T_{cu}}{L_{cu}}

But

LAL=Lcu=LL_{AL}=L_{cu}=L

AAL=Acu=AA_{AL}=A_{cu}=A

KALAΔTALL=KcuAΔTCUL\frac{K_{AL}A\Delta T_{AL}}{L}=\frac{-K_{cu}A\Delta T_{CU}}{L}

KALΔTAL=KcuΔcuK_{AL}\Delta T_{AL}=-K{cu}\Delta _{cu}

KAL(TTC)=Kcu(TTH)K_{AL}(T-T_C)=-K_{cu}(T-T_H)

kALT=KCUT+Kcu(TH)k_{AL}T=-K_{CU}T+K_{cu}(T_H)

T(kAL+Kcu)=KcuTHT(k_{AL}+K_{cu})=K_{cu}T_H

T=KcuTH(KAL+Kcu)T=\frac{K_{cu}T_H}{(K_{AL}+K_{cu})}

T=380×200380+200T=\frac{380\times200}{380+200}

T=131.034oC\boxed{T=131.034 ^oC}


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