Question

Two metal cubes with 3cm edge of copper and aluminum are arranged as shown in figure (assume no loss of heat from open surfaces) $\left(K_{CU}=385W/m-K,K_{AL}=209W/m-K\right)$The total thermal current from one reservoir to the other

• A. $1.42\times 10^{3}W$
• B. $2.53\times 10^{3}W$
• C. $1.53\times 10^{4}W$
• D. $2.53\times 10^{4}W$

Answer 1

The correct option is A $1.42\times 10^{3}W$

Given,

$K_{Cu}=385W(mK{)}^{-1}$

$K_{Al}=209W(mK{)}^{-1}$

$\frac{d{Q}_1}{dt}=\frac{K_{Cu}A\left(80\right)}{l}=\frac{385\times 9\times {10}^{-4}\times 80}{3\times {10}^{-2}}$

$\frac{d{Q}_2}{dt}=\frac{K_{Al}A\left(80\right)}{l}=\frac{209\times 9\times {10}^{-4}\times 80}{3\times {10}^{-2}}$

total Thermal Current= $\frac{d{Q}_1}{dt}+\frac{d{Q}_2}{dt}$

$=80\times 3\times {10}^{-2}(385+209)=594\times 80\times {10}^{-2}$

$=594\times 2.4=1425.6$

$=1.42\times {10}^{3}W$

Hence Option $\text{A}$ is correct.