A slab consists of two identical plates of copper and brass. The free face of the brass is at $0^{\circ}…
- $20^{\circ} \mathrm{C}$
- $40^{\circ} \mathrm{C}$
- $60^{\circ} \mathrm{C}$
- $80^{\circ} \mathrm{C}$
Solution

$\begin{aligned} & \mathrm{K}_2: \mathrm{K}_1=1: 4, \\ & \mathrm{~T}_1=0^{\circ} \mathrm{C}, \mathrm{T}_2=100^{\circ} \mathrm{C}\end{aligned}$ $\begin{aligned} & H=\frac{k_1 A(100-T)}{x}=\frac{k_2 A(T-0)}{x} \\ & \Rightarrow \frac{k_2}{k_1}=\frac{100-T}{T} \\ & \Rightarrow \frac{100-T}{T}=\frac{1}{4} \Rightarrow 400-4 T=T \\ & \therefore T=80^{\circ} \mathrm{C}\end{aligned}$
Asked in: AP EAMCET 2024 (21 May Shift 1)
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