A slab of material X- 2 mm thick- and a slab of material Y - 20 mm thick and of the same area as X- are placed together in thermal contact- The outside face of X is maintained at 100- C- the outside face of Y at 0- C- If Y has 40 times the thermal conductivity of X- their common junction temperature isA- 50- CB- 40- CC- 20- CD- 10- C

The correct option is C 20^∘ CHeat current is constant throughout, dHdt=kA(100 T)2 × 10^ 3=40kAT20 × 10^ 3 or, T=20^∘ C ...

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Standard XII

Physics

New Capacitance in Dielectric

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Question

A slab of material $X, 2 mm$ thick, and a slab of material $Y, 20 mm$ thick and of the same area as $X$ , are placed together in thermal contact. The outside face of $X$ is maintained at $100^{\circ} C$ , the outside face of $Y$ at $0^{\circ} C$ . If $Y$ has 40 times the thermal conductivity of $X$ , their common junction temperature is

50^{\circ} C

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40^{\circ} C

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20^{\circ} C

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10^{\circ} C

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Solution

The correct option is C $20^{\circ} C$
Heat current is constant throughout,

$\frac{d H}{d t} = \frac{k A (100 - T)}{2 \times 10^{- 3}} = \frac{40 k A T}{20 \times 10^{- 3}}$
or, $T = 20^{\circ} C$

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The correct option is C 20∘CHeat current is constant throughout, dHdt=kA(100−T)2×10−3=40kAT20×10−3 or, T=20∘C

The correct option is C $20^{\circ} C$
Heat current is constant throughout,

$\frac{d H}{d t} = \frac{k A (100 - T)}{2 \times 10^{- 3}} = \frac{40 k A T}{20 \times 10^{- 3}}$
or, $T = 20^{\circ} C$