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Standard XIII

Physics

Conduction Law

A plane wall ...

Question

A plane wall is $25 cm$ thick with an area of $1 m^{2}$ and has a thermal conductivity of $0.5 W/mK$ . If a temperature difference of $60^{\circ} C$ is maintained across it, then find the rate of heat flow.

120 W

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140 W

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160 W

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180 W

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Solution

The correct option is A $120 W$
We know that, rate of heat flow
$\frac{Δ Q}{Δ T} = k A \frac{T_{2} - T_{1}}{l}$
From the data given in the question,
$\frac{Δ Q}{Δ T} = 0.5 \times 1 \times \frac{60}{0.25}$
$\Rightarrow \frac{Δ Q}{Δ T} = 120 W$
Hence, option (a) is the correct answer.

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Similar questions

Q. A plane wall is

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and has a thermal conductivity of

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. If a temperature difference of

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Conduction Law

Standard XIII Physics

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A plane wall is 25 cm thick with an area of 1 m2 and has a thermal conductivity of 0.5 W/mK. If a temperature difference of 60∘C is maintained across it, then find the rate of heat flow.

The correct option is A 120 WWe know that, rate of heat flow ΔQΔT=kAT2−T1l From the data given in the question, ΔQΔT=0.5×1×600.25 ⇒ΔQΔT=120 W Hence, option (a) is the correct answer.

A plane wall is $25 cm$ thick with an area of $1 m^{2}$ and has a thermal conductivity of $0.5 W/mK$ . If a temperature difference of $60^{\circ} C$ is maintained across it, then find the rate of heat flow.

The correct option is A $120 W$
We know that, rate of heat flow
$\frac{Δ Q}{Δ T} = k A \frac{T_{2} - T_{1}}{l}$
From the data given in the question,
$\frac{Δ Q}{Δ T} = 0.5 \times 1 \times \frac{60}{0.25}$
$\Rightarrow \frac{Δ Q}{Δ T} = 120 W$
Hence, option (a) is the correct answer.