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One end of a rod of length L and cross-sectional area A is kep

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One end of a rod of length $L$ and cross-sectional area $A$ is kept in a furnace of temperature $T_{1}$ . The other end of the rod is kept at a temperature $T_{2}$ . The thermal conductivity of the material of the rod is $K$ and emissivity of the rod is $e$ .

It is given that $T_{2} = T_{s} + Δ T$ , where $Δ T <$ being the temperature of the surroundings. If $Δ T \propto (T_{1} - T_{s})$ , find the proportionality constant. Consider that heat is lost only by radiation at the end where the temperature of the rod is $T_{2} .$

Name: Video solution 1: One end of a rod of length L and cross-sectional area A is kept in a furnace of temperature T1. The other end of the rod is kept at a temperature T2. The thermal conductivity of the material of the rod is K and emissivity of the rod is e.It is given that T2=Ts+ΔT, where ΔT< being the temperature of the surroundings. If ΔT∝(T1−Ts), find the proportionality constant. Consider that heat is lost only by radiation at the end where the temperature of the rod is T2.
Uploaded: 2022-05-06T10:45:34.000Z
Duration: 7 min

Views: 5,988 students

Updated on: May 6, 2022

Exam:

JEE Advanced 2004

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Jack Discussed

One end of a rod of length

L

and cross-sectional area

A

is kept in a furnace of temperature

T_{1}

. The other end of the rod is kept at a temperature

T_{2}

. The thermal conductivity of the material of the rod is

K

and emissivity of the rod is

e

It is given that

T_{2} = T_{s} + Δ T

, where

Δ T <

being the temperature of the surroundings. If

Δ T \propto (T_{1} - T_{s})

, find the proportionality constant. Consider that heat is lost only by radiation at the end where the temperature of the rod is

T_{2} .

5 mins ago

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5 mins ago

Text solutionVerified

Rate of heat conduction through rod

=

rate of the heat lost from right end of the rod.

∴ \frac{K A ( T _{1} - T _{2} )}{L} = e A σ (T_{2}^{4} - T_{s}^{4})

Given that

∴ ∴ T_{2}^{4} = (T_{s} + Δ T)^{4} = T_{s}^{4} (1 + \frac{Δ T}{T _{s}})^{4}

Using binomial expansion, we have

T_{2}^{4} T_{2}^{4} - T_{s}^{4} = T_{s}^{4} (1 + 4 \frac{Δ T}{T _{s}}) (as Δ T ≪ T_{s}) = 4 (Δ T) (T_{s}^{3})

Substituting in Eq. (i), we have

or ∴ \frac{K ( T _{1} - T _{s} - Δ T )}{L} = 4 e σ T_{s}^{3} \cdot Δ T \frac{K ( T _{1} - T _{s} )}{L} = (4 e σ T_{s}^{3} + \frac{K}{L}) Δ T Δ T = \frac{K ( T _{1} - T _{s} )}{( 4 e σ L T _{s}^{3} + K )}

Comparing with the given relation, the proportionality constant

= \frac{K}{4 e σ L T _{x}^{3} + K}

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Question Text	One end of a rod of length $L$ and cross-sectional area $A$ is kept in a furnace of temperature $T_{1}$ . The other end of the rod is kept at a temperature $T_{2}$ . The thermal conductivity of the material of the rod is $K$ and emissivity of the rod is $e$ . It is given that $T_{2} = T_{s} + Δ T$ , where $Δ T <$ being the temperature of the surroundings. If $Δ T \propto (T_{1} - T_{s})$ , find the proportionality constant. Consider that heat is lost only by radiation at the end where the temperature of the rod is $T_{2} .$
Updated On	May 6, 2022
Topic	Thermal properties of Matter
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 1
Upvotes	214
Avg. Video Duration	7 min

Text solutionVerified

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Text solutionVerified

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