Heat and Thermodynamics 6 Question 26

31. 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«Ts,Ts being the temperature of the surroundings. If ΔT(T1Ts), find the proportionality constant. Consider that heat is lost only by radiation at the end where the temperature of the rod is T2.

(2004, 4M)

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Answer:

Correct Answer: 31. Proportionality constant =K4eσLTs3+K

Solution:

  1. Rate of heat conduction through rod = rate of the heat lost from right end of the rod.

KA(T1T2)L=eAσ(T24Ts4)(i)

Given that

T2=Ts+ΔTT24=(Ts+ΔT)4=Ts4(1+ΔTTs)4

T24=(Ts+ΔT)4

Using binomial expansion, we have

T24=Ts4(1+4ΔTTs)( as ΔT<Ts)T24Ts4=4(ΔT)(Ts3)

Substituting in Eq. (i), we have

K(T1TsΔT)L=4eσTs3ΔT or K(T1Ts)L=(4eσTs3+KL)ΔTΔT=K(T1Ts)(4eσLTs3+K)

Comparing with the given relation, proportionality constant =K4eσLTs3+K



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