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Heat and Thermodynamics question

2019 · 10 Jan · Shift 1 · Q64
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Heat and Thermodynamics question

2019 · 10 Jan · Shift 1 · Q64

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A heat source at T = 103 K is connected to another heat reservoir at T = 102 K by a copper slab which is 1 mthick. Given that the thermal conductivity of copper is 0.1 WK–1m–1, the energy flux through it in the steady state is -
  1. A
    200 Wm −-− 2
  2. B
    65 Wm −-− 2
  3. C
    120 Wm −-− 2
  4. D
    90 Wm −-− 2
View written solutionFree

Correct answer: D

  1. Use Fourier’s law of heat conduction

For steady heat conduction through a slab, the heat current per unit area (energy flux) is

QAt=kΔTL\frac{Q}{A t} = k\frac{\Delta T}{L}AtQ​=kLΔT​

where:

  • k=0.1 W m−1K−1k = 0.1\ \text{W m}^{-1}\text{K}^{-1}k=0.1 W m−1K−1
  • L=1 mL = 1\ \text{m}L=1 m
  • T1=103 KT_1 = 10^3\ \text{K}T1​=103 K
  • T2=102 KT_2 = 10^2\ \text{K}T2​=102 K
  1. Find temperature difference
ΔT=103−102=1000−100=900 K\Delta T = 10^3 - 10^2 = 1000 - 100 = 900\ \text{K}ΔT=103−102=1000−100=900 K
  1. Substitute in the formula
Flux=0.1×9001=90 W m−2\text{Flux} = 0.1 \times \frac{900}{1} = 90\ \text{W m}^{-2}Flux=0.1×1900​=90 W m−2
  1. Match with options
90 W m−290\ \text{W m}^{-2}90 W m−2

So the correct option is D.

  1. Comparison with stored answer

Stored correct answer: D

This matches the derived answer.

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