JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Heat & Thermodynamics

A heat source at

T = 1 0^{3} K

is connected to another heat reservoir at

T = 1 0^{2} K

by a copper slab which is 1 m thick. Given that the thermal conductivity of copper is

0.1 WK^{- 1} m^{- 1}

, the energy flux through it in the steady state is:

(A) $90 Wm^{- 2}$ ✓ Correct
(B) $120 Wm^{- 2}$
(C) $65 Wm^{- 2}$
(D) $200 Wm^{- 2}$

Answer: (A)

Solution

In the steady state the heat flux through the slab is

\frac{1}{A} \frac{Δ Q}{Δ t} = \frac{k}{ℓ} (T_{1} - T_{2})

. With

k = 0.1

W K

^{- 1}

^{- 1}

ℓ = 1

m and a temperature drop of

1000 - 100 = 900

K, the flux is

\frac{0.1}{1} \times 900 = 90

W m

^{- 2}

. The large

900

K difference is tamed by the small conductivity of this one-metre slab to leave a modest flux.

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option

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