Question

The thermal conductivity of a material in the CGS system is $0.4$ . In steady-state, the rate of flow of heat $10 \, cal \, s^{- 1} \, cm^{- 2}$ , then the thermal gradient will be

• A. $10^\circ \, C \, cm^{- 1}$
• B. $12 \,{}^\circ C \, cm^{- 1}$
• C. $25 \,{}^\circ C \, cm^{- 1}$
• D. $20 \,{}^\circ C \, cm^{- 1}$

Answer 1

Answer: (C)

$\frac{ΔQ}{Δt}=\frac{KAΔ\theta }{Δx}$
$\Rightarrow $ Thermal gradient
$\Rightarrow \frac{Δ\theta }{Δx}=\frac{\left(ΔQ / Δt\right) \, }{K}$
$\Rightarrow \, \frac{Δ\theta }{Δx}=\frac{10}{0 .4}=25 \,{}^\circ C \, cm^{- 1}$