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

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° C cm- 1 (B) 12 ° C cm- 1 (C) 25 ° C cm- 1 (D) 20 ° C cm- 1

Tardigrade Admin · Accepted Answer

(ÎQ/Ît)=(KAÎθ /Îx) ⇒ Thermal gradient ⇒ (Îθ /Îx)=((ÎQ / Ît) /K) ⇒ (Îθ /Îx)=(10/0 .4)=25 ° C cm- 1

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

$1 0^{\circ} C c m^{- 1}$

$12^{\circ} C c m^{- 1}$

$25^{\circ} C c m^{- 1}$

$20^{\circ} C c m^{- 1}$

$\frac{Δ Q}{Δ t} = \frac{K A Δ θ}{Δ x}$
$\Rightarrow$ Thermal gradient
$\Rightarrow \frac{Δ θ}{Δ x} = \frac{( Δ Q /Δ t )}{K}$
$\Rightarrow \frac{Δ θ}{Δ x} = \frac{10}{0.4} = 25^{\circ} C c m^{- 1}$

Q. The thermal conductivity of a material in the CGS system is 0.4 . In steady-state, the rate of flow of heat 10cals−1cm−2 , then the thermal gradient will be

10∘Ccm−1

12∘Ccm−1

25∘Ccm−1

20∘Ccm−1