
If the coefficient of conductivity of aluminium is $ 0.5 cal/cm-sec-^0C $ , then in order to conduct $ 10 cal/sec – cm^2 $ in the steady state, what is the temperature gradient in aluminium?
(A) $ 5 {^0C}cm $
(B) $ 10 {^0C}cm $
(C) $ 20 {^0C}cm $
(D) $ 10.5 {^0C}cm $
Answer
572.1k+ views
Hint
Conductivity is a measure of the ability of a substance to conduct electricity; the reciprocal of resistivity. in the case of a solution, the electrolytic conductivity is the current density divided by the electric field strength, measured in siemens per metreFormerly called: specific conductance Symbol: κ
Complete step by step answer
According to the formula we know,
$ \dfrac{{dQ}}{{dt}} = KA(\dfrac{{dT}}{{dx}}) $ … equation 1
Now, $ \dfrac{{dQ}}{{dt}} $ is also known as thermal current which represent as i, and $ \dfrac{{dT}}{{dx}} $ is the temperature gradient.
We know the value of k = 0.5 cal/cm-sec-0C
Now, continuing equation 1 we get,
$ \Rightarrow \dfrac{{dQ}}{{dt}} \times \dfrac{1}{A} = K(\dfrac{{dT}}{{dx}}) $
$ \therefore \dfrac{{dT}}{{dx}} = \dfrac{{10}}{{0.5}} = {20^\circ }c/m $
So, the temperature gradient in aluminium is $20^0 c/m$.
Option (C) is correct.
Note
A temperature gradient is just the change in temperature over a specified distance between two locations. The difference in temperature causes differences in air pressure between the two spots. Temperature gradients between water and land can also cause local atmospheric circulations which affect winds.
Conductivity is a measure of the ability of a substance to conduct electricity; the reciprocal of resistivity. in the case of a solution, the electrolytic conductivity is the current density divided by the electric field strength, measured in siemens per metreFormerly called: specific conductance Symbol: κ
Complete step by step answer
According to the formula we know,
$ \dfrac{{dQ}}{{dt}} = KA(\dfrac{{dT}}{{dx}}) $ … equation 1
Now, $ \dfrac{{dQ}}{{dt}} $ is also known as thermal current which represent as i, and $ \dfrac{{dT}}{{dx}} $ is the temperature gradient.
We know the value of k = 0.5 cal/cm-sec-0C
Now, continuing equation 1 we get,
$ \Rightarrow \dfrac{{dQ}}{{dt}} \times \dfrac{1}{A} = K(\dfrac{{dT}}{{dx}}) $
$ \therefore \dfrac{{dT}}{{dx}} = \dfrac{{10}}{{0.5}} = {20^\circ }c/m $
So, the temperature gradient in aluminium is $20^0 c/m$.
Option (C) is correct.
Note
A temperature gradient is just the change in temperature over a specified distance between two locations. The difference in temperature causes differences in air pressure between the two spots. Temperature gradients between water and land can also cause local atmospheric circulations which affect winds.
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