The temperature of the hot and cold ends of a | vedclass

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(B) In a thermal steady state,the temperature gradient along a uniform rod is constant.The temperature gradient is given by: $\frac{dT}{dx} = \frac{T_

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) In a thermal steady state,the temperature gradient along a uniform rod is constant.The temperature gradient is given by: $\frac{dT}{dx} = \frac{T_{hot} - T_{cold}}{L} = \frac{100^\circ C - 20^\circ C}{20 \ cm} = \frac{80}{20} = 4^\circ C/cm$.The temperature at any distance $x$ from the hot end is given by $T(x) = T_{hot} - (	ext{gradient} 	imes x)$.At the centre of the rod,$x = 10 \ cm$.Therefore,$T(10) = 100^\circ C - (4^\circ C/cm 	imes 10 \ cm) = 100^\circ C - 40^\circ C = 60^\circ C$.

The temperature of the hot and cold ends of a $20 \ cm$ long rod in a thermal steady state are at $100^\circ C$ and $20^\circ C$ respectively. The temperature at the centre of the rod is ...... $^\circ C$.

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