Answer
$\frac{\Delta S}{t} = +0.141 J/K.s $
Work Step by Step
The heat conduction in a cylinder is
$H = kA \frac{(T_H - T_C)}{L} $
Where $k = 401$
$A = \pi r^2 = \pi (0.02m)^2 = 0.0012566371 m^2$
$L = 1.50 \space m$
$T_H = 300 + 273 K = 573 K$
$T_L = 30 + 273 K = 303K$
Substitute all the values into the equation
$H = kA \frac{(T_H - T_C)}{L} $
$H = (401)(0.0012566371 m^2) \frac{(573 K - 303 K )}{1.50 m} $
$H = 90.7 J/s$
From this, we find the rate-of- decrease of entropy of the high temperature reservoir (at 573 K) which is
$\frac{\Delta S}{t} = -\frac{H}{T_H}$
$\frac{\Delta S}{t} = -\frac{90.7 J/s}{573 K}$
$\frac{\Delta S}{t} = - 0.158 J/K.s$
Next, we find the rate-of- increase of entropy of the low temperature reservoir (at 303 K) which is
$\frac{\Delta S}{t} = +\frac{H}{T_L}$
$\frac{\Delta S}{t} = +\frac{90.7 J/s}{303 K}$
$\frac{\Delta S}{t} = 0.299 J/K.s$
So the net rate of entropy is
$\frac{\Delta S}{t} = - 0.158 J/K.s + 0.299 J/K.s $
$\frac{\Delta S}{t} = +0.141 J/K.s $
The positive sign indicates the rate of entropy is increasing for the rod-reservoirs system
