Question

On a cold winter's day heat leaks slowly out of a house at the rate of 20.5 kw. if the inside temperature is 22° c, and the outside temperature is −17.5° c, find the rate of entropy increase.

Answer 1

From Clausius theorem we get the formula for the change of entropy:
$dS=\frac{dQ}{T}$
dS is the change of entropy, dQ is the change of energy, and T is the temperature in Kelvin.
We need to convert those temperatures:
$T_{in}=22^\circ C= 273.15+22=295.15K\\ T_{out}=-17.5^\circ C=273.15-17.5=255.65K$
Entropy changes inside and outside:
$Inside : dS= \frac{-dQ}{T_in}=\frac{-20\cdot 10^3}{295.15K}=-67.76\frac{J}{K}$
$Outside : dS= \frac{+dQ}{T_out}=\frac{+20\cdot 10^3}{255.65K}=+78.23\frac{J}{K}$
The net entropy increase is:
$dS=+78.23\frac{J}{K}-67.76\frac{J}{K}=+10.47\frac{J}{K}$