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=(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= (-dQ)/(T_in)=(-20\cdot 10^3)/(295.15K)=-67.76(J)/(K)`

`$$Outside$: dS= (+dQ)/(T_out)=(+20\cdot 10^3)/(255.65K)=+78.23(J)/(K)`
The net entropy increase is:

`dS=+78.23(J)/(K)-67.76(J)/(K)=+10.47(J)/(K)`