If the inside temp is 22degrees C and outside temp is -14.5C fidn the rate of entropy increase. HELP!?

on a cold winter's day heat leaks slowly out of a house at the rate of 20.0kW. if the inside temperature is 22 degrees C, and the outside temperature is -14.5 degrees C, find the rate of entropy increase.





the correct answer is supposed to be : 9.6 W/K but i dont know how to find that answer. please help and show steps. thanks!

If the inside temp is 22degrees C and outside temp is -14.5C fidn the rate of entropy increase. HELP!?
Inside:


- Temperature in Celsius = 22


- Temperature in Kelvin = 22 + 273.15 = 295.15





Outside:


- Temperature in Celsius = -14.5


- Temperature in Kelvin = 273.15 - 14.5 = 258.65





d(Entropy) = dQ/Temperature


So when, in 1 second, 20.0 kJ goes from outside to inside, the inside has lost 20.0 kJ, but the outside has gained 20.0 kJ.





Inside: dQ = -20.0 kJ, dQ/Temperature(K) = -20.0/295.15


Outside: dQ = +20.0 J, dQ/Temperature(K) = 20.0/258.65





Therefore, the net change in entropy in 1 second is:


20/258.65 - 20/295.15 = 0.009562 kJ/K


or 9.562 J/K





Since that is in one second, the rate we get by dividing by 1 second:


d(Entropy)/dt = 9.562 W/K