Chapter 2 - Energy, Energy Transfer, and General Energy Analysis - Problems - Page 104: 2-92

$Q_{loss}=202176kJ$

First we calculate the are of the wall: $A=2m*2m=4m^2$ Knowing that the heat transfer in this case is by conduction: $Q_{cond}=\frac{kA\Delta T}{L}=\frac{0.78\frac{W}{m^{\circ}C}*4m^2*(15^{\circ}C-6^{\circ}C)}{0.005m}=5616W$ $10h*\frac{3600s}{1h}=36000s$ Over $10hours$ the amount of heat loss is: $Q_{loss}=5616W*36000s=202176kJ$