Answer
About 15 percent more heat would be needed.
Work Step by Step
Assume that the heat loss as conductive. Using equation 14–5, we see that we may express the heat loss as $Q=K t \Delta t$, where the constant K (not kelvin) takes into account the house’s average conductivity properties (e.g., insulation, surface area). K would have units of $J/(h \cdot C^{\circ}$. In this model, the heat loss Q is proportional to the time multiplied by the temperature difference. Assume that K is constant and independent of both time and temperature.
First, if we turn down the thermostat, calculate the heat loss in joules.
$$Q_1=K(15h)(22^{\circ}C-8^{\circ}C)+ K(9h)(16^{\circ}C-0^{\circ}C)=354 K$$
Now assume we don’t turn down the thermostat, and calculate the heat loss in joules.
$$Q_2=K(15h)(22^{\circ}C-8^{\circ}C)+ K(9h)(22^{\circ}C-0^{\circ}C)=408 K$$
How much more heat is needed if we don’t turn down the thermostat?
$$\frac{\Delta Q}{Q_1}=\frac{408K-354K}{354K}=0.1525=15\%$$
About 15 percent more heat would be needed if the thermostat isn't turned down.
