In a house achieving a heat loss rate of \(200 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) equipped a \(5,000 \mathrm{~W}\) heater, what will the internal temperature be if the outside temperature is \(-10^{\circ} \mathrm{C}\) and the heater is running \(100 \%\) of the time?

We have the following information: - Heat loss rate: \(200 \mathrm{~W} /{ }^{\circ}\mathrm{C}\) - Heater power: \(5,000 \mathrm{~W}\) - Outside temperature: \(-10^{\circ}\mathrm{C}\) - Heater running at 100% capacity Our goal is to find the internal temperature of the house.

Since the heater operates at 100% of its power, it provides \(5,000 \mathrm{~W}\) of heat to the house. The heat loss rate indicates that the house loses \(200 \mathrm{~W} /{ }^{\circ}\mathrm{C}\). Let \(T_i\) be the internal temperature we need to find and \(\Delta T\) be the difference between inside and outside temperatures: \(\Delta T = T_i - (-10^{\circ}\mathrm{C}) = T_i + 10^{\circ}\mathrm{C}\). The house loses heat at a rate proportional to this temperature difference, and the heater compensates for this heat loss. Therefore, we have: $$ 200 \frac{\mathrm{W}}{{}^{\circ}\mathrm{C}} \times \Delta T = 5,000 \mathrm{~W} $$ Substitute \(\Delta T = T_i + 10^{\circ}\mathrm{C}\) into the equation: $$ 200 \frac{\mathrm{W}}{{}^{\circ}\mathrm{C}} \times (T_i + 10^{\circ}\mathrm{C}) = 5,000 \mathrm{~W} $$

Now we need to solve for \(T_i\): $$ 200 \frac{\mathrm{W}}{{}^{\circ}\mathrm{C}} \times (T_i + 10^{\circ}\mathrm{C}) = 5,000 \mathrm{~W} $$ Simplify the equation: $$ 200 \frac{\mathrm{W}}{{}^{\circ}\mathrm{C}} \times T_i + 200 \frac{\mathrm{W}}{{}^{\circ}\mathrm{C}} \times 10^{\circ}\mathrm{C} = 5,000 \mathrm{~W} $$ Cancel units and solve for \(T_i\): $$ 200T_i + 2000 = 5000 $$ Subtract 2000 from both sides of the equation: $$ 200T_i = 3000 $$ Divide both sides by 200: $$ T_i = 15^{\circ} \mathrm{C} $$

The internal temperature of the house when the heater is running at 100% capacity and the outside temperature is \(-10^{\circ}\mathrm{C}\) will be \(15^{\circ}\mathrm{C}\).