Question

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?

Short answer

Answer: The internal temperature of the house is 15°C.

Step-by-step solution

Understand the given information and the goal

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.

Create an equation using the heat loss rate

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: $$ 200 \frac{\mathrm{W}}{{}^{\circ}\mathrm{C}} \times (T_i + 10^{\circ}\mathrm{C}) = 5,000 \mathrm{~W} $$

Solve the equation for the internal temperature \(T_i\)

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: $$ T_i = 15^{\circ} \mathrm{C} $$

Give the answer

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 \(15^{\circ}\mathrm{C}\).

Key concepts

Heat Transfer

Heat transfer is the process by which thermal energy moves from one place to another. In the context of our exercise, it defines how heat moves from a heated interior of a house to the cooler outside world.
Heat transfer can occur in three ways: conduction, convection, and radiation.

• Conduction: This is the direct transfer of heat through a material. In a house, this might occur through walls, windows, and doors.
• Convection: This involves the movement of heat through fluids, like the air inside and outside of the house.
• Radiation: Heat can also be transferred through electromagnetic waves without involving any physical contact. An example includes sunlight entering through windows.

In our exercise, the rate at which the house loses heat is quantified by the heat loss rate, which is given as 200 W per degree Celsius difference between inside and outside temperatures.
This rate shows how efficiently the material of the house prevents heat from escaping. A higher value suggests more heat is lost per degree of temperature difference, indicating less effective insulation.

Energy Efficiency

Energy efficiency refers to the ability to perform a task while consuming the minimum amount of energy possible. In this exercise, we're dealing with an electric heater rated at 5,000 W.
It consumes this power to maintain the internal temperature against the heat loss.
A heater's efficiency can be better understood by how well it compensates for heat loss while using the least energy. Here are some factors that affect energy efficiency:

• Insulation: Better insulation reduces the rate of heat loss, allowing a heater to achieve desired temperatures using less power.
• Thermal Properties: Materials used in construction impact how well a house retains heat. Materials with better thermal resistance (R-value) improve energy efficiency.

In our setup, energy efficiency could be evaluated by how effectively the 5,000 W heater can keep the indoor temperature at a comfortable 15°C, despite outside temperatures at -10°C.

Temperature Control

Temperature control is crucial for maintaining a comfortable living environment. It deals with the regulation of internal temperatures within an acceptable range, despite external temperature changes. In our case, the internal temperature is controlled by balancing the heater's output with the house's heat loss rate.
Effective temperature control optimizes energy use while ensuring comfort.

• Automatic Systems: Many modern homes use programmable thermostats that adjust settings based on time or occupancy, enhancing comfort while conserving energy.
• Manual Adjustments: Users can manually manage heater settings to control indoor temperatures, though this may not always be energy efficient.

The exercise provides an example where the heater must be used at full capacity to maintain a stable internal temperature of 15°C. This stability is achieved by calculating the heat output needed to balance the 200 W per degree Celsius heat loss, resulting in a comfortable living space regardless of outdoor conditions.