Question

The north wall of an electrically heated home is 20 \(\mathrm{ft}\) long, \(10 \mathrm{ft}\) high, and \(1 \mathrm{ft}\) thick, and is made of brick whose thermal conductivity is \(k=0.42 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}{ }^{\circ} \mathrm{F}\). On a certain winter night, the temperatures of the inner and the outer surfaces of the wall are measured to be at about \(62^{\circ} \mathrm{F}\) and \(25^{\circ} \mathrm{F}\), respectively, for a period of \(8 \mathrm{~h}\). Determine \((a)\) the rate of heat loss through the wall that night and \((b)\) the cost of that heat loss to the home owner if the cost of electricity is \(\$ 0.07 / \mathrm{kWh}\).

Short answer

Complete the calculations and find the cost of the heat loss over the 8-hour period: Step 1: Calculate the Cross-Sectional Area of the Wall $$A = 20 \,\text{ft} \times 10 \,\text{ft} = 200\,\text{ft}^2$$ Step 2: Calculate the Temperature Difference Across the Wall $$\Delta T = 62 ^\circ \text{F} - 25 ^\circ \text{F} = 37 ^\circ \text{F}$$ Step 3: Determine the Rate of Heat Loss Through the Wall $$q = 0.42 \,\text{Btu/h·ft·°F} \cdot \frac{(200\,\text{ft}^2) \cdot (37\,^\circ \text{F})}{1\,\text{ft}} = 3108\,\text{Btu/h}$$ Step 4: Determine the Total Heat Loss Over the 8-hour Period $$Q_\text{total} = 3108 \,\text{Btu/h} \times 8 \,\text{h} = 24,864 \,\text{Btu}$$ Step 5: Convert the Total Heat Loss to kWh $$Q_\text{total\,(kWh)} = \frac{24,864 \,\text{Btu}}{3412.14 \,\text{Btu/kWh}} \approx 7.28 \,\text{kWh}$$ Step 6: Calculate the Cost of the Heat Loss $$\text{Cost} = 7.28 \,\text{kWh} \times 0.07 \,\text{\$/kWh} = 0.5096 \,\text{\$}$$ The cost of the heat loss through the north wall over the 8-hour period is approximately $0.51.

Step-by-step solution

Calculate the Cross-Sectional Area of the Wall

We can find the cross-sectional area of the wall by multiplying the length and height of the wall. $$A = 20 \,\text{ft} \times 10 \,\text{ft}$$

Calculate the Temperature Difference Across the Wall

We are given the inner and outer surface temperatures of the wall, so we can find the temperature difference as: $$\Delta T = T_\text{inner} - T_\text{outer} = 62 ^\circ \text{F} - 25 ^\circ \text{F}$$

Determine the Rate of Heat Loss Through the Wall

Substitute the values for thermal conductivity (\(k\)), cross-sectional area (\(A\)), temperature difference (\(\Delta T\)), and the thickness of the wall (\(L\)) into the steady-state heat transfer equation: $$q = 0.42 \,\text{Btu/h·ft·°F} \cdot \frac{(20\,\text{ft} \times 10\,\text{ft}) \cdot (62\,^\circ \text{F} - 25\,^\circ \text{F})}{1\,\text{ft}}$$ Calculate the rate of heat loss through the wall.

Determine the Total Heat Loss Over the 8-hour Period

We need to calculate the total heat loss over an 8-hour period. To do this, multiply the heat loss rate (\(q\)) by the total time in hours. $$Q_\text{total} = q \times t = q \times 8 \,\text{h}$$ Calculate the total heat loss (in Btu).

Convert the Total Heat Loss to kWh

To find the cost of heat loss, we need to convert the total heat loss (in Btu) to kWh. The conversion factor is: $$1 \,\text{kWh} = 3412.14 \,\text{Btu}$$ So, the total heat loss in kWh can be calculated as: $$Q_\text{total\,(kWh)} = \frac{Q_\text{total\,(Btu)}}{3412.14 \,\text{Btu/kWh}}$$

Calculate the Cost of the Heat Loss

We are given the cost of electricity as \(0.07 / \text{kWh}\). To find the cost of the heat loss, multiply the total heat loss in kWh by the cost per kWh. $$\text{Cost} = Q_\text{total\,(kWh)} \times 0.07 \,\text{\$/kWh}$$ Calculate the cost of the heat loss over the 8-hour period.

Key concepts

Thermal Conductivity

Thermal conductivity is a measure of a material's ability to conduct heat. It represents how easily heat passes through a material due to a temperature difference. In the context of building materials, like brick in this example, a lower thermal conductivity means that the material is a better insulator, reducing heat loss.

The rate at which heat transfers through a material is calculated using the formula: \[ q = k \cdot A \cdot \frac{\Delta T}{L} \] where \( q \) is the heat transfer rate in Btu/h, \( k \) is the thermal conductivity of the material (in Btu/h·ft·°F), \( A \) is the cross-sectional area through which heat is flowing (in square feet), \( \Delta T \) is the temperature difference between the hot and cold surfaces (in °F), and \( L \) is the thickness of the material (in feet).

This concept is crucial to engineers and architects when designing homes to ensure they are energy efficient. The goal is to minimize heat loss, keeping heating costs low and conserving energy.

Steady-State Heat Transfer

Steady-state heat transfer occurs when the temperature difference driving the heat flow does not change with time. In practical terms, this means that the inner and outer surfaces of the wall mentioned in the exercise have consistent temperatures, and the rate of heat loss is constant over time.

The concept of steady-state is an idealization that simplifies calculations; in reality, temperatures fluctuate, but over short periods, it's a valid approximation. The calculation for the steady-state heat loss through the wall, as shown in the step by step solution, assumes that the temperatures remain at \(62^\circ \text{F}\) and \(25^\circ \text{F}\) and that the properties of the wall stay constant too.

Determining the heat loss at a steady state is essential for predicting heating requirements and costs for a home or building. This helps in making informed decisions regarding insulation materials and thicknesses to use for efficient thermal management.

Energy Conversion

Energy conversion in the context of heat loss is the process of changing one form of energy to another. In our exercise, the heat energy lost through the wall is ultimately being converted to electrical energy when we calculate the cost of the lost heat in terms of electricity consumption.

Heat energy can be measured in British Thermal Units (Btu), which can then be converted to kilowatt-hours (kWh), the standard unit for electrical energy usage. The conversion factor used in the solution step is \(1 \text{kWh} = 3412.14 \text{Btu}\), which allows us to calculate the monetary cost of heat loss.

Understanding energy conversion is essential for comprehending how different forms of energy equate to one another, enabling one to comprehend energy efficiency and conservation from both a physical and economic perspective.