Problem 28 - Physics

Question

The temperature in an electric oven is \(160{ }^{\circ} \mathrm{C}\). The temperature at the outer surface in the kitchen is \(50^{\circ} \mathrm{C}\). The oven (surface area \(=1.6 \mathrm{~m}^{2}\) ) is insulated with material that has a thickness of \(0.020 \mathrm{~m}\) and a thermal conductivity of \(0.045 \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m} \cdot \mathrm{C}^{\circ}\right) .\) (a) How much energy is used to operate the oven for six hours? (b) At a price of \(\$ 0.10\) per kilowatt. hour for electrical energy, what is the cost of operating the oven?

Step-by-Step Solution

Answer

The oven uses 8.55 MJ of energy in six hours, costing $0.24.

Step 1: Identify the Given Values

The problem provides the following data:- Inner temperature, \( T_1 = 160^{\circ} \mathrm{C}\)- Outer temperature, \( T_2 = 50^{\circ} \mathrm{C}\)- Surface area, \( A = 1.6 \; \mathrm{m}^2\)- Thickness of insulation, \( d = 0.020 \; \mathrm{m}\)- Thermal conductivity, \( k = 0.045 \; \mathrm{J/(s \cdot m \cdot C^{\circ})}\)- Time duration, \( t = 6 \; \text{hours} = 21600 \; \text{seconds}\)

Step 2: Use the Heat Transfer Formula to Find Heat Loss per Second

For steady-state heat conduction, the rate of heat loss \( Q/t \) can be calculated using the formula:\[\frac{Q}{t} = \frac{k \cdot A \cdot (T_1 - T_2)}{d}\]Substitute the given values:\[\frac{Q}{t} = \frac{0.045 \; \mathrm{J/(s \cdot m \cdot C^{\circ})} \cdot 1.6 \; \mathrm{m}^2 \cdot (160^{\circ} C - 50^{\circ} C)}{0.020 \; \mathrm{m}}\]Simplifying, we find:\[\frac{Q}{t} = \frac{0.045 \cdot 1.6 \cdot 110}{0.020} \approx 396 \; \mathrm{J/s}\]

Step 3: Calculate the Total Energy Used Over Six Hours

Since we know the energy loss per second, we can multiply by the total time in seconds:\[Q = \left( \frac{Q}{t} \right) \cdot t = 396 \; \mathrm{J/s} \cdot 21600 \; \mathrm{s} \]Calculating this, we get:\[Q = 396 \cdot 21600 = 8,553,600 \; \mathrm{J}\]

Step 4: Convert Energy to Kilowatt-Hours

We need to convert joules to kilowatt-hours for the cost calculation. Note that:\[1 \; \mathrm{kWh} = 3,600,000 \; \mathrm{J}\]Therefore, the energy in kilowatt-hours is:\[Q_{\text{kWh}} = \frac{8,553,600 \; \mathrm{J}}{3,600,000 \; \mathrm{J/kWh}} \approx 2.38 \; \mathrm{kWh}\]

Step 5: Calculate the Cost of Operating the Oven

Given that the electricity rate is \\(0.10 per kWh, the cost is:\[\text{Cost} = 2.38 \; \mathrm{kWh} \times \\)0.10/\mathrm{kWh} = \$0.238\]

Step 6: Conclusion

The total energy used by the oven over six hours is approximately \(8.55 \times 10^6\) joules. The cost to operate the oven for this time at the given rate is approximately \$0.24.

Key Concepts

Heat Transfer

Heat transfer is a fundamental concept in physics that describes how energy moves from one place to another due to a temperature difference. In the context of an oven, heat is transferred from the hotter interior to the cooler exterior. This movement occurs because molecules vibrate more rapidly at higher temperatures and pass their energy to neighboring cooler molecules.

This process can be understood through three main mechanisms:

• Conduction: The transfer of heat through a solid material. It occurs when particles collide and transfer energy.
• Convection: The transfer of heat by the physical movement of fluid (like air or water).
• Radiation: The transfer of heat through electromagnetic waves.

In our scenario, the primary mode of heat transfer is conduction, where heat travels through the insulation of the oven.

Steady-State Heat Conduction

Steady-state heat conduction occurs when the rate of heat entering a system is equal to the rate of heat leaving it. In simpler terms, the temperatures within the material do not change over time.

This concept is critical in solving problems where insulation is involved, such as our oven example. The formula used to calculate heat transfer in steady-state conditions is:\[\frac{Q}{t} = \frac{k \cdot A \cdot (T_1 - T_2)}{d}\]Where:

• \(Q/t\): Heat transfer rate in Joules per second (J/s).
• \(k\): Thermal conductivity of the material.
• \(A\): Surface area through which heat is being transferred.
• \(T_1\) and \(T_2\): Inner and outer temperatures.
• \(d\): Thickness of the insulating material.

Understanding this formula helps in calculating the energy efficiency and insulation quality of devices like ovens.

Energy Cost Calculation

Knowing how much it costs to run an appliance is essential for both budgeting and energy conservation. To find out the cost, we first need to know how much energy is used in kilowatt-hours (kWh), which is a common billing unit for electricity.

The cost can then be determined using:\[\text{Cost} = Q_{\text{kWh}} \times \text{Rate per kWh}\]In our example, once we have the total energy usage in kWh, it's multiplied by the cost per kWh, given as $0.10. This calculation helps homeowners understand their energy expenses and encourages them to use energy more wisely.

Joule to Kilowatt-Hour Conversion

Conversion from joules to kilowatt-hours is a useful step because it aligns the measure of energy with the way electricity usage is billed.The conversion factor is:

• 1 kilowatt-hour (kWh) = 3,600,000 joules (J).

This means that to convert energy from joules to kWh, you divide the total energy in joules by 3,600,000.For example, if an oven uses 8,553,600 J of energy, you convert it to kWh as follows:\[Q_{\text{kWh}} = \frac{8,553,600 \text{ J}}{3,600,000 \text{ J/kWh}} \approx 2.38 \text{ kWh}\]This step is vital in translating energy calculations into a format used by electricity bills.