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Chapter 13: Problem 28

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?

Short Answer

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

Step by step solution

01

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, \( t = 6 \; \text{hours} = 21600 \; \text{seconds}\)
02

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:\[\frac{Q}{t} = \frac{0.045 \cdot 1.6 \cdot 110}{0.020} \approx 396 \; \mathrm{J/s}\]
03

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:\[Q = 396 \cdot 21600 = 8,553,600 \; \mathrm{J}\]
04

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:\[Q_{\text{kWh}} = \frac{8,553,600 \; \mathrm{J}}{3,600,000 \; \mathrm{J/kWh}} \approx 2.38 \; \mathrm{kWh}\]
05

Calculate the Cost of Operating the Oven

Given that the electricity rate is \\(0.10 per kWh, the cost is:\[\text{Cost} = 2.38 \; \mathrm{kWh} \times \\)0.10/\mathrm{kWh} = \$0.238\]
06

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Two pots are identical except that the flat bottom of one is aluminum, whereas that of the other is copper. Water in these pots is boiling away at \(100.0^{\circ} \mathrm{C}\) at the same rate. The temperature of the heating element on which the aluminum bottom is sitting is \(155.0^{\circ} \mathrm{C}\). Assume that heat enters the water only through the bottoms of the pots and find the temperature of the heating element on which the copper bottom rests.

A car parked in the sun absorbs energy at a rate of 560 watts per square meter of surface area. The car reaches a temperature at which it radiates energy at this same rate. Treating the car as a perfect radiator \((e=1)\), find the temperature.

Refer to Interactive Solution \(\underline{13.13}\) at for help in solving this problem. In an aluminum pot, \(0.15 \mathrm{~kg}\) of water at \(100^{\circ} \mathrm{C}\) boils away in four minutes. The bottom of the pot is \(3.1 \times 10^{-3} \mathrm{~m}\) thick and has a surface area of \(0.015 \mathrm{~m}^{2}\). To prevent the water from boiling too rapidly, a stainless steel plate has been placed between the pot and the heating element. The plate is \(1.4 \times 10^{-3} \mathrm{~m}\) thick, and its area matches that of the pot. Assuming that heat is conducted into the water only through the bottom of the pot, find the temperature at (a) the aluminum-steel interface and (b) the steel surface in contact with the heating element.

Concept Simulation \(13.1\) at illustrates the concepts pertinent to this problem. A person's body is covered with \(1.6 \mathrm{~m}^{2}\) of wool clothing. The thickness of the wool is \(2.0 \times 10^{-3} \mathrm{~m}\). The temperature at the outside surface of the wool is \(11{ }^{\circ} \mathrm{C}\), and the skin temperature is \(36^{\circ} \mathrm{C}\). How much heat per second does the person lose due to conduction?

Concept Questions A pot of water is boiling on a stove under one atmosphere of pressure. Assume that heat enters the pot only through its bottom, which is copper and rests on a heating element. In a certain time, a mass \(m\) of water boils away. (a) What is the temperature of the boiling water and does it change during this time? (b) What determines the amount of heat needed to boil the water? (c) Is the temperature of the heating element in contact with the pot greater than, smaller than, or equal to \(100^{\circ} \mathrm{C}\) ? Explain.
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