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An ordinary egg can be approximated as a \(5.5-\mathrm{cm}-\) diameter sphere. The egg is initially at a uniform temperature of \(8^{\circ} \mathrm{C}\) and is dropped into boiling water at \(97^{\circ} \mathrm{C}\). Taking the properties of the egg to be \(\rho=1020 \mathrm{kg} / \mathrm{m}^{3}\) and \(c_{p}=3.32 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C},\) determine \((a)\) how much heat is transferred to the egg by the time the average temperature of the egg rises to \(70^{\circ} \mathrm{C}\) and \((b)\) the amount of entropy generation associated with this heat transfer process.

Short Answer

Expert verified
Question: Determine the heat transferred to an egg when its temperature rises from 8°C to 70°C and calculate the entropy generation associated with the heat transfer process. Answer: To determine the heat transferred to the egg, first, find the volume of the egg using the formula \(V = \frac{4}{3}\pi r^3\). Then, calculate the mass of the egg using the density and volume with the formula \(m = \rho V\). With the mass and specific heat, find the heat transfer using the formula \(Q = mc_{p}\Delta T\). To calculate the entropy generation, use the formula \(S_{gen} = \int \frac{dQ}{T}\), and integrate with respect to temperature after substituting the differential heat transfer, \(dQ = mc_{p}dT\).

Step by step solution

01

Determine the volume of the egg

Since the egg is approximated as a sphere, we can find its volume using the formula for the volume of the sphere, \(V = \frac{4}{3}\pi r^3\). Given the diameter of the sphere is 5.5 cm, we can calculate the radius as \(r = \frac{5.5}{2} \mathrm{cm}\).
02

Calculate the mass of the egg

Now that we have the volume of the egg, we can calculate its mass using the provided density. The mass of the egg can be found using the formula \(m = \rho V\).
03

Determine the amount of heat transfer

With the mass calculated and specific heat given, we can find the heat transferred to the egg using the formula \(Q = mc_{p}\Delta T\), where \(Q\) is the heat transfer, \(m\) is the mass, \(c_{p}\) is the specific heat, and \(\Delta T\) is the change in temperature, which is the final temperature \((70^{\circ}\mathrm{C})\) minus the initial temperature \((8^{\circ}\mathrm{C})\).
04

Calculate the entropy generation

To find the total entropy generation, we use the formula \(S_{gen} = \int \frac{dQ}{T}\). However, since the temperature is not constant, we must integrate this formula with respect to temperature. We first find the differential heat transfer, \(dQ = mc_{p}dT\), and then substitute it into the entropy generation formula and integrate with respect to temperature.

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Key Concepts

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

Specific Heat Capacity
The concept of specific heat capacity is pivotal in understanding how different materials absorb and release heat. It is defined as the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). Represented by the symbol \(c_p\), it is typically measured in joules per kilogram per degree Celsius \(\text{J/kg}\cdot^\circ\text{C}\).

In the context of the given exercise, the specific heat capacity of the egg is essential to determine how much heat needs to be transferred for the egg's temperature to rise from its initial to the desired final temperature. With the given value of \(c_p = 3.32 \text{kJ/kg}\cdot^\circ\text{C}\), we can see that eggs have a relatively high specific heat capacity, which means they require a substantial amount of energy to change temperature. This property is vital for cooking, as it ensures that heat is distributed evenly throughout the egg.

When calculating the amount of heat transferred, the formula \(Q = mc_p\Delta T\) requires the mass of the egg (\(m\)), which is derived from its volume and density, the specific heat capacity (\(c_p\)), and the temperature change (\(\Delta T\)). In simpler terms, understanding specific heat capacity enables us to predict how much energy it takes to cook an egg to a particular temperature.
Entropy Generation
Entropy can be a somewhat abstract concept; in thermodynamics, it represents the measure of disorder or randomness within a thermodynamic system. Entropy generation, particularly, is the creation of additional entropy in a system due to irreversible processes, such as heat transfer, chemical reactions, and friction. It's a key concept in the second law of thermodynamics, which states that the total entropy of a closed system can never decrease over time.

In practical terms, whenever heat flows from a hot object to a cold one (as with our egg dropped into boiling water), the process generates entropy because it's a spontaneous and irreversible event. The entropy generated, represented by \( S_{gen} \), is not always easily calculable, as it often requires complex integration when the system's temperature changes during the process, as is the case with the heating of the egg.

The step-by-step solution provided makes use of integral calculus to find the entropy generation by integrating the differential heat transfer \(dQ\) over the temperature range. In essence, through this calculation, we understand that the egg's cooking process is not just about heating; it inherently increases the universe's entropy, reflecting the fundamental tendencies described by the second law of thermodynamics.
Thermal Properties of Materials
The thermal properties of materials are crucial characteristics that determine how materials respond to changes in temperature. These properties include specific heat capacity, thermal conductivity, thermal expansion, and more. They dictate how a material stores heat, how quickly it can transfer heat, and how it physically changes dimensions with temperature variations.

In the exercise described, understanding the thermal properties of the egg helps us effectively calculate the heat transfer process. The egg's specific heat capacity, as mentioned earlier, as well as its density (\(\rho\)), affect how heat is absorbed. A higher density generally means there's more material to heat, thus requiring more energy. Materials with high specific heat and density typically warm up and cool down more slowly, which is why boiling an egg takes several minutes.

Moreover, when solving thermodynamic problems, it is useful to remember that thermal properties like \(c_p\) of materials are often temperature-dependent. In educational or simplified settings, they may be considered constant for a given range of temperatures to facilitate calculations. However, in more complex situations, these thermal properties vary with temperature, calling for more sophisticated computational methods or iterative procedures to accurately model the heat transfer.

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Most popular questions from this chapter

The compressed-air requirements of a plant are being met by a 100 -hp screw compressor that runs at full load during 40 percent of the time and idles the rest of the time during operating hours. The compressor consumes 35 percent of the rated power when idling and 90 percent of the power when compressing air. The annual operating hours of the facility are \(3800 \mathrm{h}\), and the unit cost of electricity is \(\$ 0.115 / \mathrm{kWh}\). It is determined that the compressed-air requirements of the facility during 60 percent of the time can be met by a 25 -hp reciprocating compressor that consumes 95 percent of the rated power when compressing air and no power when not compressing air. It is estimated that the 25 -hp compressor runs 85 percent of the time. The efficiencies of the motors of the large and the small compressors at or near full load are 0.90 and \(0.88,\) respectively. The efficiency of the large motor at 35 percent load is \(0.82 .\) Determine the amount of energy and money saved as a result of switching to the 25 -hp compressor during 60 percent of the time.

The compressors of a production facility maintain the compressed-air lines at a (gage) pressure of \(700 \mathrm{kPa}\) at \(1400-\mathrm{m}\) elevation, where the atmospheric pressure is \(85.6 \mathrm{kPa}\). The average temperature of air is \(15^{\circ} \mathrm{C}\) at the compressor inlet and \(25^{\circ} \mathrm{C}\) in the compressed-air lines. The facility operates \(4200 \mathrm{h} / \mathrm{yr},\) and the average price of electricity is \(\$ 0.12 / \mathrm{kWh}\). Taking the compressor efficiency to be 0.8 the motor efficiency to be \(0.93,\) and the discharge coefficient to be \(0.65,\) determine the energy and money saved per year by sealing a leak equivalent to a 3 -mm-diameter hole on the compressed-air line.

Steam expands in an adiabatic turbine from \(4 \mathrm{MPa}\) and \(500^{\circ} \mathrm{C}\) to \(0.1 \mathrm{MPa}\) at a rate of \(2 \mathrm{kg} / \mathrm{s}\). If steam leaves the turbine as saturated vapor, the power output of the turbine is \((a) 2058 \mathrm{kW}\) \((b) 1910 \mathrm{kW}\) \((c) 1780 \mathrm{kW}\) \((d) 1674 \mathrm{kW}\) \((e) 1542 \mathrm{kW}\)

The inner and outer glasses of a \(2-\mathrm{m} \times 2\) -m double-pane window are at \(18^{\circ} \mathrm{C}\) and \(6^{\circ} \mathrm{C}\), respectively. If the glasses are very nearly isothermal and the rate of heat transfer through the window is \(110 \mathrm{W}\), determine the rates of entropy transfer through both sides of the window and the rate of entropy generation within the window, in \(\mathrm{W} / \mathrm{K}\).

Air enters a two-stage compressor at \(100 \mathrm{kPa}\) and \(27^{\circ} \mathrm{C}\) and is compressed to 625 kPa. The pressure ratio across each stage is the same, and the air is cooled to the initial temperature between the two stages. Assuming the compression process to be isentropic, determine the power input to the compressor for a mass flow rate of \(0.15 \mathrm{kg} / \mathrm{s}\). What would your answer be if only one stage of compression were used?

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