A potato may be approximated as a 5.7-cm-diameter solid sphere with the properties \(\rho=910 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4.25 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), \(k=0.68 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=1.76 \times 10^{-1} \mathrm{~m}^{2} / \mathrm{s}\). Twelve such potatoes initially at \(25^{\circ} \mathrm{C}\) are to be cooked by placing them in an oven maintained at \(250^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(95 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The amount of heat transfer to the potatoes during a 30-min period is (a) \(77 \mathrm{~kJ}\) (b) \(483 \mathrm{~kJ}\) (c) \(927 \mathrm{~kJ}\) (d) \(970 \mathrm{~kJ}\) (e) \(1012 \mathrm{~kJ}\)

Short Answer

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Question: Calculate the total heat transfer to the 12 potatoes in the 30-minute period, given the following properties: - Diameter of potato: 5.7 cm - Potato density: 910 kg/m³ - Heat transfer coefficient: 95 W/m²·K - Oven temperature: 250 K - Initial potato temperature: 25 K Answer: To calculate the total heat transfer, follow these steps: 1. Calculate the radius, surface area, and volume of the approximated potato sphere. 2. Calculate the mass of each potato. 3. Use Newton's law of cooling to calculate the heat transfer per potato. 4. Calculate the total heat transfer to all 12 potatoes in the 30-minute period. 5. Convert the total heat transfer from watts to kilojoules.

Step by step solution

01

Calculate the Potato Surface Area and Volume

First, we need to find the surface area and volume of the approximated potato sphere. The diameter of the potato is given as 5.7 cm, meaning its radius is: \(r = \frac{d}{2} = \frac{5.7 \ \text{cm}}{2} = 2.85 \ \text{cm} = 0.0285 \ \text{m}\). Now calculate the surface area (S) and the volume (V) of the sphere: \(S = 4 \pi r^2\) and \(V = \frac{4}{3} \pi r^3\).
02

Calculate Potato Mass

Given the volume and density of the potato, we can calculate its mass (m) using the following equation: \(m = \rho V\), where \(\rho = 910 \ \text{kg/m}^3\) is the potato density.
03

Calculate Heat Transfer Per Potato

Next, we will use Newton's law of cooling to find the heat transfer (Q) per potato in the 30-minute period. The formula is: \(Q = hS \Delta T\). In this case, the heat transfer coefficient (h) is 95 W/m²·K, and the temperature difference (\(\Delta T\)) is: \(\Delta T = T_\text{oven} - T_\text{initial} = 250 - 25 = 225 \ \text{K}\).
04

Calculate Total Heat Transfer

To find the total heat transfer to all 12 potatoes in the 30-minute period, we should multiply the heat transfer per potato (Q) calculated in Step 3 by the total number of potatoes. Then, multiply by the duration in seconds: \(Q_\text{total} = 12Q \times 30 \ \text{min} \times 60 \ \text{s/min}\).
05

Convert Heat Transfer to kJ

The result obtained in Step 4 is in watts (W). To convert the total heat transfer to kilojoules (kJ), multiply the value by the number of seconds in the 30-minute period and divide by 1000: \(Q_\text{total(kJ)} = \frac{Q_\text{total(W)} \times 30 \ \text{min} \times 60 \ \text{s/min}}{1000}\). The final value corresponds to the closest option among the given choices for the total heat transfer.

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Key Concepts

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

Newton's Law of Cooling
Understanding Newton's law of cooling is crucial when analyzing heat transfer situations like the cooking of potatoes in an oven. It describes the rate at which an object cools down or heats up, based on the temperature difference between the object and its surrounding environment. The law can be expressed with the formula:
case {Q = hS abla T},
where Q is the heat transfer, h is the heat transfer coefficient, S is the surface area of the object, and abla T is the temperature difference between the object and surrounding environment.
In our potato example, the oven acts as the heat source with a higher temperature compared to the initial temperature of the potatoes. The heat transfer coefficient provided represents how effectively heat is transferred from the oven's air to the potatoes. Hence, by knowing the surrounding temperature (oven temperature), the object's surface area, the heat transfer coefficient, and the initial temperature, we can use Newton's law to calculate the rate of heat transfer to the potatoes.
Thermal Conductivity
Thermal conductivity, represented by the symbol , is a measure of a material's ability to conduct heat. It is an intrinsic property that indicates how quickly heat is transferred through a material via conduction when there is a temperature difference. The higher the thermal conductivity, the more efficient the material is at transferring heat.
While thermal conductivity is not directly used in the calculation for Newton’s law of cooling, it is fundamentally related to the property , the thermal diffusivity, which measures how quickly a material adjusts its temperature to the surrounding environment's temperature. In the potato scenario, thermal conductivity is crucial for understanding how effectively heat penetrates the potato once the surface temperature increases due to the heat transfer from the oven environment.
Sphere Surface Area and Volume Calculation
Calculating the surface area and volume of a sphere is an essential skill in various scientific and engineering fields, including cooking and heat transfer studies. The equations for a sphere's surface area () and volume () are derived from its radius (). The surface area is given by the formula:
n = , where is the radius of the sphere.
The volume is given by:
n = , where is the radius.
In the context of the potato, approximation of its shape as a sphere allowed us to apply these equations to calculate the surface area, S, which is then used in the heat transfer equation, as well as the volume, V, which is essential for determining the mass of the potato when multiplied by the density (). These fundamental geometric calculations are key components for finding the heat transfer in such scenarios.

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Most popular questions from this chapter

What is the physical significance of the Fourier number? Will the Fourier number for a specified heat transfer problem double when the time is doubled?

During a fire, the trunks of some dry oak trees (es) \(\left(k=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\) and \(\left.\alpha=1.28 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) that are initially at a uniform temperature of \(30^{\circ} \mathrm{C}\) are exposed to hot gases at \(520^{\circ} \mathrm{C}\) for a period of \(5 \mathrm{~h}\), with a heat transfer coefficient of \(65 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the surface. The ignition temperature of the trees is \(410^{\circ} \mathrm{C}\). Treating the trunks of the trees as long cylindrical rods of diameter \(20 \mathrm{~cm}\), determine if these dry trees will ignite as the fire sweeps through them. Solve this problem using analytical one-term approximation method (not the Heisler charts).

A steel casting cools to 90 percent of the original temperature difference in \(30 \mathrm{~min}\) in still air. The time it takes to cool this same casting to 90 percent of the original temperature difference in a moving air stream whose convective heat transfer coefficient is 5 times that of still air is (a) \(3 \mathrm{~min}\) (b) \(6 \mathrm{~min}\) (c) \(9 \mathrm{~min}\) (d) \(12 \mathrm{~min}\) (e) \(15 \mathrm{~min}\)

A long iron \(\operatorname{rod}\left(\rho=7870 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), \(k=80.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\left.\alpha=23.1 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) with diameter of \(25 \mathrm{~mm}\) is initially heated to a uniform temperature of \(700^{\circ} \mathrm{C}\). The iron rod is then quenched in a large water bath that is maintained at constant temperature of \(50^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(128 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the time required for the iron rod surface temperature to cool to \(200^{\circ} \mathrm{C}\). Solve this problem using analytical one- term approximation method (not the Heisler charts).

The chilling room of a meat plant is \(15 \mathrm{~m} \times 18 \mathrm{~m} \times\) \(5.5 \mathrm{~m}\) in size and has a capacity of 350 beef carcasses. The power consumed by the fans and the lights in the chilling room are 22 and \(2 \mathrm{~kW}\), respectively, and the room gains heat through its envelope at a rate of \(14 \mathrm{~kW}\). The average mass of beef carcasses is \(220 \mathrm{~kg}\). The carcasses enter the chilling room at \(35^{\circ} \mathrm{C}\), after they are washed to facilitate evaporative cooling, and are cooled to \(16^{\circ} \mathrm{C}\) in \(12 \mathrm{~h}\). The air enters the chilling room at \(-2.2^{\circ} \mathrm{C}\) and leaves at \(0.5^{\circ} \mathrm{C}\). Determine \((a)\) the refrigeration load of the chilling room and \((b)\) the volume flow rate of air. The average specific heats of beef carcasses and air are \(3.14\) and \(1.0 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), respectively, and the density of air can be taken to be \(1.28 \mathrm{~kg} / \mathrm{m}^{3}\).

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