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Consider a 7.6-cm-long and 3-cm-diameter cylindrical lamb meat chunk \(\left(\rho=1030 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.49 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right.\), \(\left.k=0.456 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\). Fifteen such meat chunks initially at \(2^{\circ} \mathrm{C}\) are dropped into boiling water at \(95^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The amount of heat transfer during the first 8 minutes of cooking is (a) \(71 \mathrm{~kJ}\) (b) \(227 \mathrm{~kJ}\) (c) \(238 \mathrm{~kJ}\) \(\begin{array}{ll}\text { (d) } 269 \mathrm{~kJ} & \text { (e) } 307 \mathrm{~kJ}\end{array}\)

Short Answer

Expert verified
Answer: (c) 238 kJ

Step by step solution

01

Understand the formula for heat transfer

The formula for the heat transfer (Q) between the meatchunk and boiling water is given by the equation: Q = h * A * ΔT * t where: h = heat transfer coefficient A = surface area ΔT = Temperature difference t = time. We will calculate the parameters based on the given information.
02

Find the surface area of the meat chunk

The meat is cylindrical in shape. The formula for surface area (A) of a cylinder can be written as: A = 2 * π * r * (r + l) Given: length(l) = 7.6 cm, diameter = 3 cm which means radius(r) = 1.5 cm. Let's find the surface area (A): A = 2 * π * 1.5 * (1.5 + 7.6) = 2 * π * 1.5 * 9.1 = 27.3π cm² Since we want the surface area in meters, we convert the cm² to m²: A = 27.3π × (0.01)² = 0.0273π m²
03

Find the temperature difference ΔT between the meat chunks and boiling water

Given: initial temperature of meat = 2°C, temperature of the boiling water = 95°C Temperature Difference (ΔT) = Temperature of the boiling water - initial temperature of meat ΔT = 95 - 2 = 93°C
04

Find the time in seconds

Given: time t = 8 minutes. We need to convert the time to seconds. t = 8 × 60 = 480 seconds
05

Calculate the heat transfer Q for one meat chunk

Now we have all the parameters to calculate the heat transfer (Q) for one meat chunk: h = 1200 W/m²K A = 0.0273π m² ΔT = 93°C t = 480 seconds Q = h * A * ΔT * t Q = 1200 * 0.0273π *93 * 480 Q = 1200 * 0.0273π * 93 *480 ≈ 15.872 kJ.
06

Calculate the heat transfer Q for all 15 meat chunks

Since there are 15 meat chunks dropped into the boiling water, we need to multiply the heat transfer for one chunk by 15 to get the total heat transfer: Q(total) = Q * 15 Q(total) = 15.872 * 15 ≈ 238.08 kJ Now we will compare the obtained value with the given options. a) 71 kJ b) 227 kJ c) 238 kJ d) 269 kJ e) 307 kJ The closest option to our calculated value of heat transfer is (c) 238 kJ. So, the heat transfer during the first 8 minutes of cooking is approximately 238 kJ.

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Key Concepts

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

Cylindrical Object
In this exercise, we are dealing with a cylindrical object, which is an important shape in various real-world applications and scientific problems. Cylindrical objects have straight sides and circular bases and tops, which makes their physical properties interesting and often more complex to calculate than simple flat surfaces. The cylindrical shape is characterized primarily by its radius and height (or length). In our specific problem, we have a chunk of meat shaped like a cylinder with a radius of 1.5 cm and a length of 7.6 cm. Understanding these dimensions is crucial because they impact calculations like surface area and volume, both of which are key to determining how substances like heat transfer through the object. This basic grasp helps by also figuring out how other phenomena like pressure and force distribute across a cylindrical object.
Surface Area Calculation
The calculation of the surface area of a cylindrical object plays a vital role in determining how much of an object is exposed to the outer environment. For a cylinder, the surface area includes the circular top and bottom as well as the curved surface that connects them. The formula to calculate the surface area (A) of a cylinder is given by:
  • \[ A = 2\pi r (r + l) \]
  • where \(r\) is the radius and \(l\) is the length of the cylinder
For our cylindrical meat chunk, given the radius as 1.5 cm and length as 7.6 cm, substituting in for the formula results in a surface area calculation of 27.3π cm². However, calculations often require units to be consistent; hence converting cm² to m² (the unit required for our heat transfer formula) gives us 0.0273π m². Calculating accurate surface areas is essential because it affects how much heat or other types of transfer can happen between the object and its surroundings.
Temperature Difference
Temperature difference, denoted as \(\Delta T\), is a crucial element in calculating heat transfer. It determines the potential for energy transfer - specifically, heat - from one body to another. In our scenario, we have a meat chunk initially at 2°C being submerged into boiling water at 95°C. The temperature difference drives the heat transfer process, calculated as:
  • \(\Delta T = T_{\text{boiling water}} - T_{\text{initial meat}}\)
  • which simplifies to \(\Delta T = 95 - 2 = 93°C\)
The larger the temperature difference, the more intense the rate of heat transfer, assuming all other conditions such as surface area and heat transfer coefficient remain constant. This understanding is crucial when looking into systems where thermal regulation is vital, like cooking, refrigeration, or even in industrial processes.
Time Conversion
Time conversion is a fundamental step often required in physics and engineering calculations because the standard unit for time in these contexts is seconds. Our exercise initially provides the cooking time as 8 minutes. To use time in mathematical formulas, especially those involving rates like heat transfer, we need to convert this time into seconds:
  • \(t = 8 \, \text{minutes} \times 60 \, \frac{\text{seconds}}{\text{minute}} = 480 \, \text{seconds}\)
This conversion ensures that calculations remain consistent and accurate. When using formulas that involve time, ensuring the correct unit is essential to obtaining the correct result. This step highlights the importance of attention to detail in scientific calculations, as even minor mistakes in unit conversions can lead to significant errors in outcomes.

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Most popular questions from this chapter

A man is found dead in a room at \(16^{\circ} \mathrm{C}\). The surface temperature on his waist is measured to be \(23^{\circ} \mathrm{C}\) and the heat transfer coefficient is estimated to be \(9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Modeling the body as \(28-\mathrm{cm}\) diameter, \(1.80\)-m-long cylinder, estimate how long it has been since he died. Take the properties of the body to be \(k=0.62 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=0.15 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\), and assume the initial temperature of the body to be \(36^{\circ} \mathrm{C}\).

Consider a sphere and a cylinder of equal volume made of copper. Both the sphere and the cylinder are initially at the same temperature and are exposed to convection in the same environment. Which do you think will cool faster, the cylinder or the sphere? Why?

Consider a curing kiln whose walls are made of \(30-\mathrm{cm}-\) thick concrete with a thermal diffusivity of \(\alpha=0.23 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). Initially, the kiln and its walls are in equilibrium with the surroundings at \(6^{\circ} \mathrm{C}\). Then all the doors are closed and the kiln is heated by steam so that the temperature of the inner surface of the walls is raised to \(42^{\circ} \mathrm{C}\) and the temperature is maintained at that level for \(2.5 \mathrm{~h}\). The curing kiln is then opened and exposed to the atmospheric air after the steam flow is turned off. If the outer surfaces of the walls of the kiln were insulated, would it save any energy that day during the period the kiln was used for curing for \(2.5 \mathrm{~h}\) only, or would it make no difference? Base your answer on calculations.

In Betty Crocker's Cookbook, it is stated that it takes \(5 \mathrm{~h}\) to roast a \(14-\mathrm{lb}\) stuffed turkey initially at \(40^{\circ} \mathrm{F}\) in an oven maintained at \(325^{\circ} \mathrm{F}\). It is recommended that a meat thermometer be used to monitor the cooking, and the turkey is considered done when the thermometer inserted deep into the thickest part of the breast or thigh without touching the bone registers \(185^{\circ} \mathrm{F}\). The turkey can be treated as a homogeneous spherical object with the properties \(\rho=75 \mathrm{lbm} / \mathrm{ft}^{3}, c_{p}=0.98 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\), \(k=0.26 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\), and \(\alpha=0.0035 \mathrm{ft}^{2} / \mathrm{h}\). Assuming the tip of the thermometer is at one- third radial distance from the center of the turkey, determine \((a)\) the average heat transfer coefficient at the surface of the turkey, \((b)\) the temperature of the skin of the turkey when it is done, and \((c)\) the total amount of heat transferred to the turkey in the oven. Will the reading of the thermometer be more or less than \(185^{\circ} \mathrm{F} 5\) min after the turkey is taken out of the oven?

Layers of 23 -cm-thick meat slabs \((k=0.47 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=0.13 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) initially at a uniform temperature of \(7^{\circ} \mathrm{C}\) are to be frozen by refrigerated air at \(-30^{\circ} \mathrm{C}\) flowing at a velocity of \(1.4 \mathrm{~m} / \mathrm{s}\). The average heat transfer coefficient between the meat and the air is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the size of the meat slabs to be large relative to their thickness, determine how long it will take for the center temperature of the slabs to drop to \(-18^{\circ} \mathrm{C}\). Also, determine the surface temperature of the meat slab at that time.

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