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Consider two identical 4-kg pieces of roast beef. The first piece is baked as a whole, while the second is baked after being cut into two equal pieces in the same oven. Will there be any difference between the cooking times of the whole and cut roasts? Why?

Short Answer

Expert verified
Answer: Cutting a roast into two equal pieces before baking increases the surface area to volume ratio, which results in faster cooking time for the cut roasts compared to the whole roast.

Step by step solution

01

Define the given variables and identify the formula

We are given two identical roast beefs, each with a mass of 4 kg. Let's assume that the roast beefs are of a cylindrical shape (although the results will be similar for any shape). The cooking time depends on the surface area to volume ratio (SA/V) of the roast. We need to calculate this ratio for both cases (whole and cut) and then compare them.
02

Calculate the surface area to volume ratio for the whole roast

Assuming the roast beef is cylindrical, we can use the formula for the surface area (SA) of a cylinder: SA = 2πrh + 2πr^2, where r is the radius and h is the height (or thickness). And for the volume (V), we have V = πr^2h. Now, we can calculate the ratio for the whole roast: SA/V = (2πrh + 2πr^2) / (πr^2h).
03

Calculate the surface area to volume ratio for the cut roasts

After cutting the roast beef into two equal pieces, each piece would have half the height of the original roast. Since they are equal, we can just consider one of them for the calculations. The surface area of each cut roast would now be SA' = 2πr(h/2) + 2πr^2 + πr^2. The volume would be V' = πr^2(h/2). Now, we can calculate the ratio for one-cut roast: SA'/V' = (2πr(h/2) + 2πr^2 + πr^2) / (πr^2(h/2)).
04

Compare the surface area to volume ratios

Now, we will compare the SA/V ratios between the whole roast and one of the cut roasts. SA/V = (2πrh + 2πr^2) / (πr^2h) and SA'/V' = (2πr(h/2) + 2πr^2 + πr^2) / (πr^2(h/2)). If the ratio is higher for one of the cut roasts, it means that the cooking time would be shorter for the cut roasts. Calculating and comparing these ratios would reveal that the SA'/V' ratio for the cut roast is higher than the SA/V ratio for the whole roast. That means, cut roasts cook faster because they have a higher surface area per unit of volume, allowing heat to penetrate the roast more quickly. So, there will be a difference in the cooking times of the whole and cut roasts, with the cut roasts cooking faster due to their higher surface area to volume ratio.

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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 cooking, as it describes how heat energy moves from the oven's hot environment to the food being cooked. There are three primary modes of heat transfer: conduction, convection, and radiation. In the context of cooking roast beef, all three can play a role.

Conduction occurs when heat moves through a substance, such as meat, from the hotter exterior to the cooler interior. The larger the surface area in contact with the heat source, the more efficiently heat can be conducted. Convection involves the movement of hot air around the food, promoting an even cooking temperature. Lastly, radiation refers to the transfer of heat through electromagnetic waves, such as the infrared heat from an oven's walls. The surface area to volume ratio directly affects how quickly heat is transferred to the roast: a higher ratio means more surface area is exposed to these heat transfer mechanisms, speeding up the cooking process.

Surface Area and Heat Absorption

A higher surface area to volume ratio allows the roast to absorb heat more rapidly because more of the roast's surface is in direct contact with the hot air and surfaces of the oven. This is precisely why cutting the roast into pieces can decrease cooking time.
Cooking Time Calculation
Calculating cooking time is a practical application of thermodynamics that concerns not just chefs, but anyone looking to prepare a delicious meal efficiently. Understanding the surface area to volume ratio (SA/V) is key when determining cooking times. The concept is simple: the greater the ratio, the less time it takes for the food to cook. This happens because heat has less volume to penetrate, and there's more exposed surface to absorb heat.

When cooking, understanding the geometry of the food is important. For instance, slicing the roast beef increases its surface area while keeping the volume constant, which allows heat to reach the interior more quickly. Consequently, thinner or smaller pieces with a high SA/V ratio will cook faster than a larger, intact piece of the same food.

Practical Applications in the Kitchen

By cutting the roast, a cook effectively alters the food's geometry, ensuring that the interior reaches the desired temperature faster. When planning meal preparation, considering the SA/V ratio can help in making adjustments to the cooking time to ensure the food is properly cooked without being overdone. It's a blend of science and culinary art that, when mastered, can vastly improve the quality and efficiency of cooking.
Thermodynamics of Cooking
The thermodynamics of cooking revolves around the principles governing energy transfer and its conversion from one form to another. During cooking, the goal is to transfer heat energy from the heat source into the food. Key to this process is the concept of specific heat capacity, or the amount of heat required to raise the temperature of a substance by one degree Celsius. Different foods have different specific heat capacities, affecting how they absorb heat.

The rate at which food absorbs heat (and therefore cooks) is influenced by its thermal conductivity and surface area to volume ratio. A high thermal conductivity means heat spreads throughout the food faster, whereas a higher SA/V ratio means the food is less resistant to heat penetration.

Optimizing Cooking with Thermodynamic Principles

By applying thermodynamic principles, cooks can optimize the cooking process. For example, understanding that cutting the roast increases its SA/V ratio informs them that it will cook more quickly, potentially saving time and energy. It also highlights the importance of even cutting to ensure uniform cooking. In essence, the thermodynamics of cooking offers a scientific approach to perfecting culinary techniques.

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Most popular questions from this chapter

A small chicken \(\left(k=0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=0.15 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) can be approximated as an \(11.25\)-cm-diameter solid sphere. The chicken is initially at a uniform temperature of \(8^{\circ} \mathrm{C}\) and is to be cooked in an oven maintained at \(220^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). With this idealization, the temperature at the center of the chicken after a 90 -min period is (a) \(25^{\circ} \mathrm{C}\) (b) \(61^{\circ} \mathrm{C}\) (c) \(89^{\circ} \mathrm{C}\) (d) \(122^{\circ} \mathrm{C}\) (e) \(168^{\circ} \mathrm{C}\)

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 \(1.28 \mathrm{~kg} / \mathrm{m}^{3}\).

Consider a hot semi-infinite solid at an initial temperature of \(T_{i}\) that is exposed to convection to a cooler medium at a constant temperature of \(T_{\infty}\), with a heat transfer coefficient of \(h\). Explain how you can determine the total amount of heat transfer from the solid up to a specified time \(t_{o}\).

The walls of a furnace are made of \(1.2\)-ft-thick concrete \(\left(k=0.64 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=0.023 \mathrm{ft}^{2} / \mathrm{h}\right)\). Initially, the furnace and the surrounding air are in thermal equilibrium at \(70^{\circ} \mathrm{F}\). The furnace is then fired, and the inner surfaces of the furnace are subjected to hot gases at \(1800^{\circ} \mathrm{F}\) with a very large heat transfer coefficient. Determine how long it will take for the temperature of the outer surface of the furnace walls to rise to \(70.1^{\circ} \mathrm{F}\). Answer: \(116 \mathrm{~min}\)

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