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Chapter 4: Problem 8

Consider a hot baked potato on a plate. The temperature of the potato is observed to drop by \(4^{\circ} \mathrm{C}\) during the first minute. Will the temperature drop during the second minute be less than, equal to, or more than \(4^{\circ} \mathrm{C}\) ? Why?

Short Answer

Expert verified
Answer: The temperature drop during the second minute will be less than 4°C, as the rate of heat loss decreases due to the reduced temperature difference between the potato and surroundings, according to Newton's law of cooling.

Step by step solution

01

Recall Newton's law of cooling

Newton's law of cooling states that the rate of temperature change of an object is proportional to the difference in temperature between the object and its surroundings. Mathematically, this can be expressed as: \(\frac{dT}{dt} = k(T - T_{s})\) where \(\frac{dT}{dt}\) is the rate of temperature change, \(T\) is the temperature of the object, \(T_{s}\) is the temperature of the surroundings, and \(k\) is a proportionality constant that depends on the object's thermal properties and the heat transfer medium.
02

Analyze the temperature drop in the first minute

Since the baked potato's temperature drops by \(4^{\circ} \mathrm{C}\) in the first minute, it means that the initial rate of temperature change is quite high as the temperature difference between the potato and its surroundings is large.
03

Predict the temperature drop in the second minute

As time passes, the temperature of the potato decreases, bringing it closer to the surrounding temperature. This reduces the temperature difference, leading to a lower rate of temperature change as the potato loses less heat to the surroundings. According to Newton's law of cooling, this implies that the temperature drop during the second minute will be less than the temperature drop in the first minute since the rate of heat loss decreases over time.
04

Answer the question

The temperature drop during the second minute will be less than \(4^{\circ} \mathrm{C}\). This is because, according to Newton's law of cooling, the rate of heat loss decreases as the temperature difference between the potato and the surroundings decreases.

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Most popular questions from this chapter

A thermocouple, with a spherical junction diameter of \(0.5 \mathrm{~mm}\), is used for measuring the temperature of hot air flow in a circular duct. The convection heat transfer coefficient of the air flow can be related with the diameter \((D)\) of the duct and the average air flow velocity \((V)\) as \(h=2.2(V / D)^{0.5}\), where \(D, h\), and \(V\) are in \(\mathrm{m}, \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(\mathrm{m} / \mathrm{s}\), respectively. The properties of the thermocouple junction are \(k=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=8500 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=320 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Determine the minimum air flow velocity that the thermocouple can be used, if the maximum response time of the thermocouple to register 99 percent of the initial temperature difference is \(5 \mathrm{~s}\).

A potato that may be approximated as a \(5.7-\mathrm{cm}\) 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 \(\alpha=1.76 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\). 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 by the time the center temperature reaches \(100^{\circ} \mathrm{C}\) is (a) \(56 \mathrm{~kJ}\) (b) \(666 \mathrm{~kJ}\) (c) \(838 \mathrm{~kJ}\) (d) \(940 \mathrm{~kJ}\) (e) \(1088 \mathrm{~kJ}\)

A long 35-cm-diameter cylindrical shaft made of stainless steel \(304\left(k=14.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7900 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=477 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), and \(\alpha=3.95 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) comes out of an oven at a uniform temperature of \(400^{\circ} \mathrm{C}\). The shaft is then allowed to cool slowly in a chamber at \(150^{\circ} \mathrm{C}\) with an average convection heat transfer coefficient of \(h=60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperature at the center of the shaft \(20 \mathrm{~min}\) after the start of the cooling process. Also, determine the heat transfer per unit length of the shaft during this time period. Solve this problem using analytical one-term approximation method (not the Heisler charts).

Consider a sphere of diameter \(5 \mathrm{~cm}\), a cube of side length \(5 \mathrm{~cm}\), and a rectangular prism of dimension \(4 \mathrm{~cm} \times\) \(5 \mathrm{~cm} \times 6 \mathrm{~cm}\), all initially at \(0^{\circ} \mathrm{C}\) and all made of silver \((k=\) \(\left.429 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=10,500 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=0.235 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\). Now all three of these geometries are exposed to ambient air at \(33^{\circ} \mathrm{C}\) on all of their surfaces with a heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine how long it will take for the temperature of each geometry to rise to \(25^{\circ} \mathrm{C}\).

In Betty Crocker's Cookbook, it is stated that it takes \(2 \mathrm{~h} \mathrm{} 45 \mathrm{~min}\) to roast a \(3.2-\mathrm{kg}\) rib initially at \(4.5^{\circ} \mathrm{C}\) "rare" in an oven maintained at \(163^{\circ} \mathrm{C}\). It is recommended that a meat thermometer be used to monitor the cooking, and the rib is considered rare done when the thermometer inserted into the center of the thickest part of the meat registers \(60^{\circ} \mathrm{C}\). The rib can be treated as a homogeneous spherical object with the properties \(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4.1 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=\) \(0.91 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\). Determine \((a)\) the heat transfer coefficient at the surface of the rib; \((b)\) the temperature of the outer surface of the rib when it is done; and \((c)\) the amount of heat transferred to the rib. \((d)\) Using the values obtained, predict how long it will take to roast this rib to "medium" level, which occurs when the innermost temperature of the rib reaches \(71^{\circ} \mathrm{C}\). Compare your result to the listed value of \(3 \mathrm{~h} \mathrm{} 20 \mathrm{~min}\). If the roast rib is to be set on the counter for about \(15 \mathrm{~min}\) before it is sliced, it is recommended that the rib be taken out of the oven when the thermometer registers about \(4^{\circ} \mathrm{C}\) below the indicated value because the rib will continue cooking even after it is taken out of the oven. Do you agree with this recommendation? Solve this problem using analytical one-term approximation method (not the Heisler charts).
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