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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: Less than 4°C.

Step by step solution

01

Newton's Law of Cooling states that the rate of heat transfer (or temperature change) is proportional to the temperature difference between the body and its surroundings. Mathematically it can be written as \(\frac{dT}{dt}=-k(T-T_{s})\), where \(dT\) is the rate of temperature change, \(dt\) is time, \(T\) is the temperature of the body (in this case, the potato), \(T_{s}\) is the temperature of the surroundings, and \(k\) is a constant of proportionality. #Step 2: Analyzing the first minute#

At the beginning, let's assume that the temperature of the potato was \(T_1\) and the temperature of the surroundings was constant at \(T_s\). After the first minute, the temperature of the potato dropped by \(4^{\circ}\mathrm{C}\). Therefore, the new temperature of the potato is \((T_1 - 4^{\circ}\mathrm{C})\). #Step 3: Temperature difference during the first and second minute#
02

During the first minute, the temperature difference between the potato and its surroundings was \((T_1 - T_s)\). During the second minute, the temperature difference is now reduced to \([(T_1 - 4^{\circ}\mathrm{C}) - T_s]\). #Step 4: Comparing temperature drop#

As per Newton's law of cooling, the rate of heat transfer is directly proportional to the temperature difference. Since the temperature difference during the second minute is smaller than during the first minute, the rate of temperature decrease will be lower during the second minute, and therefore, the temperature drop during the second minute will be less than 4°C.

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Most popular questions from this chapter

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.\(, \)k=0.456 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}$ ). 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 \(8 \mathrm{~min}\) of cooking is (a) \(71 \mathrm{~kJ}\) (b) \(227 \mathrm{~kJ}\) (c) \(238 \mathrm{~kJ}\) (d) \(269 \mathrm{~kJ}\) (e) \(307 \mathrm{~kJ}\)

A small chicken $(k=0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \quad \alpha=0.15 \times\( \)10^{-6} \mathrm{~m}^{2} / \mathrm{s}$ ) can be approximated as an \(11.25-\mathrm{cm}\)-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) \(168^{\circ} \mathrm{C}\)

Consider the engine block of a car made of cast iron $\left(k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\( and \)\left.\alpha=1.7 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)$. The engine can be considered to be a rectangular block whose sides are \(80 \mathrm{~cm}\), \(40 \mathrm{~cm}\), and \(40 \mathrm{~cm}\). The engine is at a temperature of \(150^{\circ} \mathrm{C}\) when it is turned off. The engine is then exposed to atmospheric air at \(17^{\circ} \mathrm{C}\) with a heat transfer coefficient of $6 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\(. Determine \)(a)$ the center temperature of the top surface whose sides are \(80 \mathrm{~cm}\) and \(40 \mathrm{~cm}\) and \((b)\) the corner temperature after 45 min of cooling. Solve this problem using the analytical one-term approximation method.

A thermocouple with a spherical junction diameter of \(0.5 \mathrm{~mm}\) is used for measuring the temperature of hot airflow in a circular duct. The convection heat transfer coefficient of the airflow can be related with the diameter \((D)\) of the spherical junction and the average airflow 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 airflow velocity that the thermocouple can use, if the maximum response time of the thermocouple to register 99 percent of the initial temperature difference is \(5 \mathrm{~s}\).

Consider a 7.6-cm-diameter cylindrical lamb meat chunk $\left(\rho=1030 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.49 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.456 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.$, \(\alpha=1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) ). Such a meat chunk intially at \(2^{\circ} \mathrm{C}\) is 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 time it takes for the center temperature of the meat chunk to rise to \(75^{\circ} \mathrm{C}\) is (a) \(136 \mathrm{~min}\) (b) \(21.2 \mathrm{~min}\) (c) \(13.6 \mathrm{~min}\) (d) \(11.0 \mathrm{~min}\) (e) \(8.5 \mathrm{~min}\)
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