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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掳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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Key Concepts

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

Heat Transfer
Understanding heat transfer is crucial when discussing Newton's Law of Cooling. Heat transfer refers to the movement of thermal energy from a hotter object to a cooler one. In our baked potato example, heat is moving from the hot potato to the cooler surrounding air. As heat leaves the potato, energy is lost, and the temperature drops.
  • Heat moves from the potato to the air because the potato is initially hotter than the air.
  • This transfer of energy continues until an equilibrium is reached where both objects are at the same temperature.
The speed and amount of heat transferred depend on several factors, including the temperature difference between the two objects. This is where Newton's Law of Cooling comes into play, offering a mathematical relationship to predict how temperatures equalize over time.
Temperature Change
Temperature change happens when an object loses or gains thermal energy. In the context of Newton's Law of Cooling, temperature change is specifically about how quickly the object's temperature (like our potato) moves towards the surrounding temperature. Initially, the temperature change is significant because of the large difference between the object and its environment.
  • The potato starts hot, leading to a rapid temperature drop since it quickly releases heat.
  • As the difference in temperature lessens, the rate at which this change occurs also reduces.
The key takeaway is that temperature change isn't constant 鈥� it's dynamic and decreases as the object approaches the surrounding temperature.
Cooling Rate
The cooling rate is all about how fast the temperature of an object decreases. According to Newton's Law of Cooling, the rate decreases as the temperature of the object approaches the surrounding temperature. Initially, in our potato example, the cooling rate is high due to the substantial temperature difference.
  • The bigger the difference in temperature, the faster the initial cooling rate.
  • Over time, as the potato's temperature comes closer to room temperature, the cooling rate slows down.
This explains why during the second minute of cooling, the potato's temperature diminishes by less than it did during the first minute.
Thermal Properties
Thermal properties are intrinsic characteristics of materials that dictate how they respond to changes in temperature. This includes aspects like thermal conductivity, specific heat capacity, and emissivity. These properties impact how effectively and quickly an object like a potato can transfer heat.
  • Thermal conductivity describes how well heat flows through a material.
  • Specific heat capacity indicates how much energy is needed to change the temperature of the material.
  • Emissivity reflects how well an object emits thermal radiation.
Such properties determine the proportionality constant, \( k \), in Newton's Law of Cooling, affecting how the cooling process unfolds.

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Most popular questions from this chapter

A person puts a few apples into the freezer at \(-15^{\circ} \mathrm{C}\) to cool them quickly for guests who are about to arrive. Initially, the apples are at a uniform temperature of \(20^{\circ} \mathrm{C}\), and the heat transfer coefficient on the surfaces is \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Treating the apples as 9 -cm-diameter spheres and taking their properties to be \(\rho=840 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.81 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.418 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\), determine the center and surface temperatures of the apples in \(1 \mathrm{~h}\). Also, determine the amount of heat transfer from each apple. Solve this problem using analytical one-term approximation method (not the Heisler charts).

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) \(168^{\circ} \mathrm{C}\)

In a production facility, 3-cm-thick large brass plates \(\left(k=110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=8530 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=380 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), and \(\left.\alpha=33.9 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) that are initially at a uniform temperature of \(25^{\circ} \mathrm{C}\) are heated by passing them through oven maintained at \(700^{\circ} \mathrm{C}\). The plates remain in the oven for a period of \(10 \mathrm{~min}\). Taking the convection heat transfer coefficient to be \(h=80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the surface temperature of the plates when they come out of the oven. Solve this problem using analytical one-term approximation method (not the Heisler charts). Can this problem be solved using lumped system analysis? Justify your answer.

Consider a cubic block whose sides are \(5 \mathrm{~cm}\) long and a cylindrical block whose height and diameter are also \(5 \mathrm{~cm}\). Both blocks are initially at \(20^{\circ} \mathrm{C}\) and are made of granite \((k=\) \(2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=1.15 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\). Now both blocks are exposed to hot gases at \(500^{\circ} \mathrm{C}\) in a furnace on all of their surfaces with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the center temperature of each geometry after 10,20 , and \(60 \mathrm{~min}\).

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) \(1012 \mathrm{~kJ}\)
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