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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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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 – 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 6-cm-diameter 13-cm-high canned drink ( \(\rho=\) \(\left.977 \mathrm{~kg} / \mathrm{m}^{3}, k=0.607 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) initially at \(25^{\circ} \mathrm{C}\) is to be cooled to \(5^{\circ} \mathrm{C}\) by dropping it into iced water at \(0^{\circ} \mathrm{C}\). Total surface area and volume of the drink are \(A_{s}=\) \(301.6 \mathrm{~cm}^{2}\) and \(V=367.6 \mathrm{~cm}^{3}\). If the heat transfer coefficient is \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine how long it will take for the drink to \(\operatorname{cool}\) to \(5^{\circ} \mathrm{C}\). Assume the can is agitated in water and thus the temperature of the drink changes uniformly with time. (a) \(1.5 \mathrm{~min}\) (b) \(8.7 \mathrm{~min}\) (c) \(11.1 \mathrm{~min}\) (d) \(26.6 \mathrm{~min}\) (e) \(6.7 \mathrm{~min}\)

A body at an initial temperature of \(T_{i}\) is brought into a medium at a constant temperature of \(T_{\infty}\). How can you determine the maximum possible amount of heat transfer between the body and the surrounding medium?

A large heated steel block \(\left(\rho=7832 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=\right.\) \(434 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=63.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\left.\alpha=18.8 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) is allowed to cool in a room at \(25^{\circ} \mathrm{C}\). The steel block has an initial temperature of \(450^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming that the steel block can be treated as a quarter-infinite medium, determine the temperature at the edge of the steel block after 10 minutes of cooling.

An electronic device dissipating \(20 \mathrm{~W}\) has a mass of \(20 \mathrm{~g}\), a specific heat of \(850 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and a surface area of \(4 \mathrm{~cm}^{2}\). The device is lightly used, and it is on for \(5 \mathrm{~min}\) and then off for several hours, during which it cools to the ambient temperature of \(25^{\circ} \mathrm{C}\). Taking the heat transfer coefficient to be \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature of the device at the end of the 5 -min operating period. What would your answer be if the device were attached to an aluminum heat sink having a mass of \(200 \mathrm{~g}\) and a surface area of \(80 \mathrm{~cm}^{2}\) ? Assume the device and the heat sink to be nearly isothermal.

Under what conditions can a plane wall be treated as a semi-infinite medium?

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