/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 79 To determine which parts of a sp... [FREE SOLUTION] | 91Ó°ÊÓ

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To determine which parts of a spider's brain are triggered into neural activity in response to various optical stimuli, researchers at the University of Massachusetts Amherst desire to examine the brain as it is shown images that might evoke emotions such as fear or hunger. Consider a spider at \(T_{i}=20^{\circ} \mathrm{C}\) that is shown a frightful scene and is then immediately immersed in liquid nitrogen at \(T_{\infty}=77 \mathrm{~K}\). The brain is subsequently dissected in its frozen state and analyzed to determine which parts of the brain reacted to the stimulus. Using your knowledge of heat transfer, determine how much time elapses before the spider's brain begins to freeze. Assume the brain is a sphere of diameter \(D_{b}=1 \mathrm{~mm}\), centrally located in the spider's cephalothorax, which may be approximated as a spherical shell of diameter \(D_{c}=3 \mathrm{~mm}\). The brain and cephalothorax properties correspond to those of liquid water. Neglect the effects of the latent heat of fusion and assume the heat transfer coefficient is \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Short Answer

Expert verified
The time it takes for the spider's brain to start freezing when immersed in liquid nitrogen at \(77\,K\) is approximately \(17.49\) seconds, using the Lumped Capacitance method.

Step by step solution

01

Calculate the Biot Number for the brain

First, we need to find the Biot number (Bi) to check if the Lumped Capacitance method is applicable. The Biot number is given by: \(Bi = \frac{hL_c}{k}\) Where \(L_c\) represents the characteristic length and is equal to \(L_c = \frac{V_{brain}}{A_{brain}}\). For a sphere, \(V_{brain} = \frac{4}{3}\pi \left(\frac{D_{b}}{2}\right)^3\) and \(A_{brain} = 4\pi \left(\frac{D_{b}}{2}\right)^2\). Therefore, the Biot Number formula becomes: \[ Bi= \frac{h}{6k} \cdot D_{b}\] We need to use the thermal conductivity of liquid water (\(k=0.6\;W/m\cdot K\)) and the given heat transfer coefficient (\(h=100\;W/m^2\cdot K\)) to find the Biot number: \(Bi = \frac{100}{6(0.6)} \cdot 0.001 \)
02

Verify if the Lumped Capacitance method is applicable

If the Biot number is less than 0.1, we can apply the Lumped Capacitance method. Calculate the Biot number value and check the condition: \(Bi = 0.0278\) Since the Biot number is less than 0.1, we can use the Lumped Capacitance method for this problem.
03

Apply the Lumped Capacitance method

Using the Lumped Capacitance method, we can find the time required for the brain to start freezing. The equation is: \(t = \frac{C \rho V(T_i - T_\infty)}{hA(T_i - T_\infty)}\) where \(C\) is the specific heat capacity, \(\rho\) is the density, and \(V\) is the volume of the brain. Using the properties of liquid water, we have \(C = 4.18\times10^3\,J/kg\cdot K\), \(\rho = 1000\,kg/m^3\), and given values \(T_i = 20^{\circ} C\) and \(T_\infty = 77\,K\). The volume of the brain is given by: \(V_{brain} = \frac{4}{3}\pi \left(\frac{D_{b}}{2}\right)^3\) The surface area of the brain is given by: \(A_{brain} = 4\pi \left(\frac{D_{b}}{2}\right)^2\) Calculate the time, t required for the brain to start freezing: \(t = \frac{(4.18\times10^3)(1000)(\frac{4}{3}\pi \left(\frac{0.001}{2}\right)^3)(293 - 77)}{100(4\pi \left(\frac{0.001}{2}\right)^2)(293 - 77)}\)
04

Calculate the time before the brain starts freezing

After solving the equation, we will get the required time: \(t= 17.49 \, sec\) So it takes 17.49 seconds for the spider's brain to start freezing when it is immersed in liquid nitrogen at \(77\,K\).

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Key Concepts

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

Understanding Biot Number in Heat Transfer
The Biot Number (Bi) is a dimensionless quantity that plays a critical role in heat transfer analysis, particularly when determining the appropriate model to use for predicting temperature changes in an object. It is defined as the ratio of the heat transfer resistances inside of an object to the heat transfer resistance at the surface of the object. Mathematically, the Biot Number is expressed as:
\[\begin{equation}Bi = \frac{hL_c}{k}\end{equation}\]
where:
  • (L_c) represents the characteristic length of the object,
  • (h) is the convective heat transfer coefficient, and
  • (k) is the thermal conductivity of the material.

In the exercise, the characteristic length for a spherical object like the spider's brain is derived from its volume and surface area. It's crucial to note that the Biot Number helps in deciding if the temperature distribution within the object can be assumed uniform. Specifically, if Bi is less than 0.1, it indicates that the temperature gradient within the object can be neglected, permitting the use of the Lumped Capacitance Method for simpler calculations. Understanding Bi is essential for correctly modeling the heat transfer process and for making predictions about the thermal behavior of materials and objects.
The Lumped Capacitance Method
The Lumped Capacitance Method is a simplified approach to solving heat transfer problems where the temperature gradient within the object is small enough to be considered negligible. This approach is valid when the Biot Number is less than 0.1, as stated earlier. With this method, the object's temperature is assumed to be uniform at any given time, which greatly simplifies the calculations of cooling or heating times.

The method uses the following formula to calculate the time (t) required for the temperature of the object to change from an initial temperature (T_i) to the surrounding temperature (T_{\text{infinity}}):\[\begin{equation} t = \frac{C \rho V(T_i - T_\infty)}{hA(T_i - T_\infty)} \end{equation}\]where
    \t
  • \t(C\t) is the specific heat capacity, \t
  • \t(\rho\t) is the density, \t
  • \t(V\t) is the volume of the object, and \t
  • \t(A\t) is the surface area.

Using the Lumped Capacitance Method allows for quick estimates of the response time of objects to changes in their thermal environment, proving immensely useful in many engineering applications, including the analysis of quick thermal stimuli responses in biological studies, as outlined in the given exercise.
Thermal Conductivity and Its Impact on Heat Transfer
Thermal conductivity ((k)) is a fundamental property of materials that measures their ability to conduct heat. It appears in several heat transfer equations, including the calculation of the Biot Number and when solving problems using the Lumped Capacitance Method. The higher the thermal conductivity of a material, the more efficiently it can transfer heat.

In the context of the exercise, the thermal conductivity of liquid water is a vital factor because the spider’s brain and cephalothorax properties correspond to those of liquid water. The overall rate of heat transfer from the spider's brain to the surrounding liquid nitrogen depends heavily on this property.

In practical applications, understanding the thermal conductivity of materials is crucial for designing efficient cooling systems, thermal insulations, and various other thermal management solutions. It affects how quickly an object can reach thermal equilibrium with its environment and also influences the choice of materials for thermal applications. Naturally, in biological systems, the thermal properties mimic those of water, making water a standard reference point for biological heat transfer analysis.

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Most popular questions from this chapter

A constant-property, one-dimensional plane slab of width \(2 L\), initially at a uniform temperature, is heated convectively with \(B i=1\). (a) At a dimensionless time of \(F o_{1}\), heating is suddenly stopped, and the slab of material is quickly covered with insulation. Sketch the dimensionless surface and midplane temperatures of the slab as a function of dimensionless time over the range \(0

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In a manufacturing process, long rods of different diameters are at a uniform temperature of \(400^{\circ} \mathrm{C}\) in a curing oven, from which they are removed and cooled by forced convection in air at \(25^{\circ} \mathrm{C}\). One of the line operators has observed that it takes \(280 \mathrm{~s}\) for a \(40-\mathrm{mm}\) diameter rod to cool to a safe-to-handle temperature of \(60^{\circ} \mathrm{C}\). For an equivalent convection coefficient, how long will it take for an 80 -mm-diameter rod to cool to the same temperature? The thermophysical properties of the rod are \(\rho=2500 \mathrm{~kg} / \mathrm{m}^{3}, c=900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Comment on your result. Did you anticipate this outcome?

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