/*! 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 2 A liquid-nitrogen container is m... [FREE SOLUTION] | 91Ó°ÊÓ

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A liquid-nitrogen container is made of a \(1-\mathrm{cm}\) thick styrofoam \(\quad\) sheet \(\quad\) having \(\quad\) thermal conductivity \(0025 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1_{0}} \mathrm{C}^{-1}\). Liquid nitrogen at \(80 \mathrm{~K}\) is kept in it. A total area of \(0.80 \mathrm{~m}^{2}\) is in contact with the liquid nitrogen. The atmospheric temperature is \(300 \mathrm{~K}\). Calculate the rate of heat flow from the atmosphere to the liquid nitrogen.

Short Answer

Expert verified
The rate of heat flow is 440 J/s.

Step by step solution

01

Identify the Given Variables

We are given:- Thickness of the styrofoam, \(d = 0.01 \; \text{m}\)- Thermal conductivity of the styrofoam, \(k = 0.025 \; \text{J/s}\cdot\text{m}\cdot\degree{C}\)- Area in contact with liquid nitrogen, \(A = 0.80 \; \text{m}^2\)- Temperature of liquid nitrogen, \(T_1 = 80 \; \text{K}\)- Atmospheric temperature, \(T_2 = 300 \; \text{K}\)
02

Temperature Difference Calculation

Calculate the temperature difference that drives the heat flow:\[\Delta T = T_2 - T_1 = 300 \; \text{K} - 80 \; \text{K} = 220 \; \text{K}\]
03

Apply Fourier's Law of Heat Conduction

Fourier's Law of Heat Conduction can be used to calculate the rate of heat flow:\[Q = \frac{k \cdot A \cdot \Delta T}{d}\]Where:\(Q\) is the rate of heat flow (in \(\text{J/s}\) or Watts),\(k\) is the thermal conductivity,\(A\) is the area,\(\Delta T\) is the temperature difference,\(d\) is the thickness.
04

Compute the Rate of Heat Flow

Substitute the values into Fourier's Law for the rate of heat flow:\[Q = \frac{0.025 \; \text{J/s} \cdot \text{m} \cdot \degree{C} \cdot 0.80 \; \text{m}^2 \cdot 220 \; \text{K}}{0.01 \; \text{m}} = 440 \; \text{J/s}\]Therefore, the rate of heat flow from the atmosphere to the liquid nitrogen is \(440 \; \text{J/s}\).

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Key Concepts

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

Understanding Thermal Conductivity
Thermal conductivity is a key concept when discussing heat transfer, especially in materials used for insulation, like styrofoam. This property describes how well a material can conduct heat.
  • Materials with high thermal conductivity transfer heat quickly. Metals are a good example, as they conduct heat rapidly.
  • Materials with low thermal conductivity, like styrofoam, are insulators. They slow down heat flow, making them perfect for applications where heat retention or insulation is needed.
In the original exercise, the thermal conductivity of styrofoam is given as 0.025 J/s·m·°C. This value indicates how much heat flows through the material when there's a unit temperature difference across it, for every meter of thickness. A lower thermal conductivity means that styrofoam is effective at reducing the rate of heat flow, keeping the liquid nitrogen inside cool.
Calculating the Rate of Heat Flow
The rate of heat flow, denoted as \(Q\), is a measure of the amount of heat energy transferred per unit time. It is crucial for understanding energy efficiency and insulation effectiveness. To calculate \(Q\), we use Fourier’s Law of Heat Conduction:\[Q = \frac{k \cdot A \cdot \Delta T}{d}\]In this formula:
  • \(k\) is the thermal conductivity of the material.
  • \(A\) represents the contact area through which heat is conducted.
  • \(\Delta T\) is the temperature difference across the material.
  • \(d\) is the thickness of the material.
Substitute the values from the problem:- \(k = 0.025 \text{ J/s}\cdot\text{m}\cdot\degree{C}\)- \(A = 0.80 \text{ m}^2\)- \(\Delta T = 220 \text{ K}\)- \(d = 0.01 \text{ m}\)This results in a heat flow rate of 440 J/s, meaning that each second, 440 joules of energy are transferred from the atmosphere to the liquid nitrogen. Lower rates are often desirable in cases where maintaining temperature control is critical.
Impact of Temperature Difference
Temperature difference is a driving force in the process of heat conduction. Without a temperature difference, there would be no heat flow. The greater the difference, the higher the rate of heat transfer becomes.
  • In the given scenario, the temperature difference \(\Delta T\) between the liquid nitrogen and the atmosphere is substantial, amounting to 220 K.
  • This large difference increases the driving force for heat to flow from the warmer atmosphere into the cooler liquid nitrogen.
Essentially, reducing the temperature difference can decrease heat transfer rates. This is why doing things like increasing insulation thickness, using materials with low thermal conductivity, or minimizing the surface area in contact can be beneficial in controlling heat flow. Maintaining a minimal temperature difference can significantly contribute to energy efficiency, especially in temperature-sensitive applications. Keeping the liquid nitrogen cool while limiting energy transfer from the environment is critical in scientific and industrial contexts.

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Most popular questions from this chapter

A copper sphere is suspended in an evacuated chamber maintained at \(300 \mathrm{~K}\). The sphere is maintained at a constant temperature of \(500 \mathrm{~K}\) by heating it electrically. A total of \(210 \mathrm{~W}\) of electric power is needed to do it. When the surface of the copper sphere is completely blackened, \(700 \mathrm{~W}\) is needed to maintain the same temperature of the sphere. Calculate the emissivity of copper.

A uniform slab of dimension \(10 \mathrm{~cm} \times 10 \mathrm{~cm} \times 1 \mathrm{~cm}\) is kept between two heat reservoirs at temperatures \(10^{\circ} \mathrm{C}\) and \(90^{\circ} \mathrm{C}\). The larger surface areas touch the reservoirs. The thermal conductivity of the material is \(0-80 \mathrm{~W} \mathrm{~m}^{-1}{ }^{-1} \mathrm{C}^{-1}\). Find the amount of heat flowing through the slab per minute.

A hot body placed in a surrounding of temperature \(\theta_{0}\) obeys Newton's law of cooling \(\frac{d \theta}{d t}=-k\left(\theta-\theta_{0}\right) .\) Its temperature at \(t=0\) is \(\theta_{1} .\) The specific heat capacity of the body is \(s\) and its mass is \(m\). Find (a) the maximum heat that the body can lose and (b) the time starting from \(t=0\) in which it will lose \(90 \%\) of this maximum heat.

An amount \(n\) (in moles) of a monatomic gas at an initial temperature \(T_{0}\) is enclosed in a cylindrical vessel fitted with a light piston. The surrounding air has a temperature \(T_{s}\left(>T_{0}\right)\) and the atmospheric pressure is \(p_{a} .\) Heat may be conducted between the surrounding and the gas through the bottom of the cylinder. The bottom has a surface area \(A\), thickness \(x\) and thermal conductivity \(K\). Assuming all changes to be slow, find the distance moved by the piston in time \(t\).

A room has a window fitted with a single \(1-0 \mathrm{~m} \times 2 \cdot 0 \mathrm{~m}\) glass of thickness \(2 \mathrm{~mm}\). (a) Calculate the rate of heat flow through the closed window when the temperature inside the room is \(32^{\circ} \mathrm{C}\) and that outside is \(40^{\circ} \mathrm{C} .\) (b) The glass is now replaced by two glasspanes, each having a thickness of \(1 \mathrm{~mm}\) and separated by a distance of \(1 \mathrm{~mm}\). Calculate the rate of heat flow under the same conditions of temperature. Thermal conductivity of window glass \(=1 \cdot 0 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-10} \mathrm{C}^{-1}\) and that of air \(=0 \cdot 025 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-10} \mathrm{C}^{-1}\).

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