/*! 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 45 A cubical block of mass \(1 \cdo... [FREE SOLUTION] | 91Ó°ÊÓ

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A cubical block of mass \(1 \cdot 0 \mathrm{~kg}\) and edge \(5 \cdot 0 \mathrm{~cm}\) is heated to \(227^{\circ} \mathrm{C}\). It is kept in an evacuated chamber maintained at \(27^{\circ} \mathrm{C}\). Assuming that the block emits radiation like a blackbody, find the rate at which the temperature of the block will decrease. Specific heat capacity of the material of the block is \(400 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\).

Short Answer

Expert verified
The rate at which the temperature of the block decreases is approximately \(0.0237 \text{ K/s}\).

Step by step solution

01

Understanding the Given Values

Convert the edge of the cubical block to meters: \(5.0 \text{ cm} = 0.05 \text{ m}\). Convert the initial and ambient temperatures to Kelvin: \(T_1 = 227^{\circ} C + 273.15 = 500.15 \text{ K}\) and \(T_2 = 27^{\circ} C + 273.15 = 300.15 \text{ K}\). Mass of the block \(m = 1.0 \text{ kg}\), specific heat capacity \(C = 400 \text{ J kg}^{-1} \text{ K}^{-1}\).
02

Calculate the Surface Area of the Block

The block is a cube, so its surface area \(A = 6L^2\), where \(L\) is the edge length in meters. Substitute \(L = 0.05 \text{ m}\): \(A = 6 \times (0.05)^2 = 0.015 \text{ m}^2\).
03

Use Stefan-Boltzmann Law to Determine Heat Loss Rate

According to the Stefan-Boltzmann law, the power radiated \(P = \epsilon \sigma A (T_1^4 - T_2^4)\), for a blackbody \(\epsilon = 1\). \(\sigma = 5.67 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}\). Substitute the values: \[P = 1 \times 5.67 \times 10^{-8} \times 0.015 \times (500.15^4 - 300.15^4)\]. Calculating this gives \(P \approx 9.47 \text{ W}\).
04

Calculate the Rate of Temperature Decrease

The rate at which temperature decreases \( \frac{\Delta T}{\Delta t} = \frac{P}{mC}\). Substitute \(P = 9.47 \text{ W}\), \(m = 1.0 \text{ kg}\), \(C = 400 \text{ J kg}^{-1} \text{ K}^{-1}\). \( \frac{\Delta T}{\Delta t} = \frac{9.47}{1.0 \times 400} \approx 0.0237 \text{ K s}^{-1}\).

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Key Concepts

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

Stefan-Boltzmann Law
The Stefan-Boltzmann Law is a fundamental principle in physics that describes how objects radiate energy. It applies significantly to objects known as "blackbodies," which absorb and emit all radiation perfectly. For these objects, the power emitted per unit area is proportional to the fourth power of the temperature, expressed in Kelvin.

Mathematically, it is given by the equation:
\[P = \epsilon \sigma A T^4\]Here, \epsilon is the emissivity of the object; for a true blackbody, this value is 1. \sigma is the Stefan-Boltzmann constant \( \approx 5.67 \times 10^{-8} \text{W m}^{-2} \text{K}^{-4} \). A is the surface area, and T is the absolute temperature in Kelvin. This law helps us understand the energy loss due to radiation in heated objects, critical in various practical applications, including thermal insulation and solar power.

In the context of the exercise, the Stefan-Boltzmann Law is used to estimate how fast the block cools by calculating the power lost as heat radiates away from the block.
Specific Heat Capacity
Specific heat capacity is a measure of the amount of heat energy required to change the temperature of a substance by a certain amount. Specifically, it is defined as the heat required to raise the temperature of one kilogram of a material by one Kelvin (or one-degree Celsius).

The formula for calculating heat energy based on specific heat capacity is:
\[q = m C \Delta T\]
  • \(q\) is the heat energy (in joules),
  • \(m\) is the mass of the substance (in kilograms),
  • \(C\) is the specific heat capacity of the material (in \(\text{J kg}^{-1} \text{K}^{-1}\)),
  • \(\Delta T\) is the change in temperature (in Kelvin).
Understanding this concept is crucial because it reveals how different materials respond to heat. For instance, substances with a high specific heat capacity can absorb a lot of heat without experiencing a significant change in temperature, making them excellent for thermal energy storage.

In the example exercise, specific heat capacity is used to determine how much the temperature of the cubical block changes as it loses heat.
Blackbody Radiation
Blackbody radiation refers to the theoretical concept of a perfect blackbody, which is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. No light is reflected, so the name "blackbody" fits perfectly.

Such an object will re-radiate energy based on its temperature, and this emission is what we call blackbody radiation. The intensity and spectrum of this radiation depend solely on the temperature of the blackbody, described well by Planck's Law. At higher temperatures, a blackbody will emit more radiation and at shorter wavelengths.

One can visualize blackbody radiation through images of heated metals glowing, where they emit energy in the form of visible light. However, in space or with many laboratory applications, this concept explains how stars and other celestial bodies emit radiation.
  • Explains why hotter stars appear bluer and cooler stars appear redder.
  • Used in designing equipment like cryogenic detectors which rely on sensing infrared radiation.
In the exercise context, the block is treated as a blackbody to simplify calculations of heat loss and understand the rate of temperature decrease due to radiative emission.

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Most popular questions from this chapter

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.

The left end of a copper rod (length \(=20 \mathrm{~cm}\), area of cross section \(\left.=0-20 \mathrm{~cm}^{2}\right)\) is maintained at \(20^{\circ} \mathrm{C}\) and the right end is maintained at \(80^{\circ} \mathrm{C}\). Neglecting any loss of heat through radiation, find (a) the temperature at a point \(11 \mathrm{~cm}\) from the left end and (b) the heat current through the rod. Thermal conductivity of copper \(=385 \mathrm{~W} \mathrm{~m}^{-1}{ }^{\circ} \mathrm{C}^{-1}\).

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 hollow metallic sphere of radius \(20 \mathrm{~cm}\) surrounds a concentric metallic sphere of radius \(5 \mathrm{~cm}\). The space between the two spheres is filled with a nonmetallic material. The inner and outer spheres are maintained at \(50^{\circ} \mathrm{C}\) and \(10^{\circ} \mathrm{C}\) respectively and it is found that \(100 \mathrm{~J}\) of heat passes from the inner sphere to the outer sphere per second. Find the thermal conductivity of the material between the spheres.

A rod of negligible heat capacity has length \(20 \mathrm{~cm}\), area of cross section \(1 \cdot 0 \mathrm{~cm}^{2}\) and thermal conductivity \(200 \mathrm{~W} \mathrm{~m}^{-1_{0}} \mathrm{C}^{-1}\). The temperature of one end is maintained at \(0^{\circ} \mathrm{C}\) and that of the other end is slowly and linearly varied from \(0^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C}\) in 10 minutes. Assuming no loss of heat through the sides, find the total heat transmitted through the rod in these 10 minutes.

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