/*! 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 21 A \(10 \mathrm{~g}\) body is kep... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(10 \mathrm{~g}\) body is kept in an enclosure of \(27^{\circ} \mathrm{C}\). For body's temperature \(127^{\circ} \mathrm{C}\), the specific heat \(0.1 \mathrm{~K} \mathrm{cal} / \mathrm{kg}^{\circ} \mathrm{C}\) and surface area \(10^{-3} \mathrm{~m}^{2}\). The \(\left(\sigma=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \mathrm{k}^{4}\right]\) (A) Rate of cooling is \(0.227 \mathrm{ks}^{-1}\) (B) Rate of cooling will be zero at \(400 \mathrm{~K}\) enclosure (C) Cooling does not take place (D) Cooling will be faster at \(127^{\circ} \mathrm{C}\) enclosure

Short Answer

Expert verified
(A) Rate of cooling is \(0.227 \mathrm{ks}^{-1}\).

Step by step solution

01

Convert temperatures to Kelvin

The temperatures given are in degrees Celsius. The Stefan-Boltzmann law requires temperatures to be in Kelvin. To convert from Celsius to Kelvin, add 273 to the Celsius temperature. Therefore, the body's temperature in Kelvin is \(127^{\circ} \mathrm{C} + 273 = 400K\). Similarly, the enclosure's temperature in Kelvin is \(27^{\circ} \mathrm{C} + 273 = 300K\).
02

Apply the Stefan-Boltzmann Law

To find the rate of cooling, apply the Stefan-Boltzmann law, which is \(P = \sigma \cdot A (T^4 - T0^4)\) where P is power (rate of cooling), \(\sigma\) is the Stefan-Boltzmann constant, A is surface area, \(T\) is the temperature of the body and \(T0\) is the temperature of the environment. Substituting the given values, we find \(P = (5.67 \times 10^{-8}) \cdot (10^{-3}) ((400)^4 - (300)^4).\))
03

Calculate the equation

Solve the equation to find P, which is the rate of cooling. The result should be \(0.227 \mathrm{ks}^{-1}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

On the surface of lake when the atmospheric temperature is \(-15^{\circ} \mathrm{C}, 1.5 \mathrm{~cm}\) thick layer of ice is formed in \(20 \mathrm{~min}\), time taken to change its thickness from \(1.5 \mathrm{~cm}\) to \(3 \mathrm{~cm}\) will be (A) \(20 \mathrm{~min}\) (B) less than \(20 \mathrm{~min}\) (C) greater than \(40 \mathrm{~min}\) (D) less than \(40 \mathrm{~min}\) and greater than \(20 \mathrm{~min}\)

Assuming the sun to be a spherical body of radius \(R\) at a temperature of \(T \mathrm{~K}\). Evaluate the total radiant power, incident on earth, at a distance \(r\) from the sun (A) \(\frac{4 \pi r_{0}^{2} R^{2} \sigma T^{4}}{r^{2}}\) (B) \(\frac{\pi r_{0}^{2} R^{2} \sigma T^{4}}{r^{2}}\) (C) \(\frac{r_{0}^{2} R^{2} \sigma T^{4}}{4 \pi r^{2}}\) (D) \(\frac{R^{2} \sigma T^{4}}{r^{2}}\) where \(r_{0}\) is the radius of the earth and \(s\) is Stefan's constant.

Inner surface of a cylindrical shell of length \(l\) and of material of thermal conductivity \(k\) is kept at constant temperature \(T_{1}\) and outer surface of the cylinder is kept at constant temperature \(T_{2}\) such that \(T_{1}>T_{2}\) as shown in Fig. 11.3. Heat flows from inner surface to outer surface radially outward. Inner and outer radii of the shell are \(R\) and \(2 R\), respectively. Due to lack of space, this cylinder has to be replaced by a smaller cylinder of length \(\frac{l}{2}\), inner and outer radii \(\frac{R}{4}\) and \(R\), respectively, and thermal conductivity of material \(n k\). If rate of radial outward heat flow remains same for same temperatures of inner and outer surface, i.e., \(T_{1}\) and \(T_{2}\), then find the value of \(n\).

The rate of flow of thermal current depends on the nature of material (thermal conductivity), cross-sectional area \(A\), and temperature gradient. More the temperature difference, higher the thermal current flow. This fact identifies the thermal resistance offered by the material while conducting heat. One can find by equivalent resistance in heat flow using the same principles as for current. A body may transfer energy better by radiation. The nature of radiating surfaces play a role in the power radiated from it. On covering a surface by non-conducting/radiating media, the loss of the heat energy can be controlled. A rod of length \(l\) and conductivity \(k\) is placed between two reservoirs maintained at \(0^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\). At a distance \(\frac{l}{3}\) and \(\frac{l}{5}\) from \(100^{\circ} \mathrm{C}\) reservoir the rate of flow of thermal current are \(I_{1}\) and \(I_{2} .\) The temperatures at \(\frac{l}{5}\) and \(\frac{l}{3}\) are \(\theta_{1}\) and \(\theta_{2}\) then (A) \(I_{1}=I_{2}, 2 \theta_{1}=3 \theta_{2}\) (B) \(5 \theta_{1}-3 \theta_{2}=200\) (C) \(I_{1}=I_{2}, 5 \theta_{1}-3 \theta_{2}=200\) (D) \(I_{1} \neq I_{2}, \quad 5 \theta_{2}=3 \theta_{1}\)

One end of a thermally insulated rod is kept at a temperature \(T_{1}\) and the other end at \(T_{2}\). The rod is composed of two sections of length \(l_{1}\) and \(l_{2}\) and thermal conductivities \(K_{1}\) and \(K_{2}\), respectively. The temperature at the interface of the two sections is (A) \(\left(K_{2} l_{2} T_{1}+K_{1} l_{1} T_{2}\right) /\left(K_{1} l_{1}+K_{2} l_{2}\right)\) (B) \(\left(K_{2} l_{1} T_{1}+K_{1} l_{2} T_{2}\right) /\left(K_{2} l_{1}+K_{1} l_{2}\right)\) (C) \(\left(K_{1} l_{2} T_{1}+K_{2} l_{1} T_{2}\right) /\left(K_{1} l_{2}+K_{2} l_{1}\right)\) (D) \(\left(K_{1} l_{1} T_{1}+K_{2} l_{2} T_{2}\right) /\left(K_{1} l_{1}+K_{2} l_{2}\right)\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.