/*! 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 5 A hot body is being cooled in ai... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 5

A hot body is being cooled in air according to Newton's law of cooling, the rate of fall of temperature being \(k\) times the difference of its temperature with respect to that of surroundings. The time, after which the body will lose half the maximum heat it can lose, is (A) \(\frac{1}{k}\) (B) \(\frac{\ln 2}{k}\) (C) \(\frac{\ln 3}{k}\) (D) \(\frac{2}{k}\)

Short Answer

Expert verified
Based on the presented solution, the correct answer is option B: \( \frac{\ln 2}{k} \).

Step by step solution

01

Step 1. Express the Situation Mathematically

Let \( T \) be the temperature of the body at any time \( t \), and \( T_0 \) be the temperature of the surroundings. The rate of cooling is\( \frac{dT}{dt} = -k(T - T_0) \)
02

Step 2. Solve the Differential Equation

This is a typical first order linear differential equation. Separating variables and integrating, we obtain \( \int_{T_0}^{T} \frac{d(T)}{T - T_0} = -k \int_{0}^{t} dt \), after solving the integrals, we get \( \ln |T - T_0| = -kt + C \), where \( C \) is the constant of integration.
03

Step 3. Determine the Constant of Integration

At \( t = 0 \), \( T = T_i \) (Initial temperature), hence \( C = \ln |T_i - T_0| \). Thus the equation becomes \( \ln |T - T_0| = -kt + \ln |T_i - T_0| \) or \( T = T_0 + (T_i - T_0)e^{-kt} \)
04

Step 4. Find Time for Half Heat Loss

The heat lost by the body = \( c(T_i - T) \), where \( c \) is the specific heat capacity. Half the maximum heat means \( T = \frac{1}{2}(T_i + T_0) \), substituting this in our equation, we get \( \frac{1}{2}(T_i + T_0) = T_0 + (T_i - T_0)e^{-kt} \). Further simplification gives \( e^{kt} = 2 \), which means \( kt = \ln 2 \), therefore \( t = \frac{\ln 2}{k} \)

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 91影视!

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

A body cools from \(60^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) in \(10 \mathrm{~min}\). If the room temperature is \(25^{\circ} \mathrm{C}\) and assuming Newton's law of cooling to hold good, the temperature of the body at the end of the next 10 min will be (A) \(38.5^{\circ} \mathrm{C}\) (B) \(40^{\circ} \mathrm{C}\) (C) \(42.85^{\circ} \mathrm{C}\) (D) \(45^{\circ} \mathrm{C}\)

The temperature drop through a two-layered furnace wall is \(900^{\circ} \mathrm{C}\). Each layer is of equal area of cross-section. Which of the following actions will result in lowering the temperature \(\theta\) of the interface? (A) By increasing the thermal conductivity of outer layer (B) By increasing the thermal conductivity of inner layer (C) By increasing thickness of outer layer (D) By increasing thickness of inner layer

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) \(I_{1} \neq I_{2}, \quad 5 \theta_{2}=3 \theta_{1}\)

According to Newton's law of cooling, the rate of cooling of a body is proportional to \((\Delta \theta)^{n}\) where \(\Delta \theta\) is the difference between the temperature of the body and the surrounding. Then \(n\) is equal to (A) 2 (B) 3 (C) 4 (D) 1

Heat energy is transferred from a heat source maintained at temperature \(T_{1}=300^{\circ} \mathrm{C}\) to two lower temperature heat reservoirs maintained at temperatures \(T_{2}=50^{\circ} \mathrm{C}\) and \(T_{3}=100^{\circ} \mathrm{C}\). Three identical solid steel rods, each of length \(L=1 \mathrm{~m}\) and cross-sectional area \(0.01 \mathrm{~m}^{2}\), are used to pipe the heat, as shown. Find the ratio of power in watts delivered by the source to each of the lower temperature reservoirs and also find the temperature of the junction \(O\). Consider only axial heat flow. Thermal conductivity of each of the rod is 46 SI units.
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Translational Dynamics

Read Explanation

Physics of Motion

Read Explanation

Atoms and Radioactivity

Read Explanation

Quantum Physics

Read Explanation

Wave Optics

Read Explanation
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.