Answer: In a 25°C room, hot coffee in a vacuum flask cools | StudySoup

Textbook Solutions for Conceptual Physics

Chapter 16 Problem 3P

Question

In a \(25^{\circ} \mathrm{C}\) room, hot coffee in a vacuum flask cools from \(75^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) in 8 hours. Explain why you predict that its temperature after another 8 hours will be \(37.5^{\circ} \mathrm{C}\).

Solution

Step 1 of 2

This question can be solved using Newton’s law of cooling.

This law states that the rate of change of temperature of a body is proportional to the temperature difference between the body and its surrounding.

It can be written as, \(\frac{d T}{d t}=k\left(T_{t} \quad-T_{1}\right)\ldots(1)\)

Here, dT = temperature change

dt = change of time

k = constant of proportionality

\(T_t=\) temperature of time t

\(T_1=\) room temperature

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full solution

Title Conceptual Physics 11 
Author Paul G. Hewitt
ISBN 9780321568090

Answer: In a 25°C room, hot coffee in a vacuum flask cools

Chapter 16 textbook questions

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