# When a hot objects is placed in a water bath whose temperature is 25^{\circ}C, it cools from...

## Question:

When a hot objects is placed in a water bath whose temperature is 25{eq}^{\circ}{/eq}C, it cools from 100{eq}^{\circ}{/eq} to 50{eq}^{\circ}{/eq} in 195 s. In another bath, the same cooling occurs in 175 s. Find the temperature of the second bath.

## Newton's Law of Cooling

This problem is an application of Newton's Law of Cooling. We are going to use the equation for newton's law of cooling with the given initial conditions to determine the constants of the equation followed by determining the temperature of the second bath.

## Answer and Explanation:

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It is given that the initial temperature is, {eq}\displaystyle T_0=25^\circ {/eq}

The temperature of the object is given by,

{eq}\displaystyle...

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Chapter 16 / Lesson 3In this lesson you'll learn how to solve a first-order linear differential equation. We first define what such an equation is, and then we give the algorithm for solving one of that form. Specific examples follow the more general description of the method.