Chapter 7.1, Problem 14E

a.

To determine

To find: when the coffee cools most quickly and what happens to the rate of cooling as time goes by, explain.

Answer to Problem 14E

The coffee cools the quickest in the beginning, when the temperature difference $(T-T_a)$ is maximum. As time progresses the rate of cooling decreases as the temperature difference $(T-T_a)$ decreases.

Explanation of Solution

Given information: Suppose you have just poured a cup of freshly brewed coffee with

temperature ${96}^{\circ }C$ in a room where the temperature is ${20}^{\circ }C$ .

Formula used:

Newton’s Law of Colling states that the rate of cooling of an object is proportional to the

temperature difference and its surrounding provided that this difference is not too largeNewton’s Law of Cooling:

  $\begin{array}{l}\frac{dT}{dt}=-k(T-T_a).\\ t=\text{Time}.\\ T=\text{Temperatue}\text{of}\text{the}\text{cooling}\text{object}.\\ T_a=\text{Ambient}\text{}\text{temperature}\text{.}\end{array}$

Calculation:

Since k and $T_a$ are constant, the magnitude of $\frac{dT}{dt}$ will be greatest when Tis the largest, is of course in the beginning before cooling.

The coffee cools the quickest in the beginning, when the temperature difference $(T-T_a)$ is maximum. As time progresses the rate of cooling decreases as the temperature difference $(T-T_a)$ decreases.

b.

To determine

To write : a differential equation that expresses Newton’s Law of Colling for this particular situation and what is the initial condition and find this differential equation is an appropriate model for cooling.

Answer to Problem 14E

  $-\frac{dT}{dt}=k(T-20)$ .

Initial condition is $T(0)={95}^{\circ }C$ .

The rate of cooling is increases as T increases.

The temperature of coffee is maximum initially; hence rate of cooling is maximum initially.

Explanation of Solution

Given information: Newton’s Law of Colling states that the rate of cooling of an object is proportional to the temperature difference and its surrounding, provided that this difference is not too large.

Formula used:

Newton’s Law of Colling states that the rate of cooling of an object is proportional to the

temperature difference and its surrounding provided that this difference is not too largeNewton’s Law of Cooling:

  $\begin{array}{l}\frac{dT}{dt}=-k(T-T_a).\\ t=\text{Time}.\\ T=\text{Temperatue}\text{of}\text{the}\text{cooling}\text{object}.\\ T_a=\text{Ambient}\text{}\text{temperature}\text{.}\end{array}$

Calculation:

Let the temperature of the coffee be T .

Temperature of the surroundings $T_a$ =20 ${}^{\circ }C$ (given)

So, According to Newton’s Law of Colling, the rate of cooling of coffee is:

  $\begin{array}{l}\frac{dT}{dt}=-k(T-20).\\ \frac{dT}{dt}=-k(T-20).\\ -\frac{dT}{dt}=k(T-20)\end{array}$

Initial condition is $T(0)={95}^{\circ }C$ .

Equation of the curve is $T=20+75e^{-kt}$ .

Note the rate of cooling is increases as T increases.

The temperature of coffee is maximum initially; hence rate of cooling is maximum initially.

Also the rate of cooling decreases as T decreases, this is consistent with part (a).

c.

To determine

To make: a rough sketch of the graph of the solution of the initial- value problem in part (b).

Answer to Problem 14E

Equation of the curve is $T=20+75e^{-kt}$ .

Explanation of Solution

Given information: Newton’s Law of Colling states that the rate of cooling of an object is proportional to the temperature difference and its surrounding, provided that this difference is not too large.

Calculation:

The graph is shown below.

Temperature is plotted on the vertical axis and time on the horizontal axis.

Note that the initial temperature is 95 ${}^{\circ }C$ and is continuously decreasing.

As the time passes the temperature decreases but the rate of decrease slows down and the temperature of the coffee approaches 20 ${}^{\circ }C$

Equation of the curve is $T=20+75e^{-kt}$ .

Where k is a positive constant