Chapter 5.8, Problem 13AYU

Cooling Time of a Pizza Apizza baked at ${\text{450}}^{\text{o}}\text{F}$ is removed from the oven at $5:00$ PM and placed in a room that is a constant ${\text{70}}^{\text{o}}\text{F}$. After $5$ minutes, the pizza is at ${\text{300}}^{\text{o}}\text{F}$.

At what time can you begin eating the pizza if you want its temperature to be ${\text{135}}^{\text{o}}\text{F}$?

Determine the time that needs to elapse before the temperature of the pizza is ${\text{160}}^{\text{o}}\text{F}$.

What do you notice about the temperature as time passes?

To determine

To calculate: The time when the temperature of the pizza would be ${135}^o\text{F}$.

Answer to Problem 13AYU

Solution:

After $17.58$ minutes, the temperature of pizza is ${135}^o\text{F}$.

Explanation of Solution

Given information:

A pizza baked at ${450}^o\text{F}$ is removed from the oven at $5:00$ pm and placed in a room that is a constant ${70}^o\text{F}$, and after $5$ minutes, the temperature of pizza is ${300}^o\text{F}$.

Formula used:

The temperature $u$ at time $t$ can be modelled by the function,

$u\left(t\right)=T+\left(u_o-T\right)e^{kt}$, $k<0$

Where $T$ is the constant room temperature, $u_o$ is the initial reading of the thermometer, and $k$ is negative constant.

Calculation:

The constant room temperature is ${70}^o\text{F}$, and initial temperature of the pizza is ${450}^o\text{F}$.

Therefore, plug the values $u_o=450$ and $T=70$ in the function $u\left(t\right)=T+\left(u_o-T\right)e^{kt}$,

$\Rightarrow u\left(t\right)=70+\left(450-70\right)e^{kt}$

$\Rightarrow u\left(t\right)=70+380e^{kt}$

Since the temperature of pizzais ${300}^o\text{F}$ after $5$ minutes, $u\left(t\right)=300$ and $t=5$.

To find the value of $k$, substitute $u\left(t\right)=300$ and $t=5$ in the equation $u\left(t\right)=70+380e^{kt}$,

$\Rightarrow 300=70+380e^{5k}$

Subtract $70$ from both sides of the equation,

$\Rightarrow 230=380e^{5k}$

Divide by $380$ on both sides of the equation,

$\Rightarrow 0.6053=e^{5k}$

Taking logarithm on both sides of the equation,

$\Rightarrow \ln \left(0.6053\right)=\ln \left(e^{5k}\right)$

$\Rightarrow -0.5021=5k$

$\Rightarrow k=-0.1004$.

Substitute $k=-0.1004$ in the equation $u\left(t\right)=70+380e^{kt}$,

$\Rightarrow u\left(t\right)=70+380e^{-0.1004t}$

To find the time when temperature of pizza is ${135}^o\text{F}$, substitute $u\left(t\right)=135$ in $u\left(t\right)=70+380e^{-0.1004t}$

$\Rightarrow 135=70+380e^{-0.1004t}$

By subtracting $70$ from both sides,

$\Rightarrow 65=380e^{-0.1004t}$

By dividing both sides by $380$,

$\Rightarrow 0.1711=e^{-0.1004t}$

By taking logarithm on both sides,

$\Rightarrow \ln \left(0.1711\right)=\ln \left(e^{-0.1004t}\right)$

$\Rightarrow -1.7658=-0.1004t$

By dividing both sides by $\left(-0.1004\right)$,

$t=17.58$

Thus, after $17.58$ minutes, the temperature of pizza is ${135}^o\text{F}$.

To determine

The time that needs to elapse before the temperature of the pizza is ${160}^o\text{F}$.

Answer to Problem 13AYU

Solution:

About $14.34$ minutes needs to elapse before the temperature of pizza is ${160}^o\text{F}$.

Explanation of Solution

Given information:

A pizza baked at ${450}^o\text{F}$ is removed from the oven at $5:00$ pm and placed in a room that is a constant ${70}^o\text{F}$, and after $5$ minutes, the temperature of pizza is ${300}^o\text{F}$.

Explanation:

From part (a), the equation of Colling is $u\left(t\right)=70+380e^{-0.1004t}$.

To find the time $t$, substitute $u\left(t\right)=135$ in $u\left(t\right)=70+380e^{-0.1004t}$,

$\Rightarrow 160=70+380e^{-0.1004t}$

By subtracting $70$ from both sides,

$\Rightarrow 90=380e^{-0.1004t}$

By dividing by $380$ on both sides,

$\Rightarrow 0.2368=e^{-0.1004t}$

By taking logarithm on both sides,

$\Rightarrow \ln \left(0.2368\right)=\ln \left(e^{-0.1004t}\right)$

$\Rightarrow -1.4404=-0.1004t$.

By dividing both sides by $\left(-0.1004\right)$,

$\Rightarrow t=14.34$

Thus, about $14.34$ minutesneeds to elapse before the temperature of pizza is ${160}^o\text{F}$.

To determine

The behavior of the temperature as time passes.

Answer to Problem 13AYU

Solution:

As time passes, the temperature of the pizza decreases to constant room temperature ${70}^o\text{F}$.

Explanation of Solution

Given information:

A pizza baked at ${450}^o\text{F}$ is removed from the oven at $5:00$ pm and placed in a room that is a constant ${70}^o\text{F}$, and after $5$ minutes, the temperature of pizza is ${300}^o\text{F}$.

Explanation:

From part (a), the equation of Colling is $u\left(t\right)=70+380e^{-0.1004t}$.

As time $t$ increases, the exponent $-0.1004t$ approaches to negative direction.

That implies, $\Rightarrow e^{-0.1004t}\approx 0$.

Therefore, $\Rightarrow u\left(t\right)=70+0=70$

Thus, as time passes, the temperature of the pizza decreases to constant room temperature ${70}^o\text{F}$.