Chapter 5.8, Problem 15AYU

Newton’s Law of Heating Athermometer reading ${\text{8}}^{\text{o}}\text{C}$ is brought into a room with a constant temperature of ${\text{35}}^{\text{o}}\text{C}$. If the thermometer reads ${\text{15}}^{\text{o}}\text{C}$ after $3$ minutes, what will it read after being in the room for $5$ minutes? For $10$ minutes?

[Hint: You need to construct formula similar to equation $(4).$]

Equation (4) is,

$u(t)=T+(u_0-T)e^{kt}k<0$

To determine

To calculate: The reading of athermometer being in the room after $5$ minutes and after $10$ minutes where the thermometer reading is $8^{o}c$ with constant room temperature ${35}^{o}c$, and if the thermometer reads ${15}^{o}c$ after $3$ minutes.

Answer to Problem 15AYU

Solution:

The thermometer will read ${18.63}^{o}c$ after $5$ minutes and ${25.07}^{o}c$ after $10$ minutes.

Explanation of Solution

Given Information:

The thermometer reading is $8^{o}c$ with constant room temperature ${35}^{o}c$, and the thermometer reads ${15}^{o}c$ after $3$ minutes.

Formula used:

The thermometer reading $u$ 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.

Explanation:

The constant room temperature is ${35}^{o}c$, and initial reading of the thermometer is $8^{o}c$.

Therefore, plug the values $u_o=8$ and $T=35$ in the function $u\left(t\right)=T+\left(u_o-T\right)e^{kt}$,

$\Rightarrow u\left(t\right)=35+\left(8-35\right)e^{kt}$

$\Rightarrow u\left(t\right)=35-27e^{kt}$

Since the thermometer reads ${15}^{0}c$ in after $3$ minutes, $u\left(t\right)=15$ and $t=3$.

To find the value of $k$, substitute $u\left(t\right)=15$ and $t=3$ in the equation $u\left(t\right)=35-27e^{kt}$,

$\Rightarrow 15=35-27e^{3k}$.

Subtract $35$ from both sides of the equation,

$\Rightarrow -20=-27e^{3k}$.

Divide by $-27$ on both sides of the equation,

$\Rightarrow 0.741=e^{3k}$.

Taking logarithm on both sides of the equation,

$\Rightarrow \ln \left(0.741\right)=\ln \left(e^{3k}\right)$

$\Rightarrow -0.3=3k$

$\Rightarrow k=-0.1$.

Plug $k=-0.1$ the equation $u\left(t\right)=35-27e^{kt}$,

$\Rightarrow u\left(t\right)=35-27e^{-0.1t}$

To find the thermometer reading after $5$ minutes, plug $t=5$ in $u\left(t\right)=35-27e^{-0.1t}$,

$\Rightarrow u\left(t\right)=35-27e^{\left(-0.1\right)\left(5\right)}$

$\Rightarrow u\left(t\right)=35-27e^{-0.5}$

$\Rightarrow u\left(t\right)=35-27\left(0.6065\right)$

$\Rightarrow u\left(t\right)=35-16.3755$

$\Rightarrow u\left(t\right)=18.63$.

Hence, the thermometer will read ${18.63}^{o}c$ after $5$ minutes.

Now, to find the thermometer reading after $10$ minutes, plug $t=10$ in $u\left(t\right)=35-27e^{-0.1t}$,

$\Rightarrow u\left(t\right)=35-27e^{\left(-0.1\right)\left(10\right)}$

$\Rightarrow u\left(t\right)=35-27e^{-1.0}$

$\Rightarrow u\left(t\right)=35-27\left(0.3679\right)$

$\Rightarrow u\left(t\right)=35-9.9327$

$\Rightarrow u\left(t\right)=25.07$.

Hence, the thermometer will read ${25.07}^{o}c$ after $10$ minutes.