Question

A body at a temperature of $50{}^\circ F$ is placed in an oven whose temperature is kept at $150{}^\circ F$. If after $10$ min the temperature of the body is $75{}^\circ F$, find the time required for the body to reach a temperature of $100{}^\circ F$(in minute).
$\begin{aligned} & A)\,24.04 \\\ & B)\,34.09 \\\ & C)\,44.09 \\\ & D)\,35.09 \\\ \end{aligned}$

Answer 1

We are given that a body of temperature $50{}^\circ F$ is placed in oven which is fixed at temperature $150{}^\circ F$, we are asked to find temperature when it is $100{}^\circ F$ we first use the rate of heat gain is directly proportional to be difference between surrounding and the body temperature, we will then interpret to get relation between body temperature (T) and time (t), we will use condition like at time $10$ min temperature was $75{}^\circ F$, to find constant and then simplify we get required solution.

Complete step by step anwer:
We are given that a body at a temperature of $50{}^\circ F$ is placed in an oven whose temperature is kept at $150{}^\circ F$. We have that after $10$ minutes its temperature is $75{}^\circ F$. We are asked to find the time when its temperature is $100{}^\circ F$.
As we know that the rate of heat gained is proportional to the difference between the surrounding temperature and the cloudy temperature. We have that surrounding temperature is fixed $150{}^\circ F$ inside the oven but the temperature of the body is changing with time ‘t’
So, let T denote body temperature at time t.
So, we have $\dfrac{dT}{dt}=k\left( 150-T \right)$
Now we integrate both sides to find the value of the unknown constant and the required value of f.
So we get,
$\Rightarrow \int{\dfrac{dT}{dt}=k\left( 150-T \right)}$
Separating variable, we get,
$\Rightarrow \int{\dfrac{dT}{\left( 150-T \right)}=\int{kdt}}$
So we get,
$\Rightarrow \left( \dfrac{1}{150-T} \right)-k+tc$ where c is constant.
Now, we have that after $10$ min body temperature was $75{}^\circ F$.
It means at $t=10$, we have $T=75{}^\circ$.
Using this we get
$\Rightarrow in\left( \dfrac{1}{15-75} \right)=10k+c$
Simplifying we get
$\begin{aligned} & \Rightarrow in\left( \dfrac{1}{75} \right)-10k=c \\\ & \dfrac{1}{a}=-in(a) \\\ \end{aligned}$
So, we get,
$\Rightarrow c=in(75)-10k$
Using this in (ii) we get,
$in\left( \dfrac{1}{150-T}=kt=10k-in(75) \right)$
Simplifying we get,
$\begin{aligned} & \Rightarrow in\left( \dfrac{1}{150-T}+in(75)-k(t-10) \right) \\\ & \Rightarrow in\left( \dfrac{75}{150-T} \right)-k\left( 10-t) \right)........(iii) \\\ \end{aligned}$
Now we also have the initial, we have that body temperature was $50{}^\circ F$.
So we get
$t=0,\,\,\,T=50$
So, using this in equation (iii), we get
$\Rightarrow in\left( \dfrac{75}{15050} \right)=k(10-0$
So, simplifying we get
$\begin{aligned} & in(0,75)=10k \\\ & \Rightarrow k=\dfrac{in(0.75)}{10} \\\ \end{aligned}$
Now we have to find value of k as c we put
$\Rightarrow k=\dfrac{in(0.75)}{10}$
We get our required equation as:
$\Rightarrow in\left( \dfrac{75}{150-T} \right)=\dfrac{in(0.75)}{10}(10-t)$
Now, as we head time when temperature is $100{}^\circ F$ so we put $100=T$, and solve for t.
So,
$\Rightarrow in\left( \dfrac{75}{150-100} \right)=\dfrac{in(0.75)}{10}(10-t)$s
Simplifying we get
$\Rightarrow in\left( \dfrac{75}{50} \right)=\dfrac{in(0.75)}{10}(10-t)$
So we get,
$\Rightarrow in\left( 1.5 \right)=\dfrac{in\left( 0.75 \right)}{10}(10-t)$
So, we get,
$\Rightarrow t=10-\dfrac{10\left( in\left( 1.5 \right) \right)}{0.75}$
So, we get,
$t=24.09$

Our required answer is $24.09$. Option A) .

Note:
While solving term using log then basic properties of log are handful like log(ab)=logatlogb
Log(a)=loga-logb
Also, remember that $$$$
$\int{\dfrac{1}{t}at=\log t}$
And
$\int{\dfrac{1}{-t}at=\log t}\left( \dfrac{1}{t} \right)$
As we take negative out, we get
$\int{\dfrac{1}{-t}at=-\int{\dfrac{1}{t}dt}}$
Which give -logt and we know
$-\log a\left( \dfrac{1}{a} \right)$
So we get $\log \left( \dfrac{1}{f} \right)$