Question

A substance takes three minutes in cooling from 50°C to 45°C and takes five minutes and cooling from 45°C to 40°C what is the temperature of the surrounding how much time will it take to cool the substances from 40°C to 35°C

Answer 1

The ambient temperature is 35°C.

It takes 15 minutes to cool the substance from 40°C to 35°C.

Step-by-step explanation:

Using Newton's Law of Cooling to answer the given problem.

`\boxed{\left\begin{array}{ccc}\text{\underline{Newton's Law of Cooling:}}\\\\ (dT)/(dt) =-k(T-T_a)\end{array}\right}`

Given:

The time it takes to cool from 50°C to 45°C = 3 minutes

The time it takes to cool from 45°C to 40°C = 5 minutes

Find:

Time ambient temperature and the time it takes to cool the substance from 40°C to 35°C

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

(1) - Using first initial condition:

`\text{Avg temp =} \ (50+45)/(2)=\boxed{47.5 \textdegree C}\\\\\Longrightarrow (50-45)/(3)=-k(47.5-T_a) \\\\\Longrightarrow \boxed{ (5)/(3)=-k(47.5-T_a)}`

(2) - Using the second initial condition:

`\text{Avg temp =} \ (45+40)/(2)=\boxed{42.5 \textdegree C}\\\\\Longrightarrow (45-40)/(5)=-k(42.5-T_a) \\\\\Longrightarrow \boxed{1=-k(42.5-T_a)}`

(3) - Now we have a system of equations.

`\left \{ (5)/(3)=-k(47.5-T_a)}} \atop {1=-k(42.5-T_a)}}} \right.`

(4) - Solve the system by dividing the top equation by the bottom equation.

`\Longrightarrow ((5)/(3)=-k(47.5-T_a))/(1=-k(42.5-T_a)) \\\\\Longrightarrow(5)/(3)=(47.5-T_a)/(42.5-T_a)\\ \\ \Longrightarrow 5(42.5-T_a)=3(47.5-T_a)\\\\\Longrightarrow 212.5-5T_a=142.5-3T_a\\\\\Longrightarrow 2T_a=70\\\\\therefore \boxed{T_a=35 \textdegree C}`

Thus, the ambient temperature is 35°C.

(5) - Find the value of "k" using either of the two previous equations

`1=-k(42.5-T_a)\\\\\Longrightarrow 1=-k(42.5-35)\\\\\Longrightarrow 1=-7.5k\\\\\Longrightarrow \boxed{ k \approx -0.133}`

(6) - Now finding "dt"

`\text{Avg temp =} \ (40+35)/(2)=\boxed{37.5 \textdegree C}\\\\\Longrightarrow (40-35)/(dt)=0.133(37.5-35) \\\\\Longrightarrow (5)/(dt)=0.3325\\\\\therefore \boxed{dt \approx15 \ min}`

Thus, it take 15 minutes to cool the substance from 40°C to 35°C.