Answer:
The ambient temperature is 35°C.
It takes 15 minutes to cool the substance from 40°C to 35°C.
Explanation:
Using Newton's Law of Cooling to answer the given problem.
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Newton's Law of Cooling:}}\\\\ \frac{dT}{dt} =-k(T-T_a)\end{array}\right}[/tex]
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
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(1) - Using first initial condition:
[tex]\text{Avg temp =} \ \frac{50+45}{2}=\boxed{47.5 \textdegree C}\\\\\Longrightarrow \frac{50-45}{3}=-k(47.5-T_a) \\\\\Longrightarrow \boxed{ \frac{5}{3}=-k(47.5-T_a)}[/tex]
(2) - Using the second initial condition:
[tex]\text{Avg temp =} \ \frac{45+40}{2}=\boxed{42.5 \textdegree C}\\\\\Longrightarrow \frac{45-40}{5}=-k(42.5-T_a) \\\\\Longrightarrow \boxed{1=-k(42.5-T_a)}[/tex]
(3) - Now we have a system of equations.
[tex]\left \{ \frac{5}{3}=-k(47.5-T_a)}} \atop {1=-k(42.5-T_a)}}} \right.[/tex]
(4) - Solve the system by dividing the top equation by the bottom equation.
[tex]\Longrightarrow \frac{\frac{5}{3}=-k(47.5-T_a)}{1=-k(42.5-T_a)} \\\\\Longrightarrow\frac{5}{3}=\frac{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}[/tex]
Thus, the ambient temperature is 35°C.
(5) - Find the value of "k" using either of the two previous equations
[tex]1=-k(42.5-T_a)\\\\\Longrightarrow 1=-k(42.5-35)\\\\\Longrightarrow 1=-7.5k\\\\\Longrightarrow \boxed{ k \approx -0.133}[/tex]
(6) - Now finding "dt"
[tex]\text{Avg temp =} \ \frac{40+35}{2}=\boxed{37.5 \textdegree C}\\\\\Longrightarrow \frac{40-35}{dt}=0.133(37.5-35) \\\\\Longrightarrow \frac{5}{dt}=0.3325\\\\\therefore \boxed{dt \approx15 \ min}[/tex]
Thus, it take 15 minutes to cool the substance from 40°C to 35°C.