You place a cup of $210^{\circ} F$ coffee on a table in a room that is $68^{\circ} F$, and 10 minutes later, it is $200^{\circ} F$. Approximately how long will it be before the coffee is $180^{\circ} F$ ? Use Newton's law of cooling:
$T(t)=T_A+\left(T_o-T_A\right) e^{-k t}$
A. 1 hour
B. 15 minutes
C. 33 minutes
D. 45 minutes
Answer (2)
Use Newton's Law of Cooling to model the temperature change: T ( t ) = T A + ( T o − T A ) e − k t .
Find the constant k using the given data: 200 = 68 + ( 210 − 68 ) e − 10 k , which gives k = − 10 1 ln ( 71 66 ) .
Set T ( t ) = 180 and solve for t : 180 = 68 + ( 210 − 68 ) e − k t , resulting in t = 10 l n ( 71 66 ) l n ( 71 56 ) .
Calculate t to be approximately 33 minutes: 33 minutes .
Explanation
Understanding the Problem and Given Information We are given Newton's Law of Cooling: T ( t ) = T A + ( T o − T A ) e − k t , where: T ( t ) is the temperature at time t ,
T A is the ambient temperature, T o is the initial temperature, and k is a constant.
We are given: T o = 21 0 ∘ F T A = 6 8 ∘ F T ( 10 ) = 20 0 ∘ F
We want to find the time t when T ( t ) = 18 0 ∘ F .
Finding the Constant k First, we need to find the constant k using the information given for T ( 10 ) = 20 0 ∘ F :
200 = 68 + ( 210 − 68 ) e − 10 k 200 = 68 + 142 e − 10 k 132 = 142 e − 10 k e − 10 k = 142 132 = 71 66 − 10 k = ln ( 71 66 ) k = − 10 1 ln ( 71 66 )
Setting up the Equation to Solve for t Now we want to find the time t when T ( t ) = 18 0 ∘ F . We substitute the value of k we just found: 180 = 68 + ( 210 − 68 ) e − k t 180 = 68 + 142 e − ( − 10 1 l n ( 71 66 )) t 112 = 142 e 10 t l n ( 71 66 ) 142 112 = 71 56 = e 10 t l n ( 71 66 )
Isolating t Take the natural logarithm of both sides: ln ( 71 56 ) = 10 t ln ( 71 66 ) t = 10 l n ( 71 66 ) l n ( 71 56 )
Calculating t Now, we calculate the value of t :
t = 10 l n ( 66/71 ) l n ( 56/71 ) ≈ 10 × − 0.0725 − 0.2369 ≈ 32.4995 So, t ≈ 32.5 minutes.
Final Answer The closest answer from the given options is 33 minutes.
Examples
Newton's Law of Cooling is used in various real-world applications, such as determining the time of death in forensic science. By measuring the body's temperature and knowing the ambient temperature, investigators can estimate how long ago the person died. It's also used in cooking to predict how long it will take for food to cool down to a safe eating temperature, or in engineering to design cooling systems for electronic devices.
By using Newton's Law of Cooling, the time it takes for the coffee to cool from 200°F to 180°F is approximately 33 minutes. This was calculated using the formula and values provided. Therefore, the correct answer is option C.
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