By Newton's Law of Cooling, an object in air cools down at a rate proportional to the difference between the object's temperature and the air temperature. When the air temperature is C, it takes 10 min for the object to cool down from C to C. The temperature of the object decreases from C to C during:
A) 30.21 min
B) 22.37 min
C) 16.67 min
D) 19.10 min
E) 6.66 min

Question

By Newton's Law of Cooling, an object in air cools down at a rate proportional to the difference between the object's temperature and the air temperature. When the air temperature is  C, it takes 10 min for the object to cool down from  C to  C. The temperature of the object decreases from  C to  C during:
A) 30.21 min 
B) 22.37 min 
C) 16.67 min 
D) 19.10 min 
E) 6.66 min

Quizplus · Accepted Answer

Let $T$ be the temperature of the object at time $t$ and let $A$ be the air temperature. Then the differential equation that models this situation is $$\frac{d T}{dt} = k(T-A).$$

We know that $T(0) = 11eaa655_3b16_2838_a696_3320595a6155_TB5596_11$ and $T(10) = 11eaa655_3b16_2839_a696_eb9226b7bc57_TB5596_11$. Plugging in these values, we can solve for $k$: $$k = \frac{1}{T(0)-A}\ln\left(\frac{T(0)-T(10)}{T(0)-A}ight) \approx \frac{1}{8.7}\ln(0.25) \approx -0.029.$$

Thus, the solution to the differential equation is $$T(t) = (T(0)-A)e^{kt} + A.$$

Using this formula, we can find how long it takes for the object to cool from $11eaa655_3b16_2838_a696_3320595a6155_TB5596_11$ to $11eaa655_3b16_2839_a696_eb9226b7bc57_TB5596_11$: $$11eaa655_3b16_2839_a696_eb9226b7bc57_TB5596_11 = (11eaa655_3b16_2838_a696_3320595a6155_TB5596_11-A)e^{-0.029t} + A.$$ Solving for $t$, we get $$t \approx 22.37.$$

Therefore, the answer is $\boxed{	extbf{(B)}}$.

By Newton's Law of Cooling, an Object in Air Cools