Question

You are driving on a hot day when your car overheats and stops running. The car overheats at 280°F and can be driven again at 230°F. When it is 80°F outside, the cooling rate of the car is r=0.0058​ . How long do you have to wait until you can continue driving? Use Newton's Law of Cooling to solve the problem. Round your answer to the nearest whole minute.

Answer 1

Answer:

50 minutes

Step-by-step explanation:

Given:

Temperature, T = 230°F

Overheat temp, To = 280°F

Ts = 80°F

Cooling rate, r = 0.0058

Since the car can be driven again at 230°F, and when it is 80°F outside, the cooling rate of the car is 0.0058, using Newton's law of cooling, we have:

$T = (T_o - T_s)​ e^-^r^t + T_s$

Substituting figures, we have:

$230 = (280 - 80)​ e^-^0^.^0^0^5^8^t + 80$

$230 = 200e^-^0^.^0^0^5^8^t + 80$

Solving further, let's subtract 80 from both sides, we now have:

$230 - 80= 200e^-^0^.^0^0^5^8^t + 80 - 80$

$150 = 200e^-^0^.^0^0^5^8^t$

Divide both sides by 200:

$\frac{150}{200}= \frac{200e^-^0^.^0^0^5^8^t}{200}$

$= \frac{150}{200}=e^-^0^.^0^0^5^8^t$

Let's take the natural log of both sides, we have:

$In 0.75 = In e^-^0^.^0^0^5^8^t$

Using logarithm power rule, we have:

= - 0.2877 = - 0.0058t

To find t, let's now divide both sides by -0.0058

$\frac{-0.2877}{-0.0058} = \frac{-0.0058t}{-0.0058}$

50 = t

Therefore, you have to wait for 50 minutes before you can continue driving.