Question

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!

Newton’s law of cooling states that for a cooling substance with initial temperature T^0 , the temperature T(t) after t minutes can be modeled by the equation T(t) = T^s+(T^0−T^s)^e−kt , where T^s is the surrounding temperature and k is the substance’s cooling rate.

A liquid substance is heated to 80°C . Upon being removed from the heat, it cools to 60°C in 15 min.

What is the substance’s cooling rate when the surrounding air temperature is 50°C ?

Round the answer to four decimal places.

A. 0.0687

B. 0.0732

C. 0.0813

D. 0.0872

Answer 1

$T(t) = T_s + (T_0 - T_s)e^{-kt}$

$T(t) - T_s = (T_0 - T_s)e^{-kt}$

$\dfrac{ T(t) - T_s}{T_0 - T_s} = e^{-kt}$

$\ln \dfrac{ T(t) - T_s}{T_0 - T_s} = -kt$

$k = - \dfrac{ \ln \dfrac{ T(t) - T_s}{T_0 - T_s} }{t}$

OK now let's plug in the numbers.

We have t=15 minutes, T_0 = 80 C, Ts = 50, T(15)=60 C,

$k = - \dfrac{ \ln \dfrac{ 60 - 50}{80 - 50} }{15 \textrm{ min}}$

$k = - \frac{1}{15} \ln \frac 1 3$

$k = \frac{1}{15} \ln 3$

$k \approx 0.07324$ per minute

Answer: B

Answer 2

Answer: B

Step-by-step explanation:

Just a verification of the anwser above.