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

Answer: B

Explanation:

Just a verification of the anwser above.

Answer 2

`T(t) = T_s + (T_0 - T_s)e^(-kt)`

`T(t) - T_s = (T_0 - T_s)e^(-kt)`

`( T(t) - T_s)/(T_0 - T_s) = e^(-kt)`

`\ln ( T(t) - T_s)/(T_0 - T_s) = -kt`

`k = - ( \ln ( 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 = - \frac{ \ln ( 60 - 50)/(80 - 50) }{15 \textrm{ min}}`

`k = - (1)/(15) \ln \frac 1 3`

`k = (1)/(15) \ln 3`

`k \approx 0.07324` per minute

Answer: B