Question

For the following exercises, use this scenario: A soup with an internal temperature of 350° Fahrenheit was taken off the stove to cool in a 71°F room. After fifteen minutes, the internal temperature of the soup was 175°F.

How many minutes will it take the soup to cool to 85°F?

Answer 1

Consider the temperature will decay exponentially towards 71 degree, that is;

T(t) = Aekt + 71

 

The initial temperature is 350. So, T(0) = 350 and substitute (0, 350) as follows:

350 = Aek(0) + 71

350 = Aek(0) + 71

    A = 279

 

After 15 minutes the temperature is risen to 175 degrees, T(15) = 175,

Then;

                   175 = 279ek(15) + 71

(175 – 71)/279 = ek(15)

               28/36 = ek(10)

             0.3728 = ek(15)

 

Take natural log on both sides as follows:

ln(0.3728) = ln{ek(15)}

ln(0.3728) = k(15)

                k = ln(0.3728)/15

                   = -0.066

Thus the equation for cooling of water is T(t) = 279e(-0.066)t + 71.

Time taken for the soup to cool to 85 degree is:

                   85 = 279e(-0.066)t + 71

(85 – 71)/279 = e(-0.066)t

     e(-0.066)5 = 0.05

Take natural log on both sides as follows:

ln(0.05) = ln(e-0.066t)

 -0.066t = ln(0.05)

            t = ln(0.05)/-0.066

              = 45

 

Therefore, it takes 45 minutes for the soup to cool.

Therefore, it takes 45 minutes for the soup to cool.