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.

Use Newton’s Law of Cooling to write a formula that models this situation.

Answer 1

Consider the surrounding air temperature in the refrigerator is 350 degrees and the water 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 – 71 = A

            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.

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