Question

a bottle of water with temperature 70 F is placed in a refrigerator with an internal temperature of 34 F. After 30 minutes, the temperature of the water has decreased to 52 F

a)Give the cooling constant k(6th decimal place), and find an expression for the temperature of the water as a function of time.

b)What is the temperature of the water after 40 minutes.

c)After what amount of time will the temperature of the water reach 38 F.

I just wanna know the steps of each part and what to plug in.

I think the first part(a) uses :
52-34=(70-34)e^(k*30)

Although, I think "52-34" is wrong.

please help

Answer 1

for part a I got k = -.023105, if this isn't correct, can someone explain what I did wrong?

how do I now find p(t)?

Answer 2

I strongly suggest you listen to the two videos by Khan

In the first he establishes the formula:
T(t) = c e^(-kt) + T_a , where T_a is the ambiant temperature.

your given data will be:
T_a = 34
T(0) = 70
T(30) = 52
your t will be in minutes

The first video results in the basic equation, and you may actually skip it
https://www.khanacademy.org/math/differential-equations/first-order-differential-equations/modeling-with-differential-equations/v/newtons-law-of-cooling

In the second video, it follows your problem quite well, all you have to do is plug in your values
https://www.khanacademy.org/math/differential-equations/first-order-differential-equations/modeling-with-differential-equations/v/applying-newtons-law-of-cooling-to-warm-oatmeal