newton's law of cooling: story problem with decreasing ambient temperature

[kangsang24]

I have been ripping my hair out over this problem!! Please help someone!

A cup of hot coffee initially at 90 C cools to 80 C in five minutes while sitting in a room of temperature 20 C. Determine when the temperature of the coffee is 50 C given that the room temperature is decreasing at the rate of 1 C.

How do you alter T = (To-M) e^(-kt) + M to accomodate the decreasing room temperature??? I was given the hint dT/dt = K[20 - t -T]. Why dT/dt? I figured you would need to find dM/dt since its the rate of the ambient temperature M that is decreasing.

 

[stapel]

I have been ripping my hair out over this problem!! Please help someone!

A cup of hot coffee initially at 90 C cools to 80 C in five minutes while sitting in a room of temperature 20 C. Determine when the temperature of the coffee is 50 C given that the room temperature is decreasing at the rate of 1 C.

How do you alter T = (To-M) e^(-kt) + M to accomodate the decreasing room temperature??? I was given the hint dT/dt = K[20 - t -T]. Why dT/dt? I figured you would need to find dM/dt since its the rate of the ambient temperature M that is decreasing.

What are the definitions of the variables (and constants?) listed in your post; namely, T, T₀, K, k, t, and M? Thank you!

 

[kangsang24]

T is the target temperature, To is the initial temperature, 1/k is the time constant (i got k to equal .0308 from the first part of this question that didnt have the decreasing ambient temperature. I am assuming that is able to be plugged into this problem) t is time in minutes, M is ambient temperature. The answer is 22.39 minutes. Big K meant to be little k btw.

1) T=(To-M)e^-kt +M is derived from 2) dT/dt=k(T-M). Then we were given the hint 3) dT/dt = k(20-t-T). I am assuming equation 3) is modified from equation 2).

 

[Ishuda]

I have been ripping my hair out over this problem!! Please help someone!

A cup of hot coffee initially at 90 C cools to 80 C in five minutes while sitting in a room of temperature 20 C. Determine when the temperature of the coffee is 50 C given that the room temperature is decreasing at the rate of 1 C.

How do you alter T = (To-M) e^(-kt) + M to accomodate the decreasing room temperature??? I was given the hint dT/dt = K[20 - t -T]. Why dT/dt? I figured you would need to find dM/dt since its the rate of the ambient temperature M that is decreasing.

Newton's law basically states the equivalent of "The rate of cooling (dT/dt) is proportional to the temperature (T) difference between an object and the ambient temperature M". Thus
dT/dt = K (M-T)
where K is a positive constant. Assuming T in degrees C and t in hours, then if M is initially at M₀ and decreases at a rate of a degree per hour and the objects temperature at time zero is T₀ then
M(t) = M₀ - a t,
dT/dt + K T = K M(t)
T(0) = T₀