How Does Newton's Law of Cooling Relate to Temperature Differences in Iced Tea?

[phil ess]

Homework Statement

If you remove a glass of iced tea from the refrigerator, and leave it sitting on the table
in a room with temperature 25 C, then Newton’s Law of Temperature Change gives
the temperature of the tea after t minutes as

f(t) = 25 − Ce^−kt

where C and k are constants. Show that the rate of change of f(t) is proportional to
the difference between the temperature of the tea, and that of the room, at every time t.

Homework Equations

That one

The Attempt at a Solution

OK I know the rate of change is the derivative. So f'(t) = Cke^-kt right? But how am I supposed to show that f'(t) is proportional to the difference mathematically?

 

[Mark44]

You need to show that this ratio is constant:

f'(t) / [(Temp of tea at time t) - (Temp of room)] = a constant

You know what the temperature of the tea at any time is, right?
You are given the room temperature.
For the constant, don't use something other than C or k, since you're already using them.

 

[HallsofIvy]

Homework Statement

If you remove a glass of iced tea from the refrigerator, and leave it sitting on the table
in a room with temperature 25 C, then Newton’s Law of Temperature Change gives
the temperature of the tea after t minutes as

f(t) = 25 − Ce^−kt

where C and k are constants. Show that the rate of change of f(t) is proportional to
the difference between the temperature of the tea, and that of the room, at every time t.

Homework Equations

That one

The Attempt at a Solution

OK I know the rate of change is the derivative. So f'(t) = Cke^-kt right? But how am I supposed to show that f'(t) is proportional to the difference mathematically?

The temperature of the tea is 25- Ce^-kt and the temperature of the room is 25. What IS the difference in temperatures?