Question

A cup of coffee cools at rate proportional to the difference between the constant room temperature of 20.0 degree C and the temperature of the coffee. If the cooling rate is determined to be 2.1% per minute and the current temperature of the coffee is 74.4 degree C, what will the temperature of the coffee be 28.0 minutes from now?

Answer 1

50.0 degrees

Explanation:

The temperature of the coffee for any given instant "t", in minutes, can be described by the following expression:

`T = R+(I-R)*(1-r)^t`

Where 'R' is the room temperature, 'I' is the initial temperature of the coffee, and 'r' is the cooling rate. Applying the given values, for t =28 minutes:

`T = 20+(74.4-20)*(1-0.021)^t\\T= 50.0`

The temperature of the coffee will be 50 degrees 28.0 minutes from now