Question in description

Question

By Newton’s law of cooling, the instantaneous rate of cooling is proportional to the difference

between the temperature of the coffee and the outside temperature. That is, dT/dm = -k(T-T_O)

where k is a positive constant, T_O is the outside temperature, dT/dm is the instantaneous rate of

cooling, and m is the time in minutes. Using the formula above and your results from

Question 5, calculate the outside temperature and the value of k.

In question 5 I was asked to find the instantaneous rate of change for two different slopes at the same point in time that are on the same graph. For slope 1, the instantaneous rate of change was -1.24 at a temperature of 60 degrees and at a time of 5 minutes. For slope 2, the instantaneous rate of change was -2.07 at a temperature of 65 degrees and at a time of 5 minutes.

Answer 1

By Newton’s law of cooling, the instantaneous rate of cooling is proportional to the difference

between the temperature of the coffee and the outside temperature. That is, dT/dm = -k(T-T_O)

where k is a positive constant, T_O is the outside temperature, dT/dm is the instantaneous rate of

cooling, and m is the time in minutes. Using the formula above and your results from

Question 5, calculate the outside temperature and the value of k.

In question 5 I was asked to find the instantaneous rate of change for two different slopes at the same point in time that are on the same graph. For slope 1, the instantaneous rate of change was -1.24 at a temperature of 60 degrees and at a time of 5 minutes. For slope 2, the instantaneous rate of change was -2.07 at a temperature of 65 degrees and at a time of 5 minutes.

Using all the results above

-1.24 = -K ( 60 - T₀ ) and

-2.07 = -k ( 65-T₀ )

or 1.24 = K (60-T₀) and

2.07 = K (65 -T₀ )

Which gives

124/207 = (60-T₀)/ (65-T₀)

Gives T₀ = (60*207 -124*65)/ (207-124) =4360/83 = 52.53

So 1.24 = K (60 -52.53) ( By plugging in above Eq)

Gives K = 1.24/17.47