Newtons law of cooling

Question

Sir Isaac Newton investigated the cooling process of water temperature and showed that the temperature change is exponential in time and can be predicted by
Tdiff=T0e^-kt
Where Tdiffis the difference in temperature between the water and the air in the room,
T0 is the initial difference in the temperatures (Tt=0 – T room)
So if T is temperature of he water any time, the equation can be re-written as
T-T room=(Tt=0-Troom)e^-kt

Where k is constant of proportionality and t is time.

Use the equation to solve this problem:

A hard boiled egg at 98°C is put in a sink of 18°C water to cool. After 5 minutes, the eggs temperature is found to be at 38°C. Assuming that the water has not warmed up, how much longer will it take the egg to reach 20°C?

 

PLEASE HELP ME :D

Answer 1

use the information at 5 minutes to calculate the value for k (constant).

 

Then use the value for k to determine the time it will take to reach the final temp (20 C)

 

38=18+(98-18)e^-k(5)

20=80e^-k(5)

ln(20/80)=-k(5)

k=.2773

 

20=18+(98-18)e^-.2773(t)

2=80e^-.2773(t)

ln(2/80)=-.2773(t)

t=13.303

 

Hence 13.3-5=8.3

 

So it will take an additional 8.3 minutes for the egg to reach 20C

Answer 2

(T-T_water) = (T_0-T_water)e^-kt   

 

The first step is to use the data about how much the egg has cooled after 5 minutes to find the value of k, which is unknown.

 

(38-18) = (98-18)e^-5k

20 = 80e^-5k

20/80 = 1/4 = e^-5k

ln(1/4) = -5k

ln(1/4)/(-5) = k

0.277 ≅ k

 

Now that we have the value of k, we can find how much longer it will take the egg to cool to 20^oC.  In this case, the egg has already cooled to 38^oC, so the starting or initial temp is 38^oC, not 98^oC.

 

(20-18) = (38-18)e^-(0.277)t

2 = 20e^-(0.277)t

2/20 = 1/10 = e^-(0.277)t

ln(1/10) = -(0.277)t

ln(1/10)/(-0.277) = t

8.3 minutes ≅ t