Newton's Cooling Law
Newton’s Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, If we let be the temperature of the object at time and be the temperature of the surroundings, then we can formulate Newton’s Law of Cooling as a differential equation:
If a bottle of soda pop at room temperature is placed in a refrigerator where the temperature is and after half an hour the soda pop has cooled to , then what is the temperature of the soda pop after another half hour?
*Give answer in Fahrenheit *
Use Floor function to round answer
The answer is 54.
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1 solution
Initial Temperature difference
is Temp. of refrigerator - Temp. of pop= 28.
Then, initially at t=0
Let
Ts=temp. of refrigerator,
Ti =initial temp. of pop,
To be the temp. difference.
T i − T s = T o = 2 8 ∗ e ( k ∗ 0 ) = 2 8 .
Tf-Ts=To e^(k t)
Tf=61
Tf-Ts=61-44=17
Tf=To e^(k 30) , I took time in minutes.
solving for k we get k = − 0 . 0 1 6 6 3 approx
Then let T1 be the temperature after 1 hour,
It follows T 1 − T s = T o ∗ e ( − 0 . 0 1 6 6 3 ∗ 6 0 )
T 1 = T s + T o ∗ e ( − . 9 9 7 8 )
T1=44+ 28*0.3686
T1=44+10.3233
T1=54 approx.