Differential Equations word problem help?
I think this one has to do with Newton’s Law of Cooling, but here it is:
A red wine is brought up from the wine cellar, which is a cool 10 degrees Celsius, and left to breathe at room temperature 23 degrees Celsius. If it takes 10 minutes for the wine to reach 15 degrees Celsius, when will the temperature of the wine reach 18 degrees Celsius?
Newtons law of colling states that the change of temperature of an object is proportional to the temperature difference to surrounding, i.e.:
dT/dt = k?(T_s – T)
For the wine T_s = 23
dT/dt = k?(23 – T)
with initial condition T(t=0) = 10
Proportionality constant k is not known, but you can calculate it the the boundary condition:
T(t=10) = 15
The differential is separable. So separate variables and integrate:
dT/dt = k?(23 – T)
<=>
1/(23 – T) dT dT = k dt
=>
? 1/(23 – T) dT = ? k dt
<=>
-ln(23 – T) = k?t + c
<=>
23 – T = e^(-k?t – c)
<=>
T = 23 – e^(-c)?e^(-k?t)
set C = e^(-c)
T = 23 – C?e^(-k?t)
Apply initial condition to find constant c
T(0) = 10
<=>
23 – C?e^(-k?0) = 10
<=>
23 – C = 10
<=>
C = 13
=>
T = 23 – 10?e^(-k?t)
Apply 2nd condition to find constant k
T(t=10) = 15
<=>
15 = 23 – 10?e^(-k?10)
<=>
e^(-k?10) = (23 – 15)/10
<=>
k = -ln(0.8)/10 = 0.0223 ( unit min?¹)
=>
T = 23 – 10?e^(-0.0223?t)
To get the time at which the temperature of the wine reach 18°C, set T=18 and solve for t:
18 = 23 – 10?e^(-0.0223??t)
<=>
e^(-0.0223??t) = (23 – 18)/10
<=>
t = -ln(0.5) / -0.0223 = 31.08…
The wine reaches 18°C after about 31 minutes.
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