Water Cooling
What would be the equation for the cooling of water using calculus?
I am doing a calculus project and i have to make a safety video for how to drink tea at a safe temperature. I have no idea where to start to find an equation or formula for the how fast water cools. Please help me!
Newton’s Cooling Law is a simple differential equation that can express temperature as a function of time, T(t), once the initial temperature of the object T_0 is known (in this case tea), the ambient temperature (temp of room) Ta is known, and a temperature reading later on is known.
Newton’s law of cooling says that the rate of temperature decrease is proportional to the difference of the ambient temp and object temp:
dT/dt is proportional to (T-Ta) where room temp Ta is constant and object temp T is a variable. Rewrite as:
dT/dt = -k(T-Ta), k is the constant of proportionality (negative since water is cooling).
Multiply eachside by dt:
dT = -k(T-Ta) dt
dT/(T-Ta) = -k dt
∫dT/(T-Ta) = ∫-k dt
ln(T-Ta) = -kt + c
e^ln(T-Ta) = e^(-kt + c)
T-Ta = e^c*e^(-kt)
T-Ta = c*e^(-kt) [e^c can be any constant c]
T(t) = Ta + c*e^(-kt)
Notice there are 2 unkown constants c and k. You must use boundary conditions to solve for them.
At t=0, the Temperature of the object is some temperature T0
T(0) = T0 = Ta + c*e^(-k*0)
T(0) = T0 = Ta + c
c = T0-Ta
The final equation is:
•••••••••••••••••••••••••••••
T(t) = Ta + (T0-Ta)*e^(-kt)
•••••••••••••••••••••••••••••
Ex. Say initial tea temp = 150 degrees and room temp = 60 deg.
To solve for k, take the temperature of the tea at some time. For example at t = 100 sec the tea temp is 115 degrees.
T(100) =115 = 60 + (150-60)*e^(-k*100)
Now you can solve for k and graph T(t) which can predict the temperature of the tea.
55 = 90*e^(-k*100)
55/90 = e^(-k*100)
ln(55/90) = -k*100
ln(55/90)/-100 = k
k = .004925
T(t) = 60 + 90*e^(-.004925t)
You can see at t=0, T(t) = 60+90=150
and as t approaches infinity T(t) = 60+90*0 = 60, which physically makes sense – the temperature of the tea can never fall below the room temperature.
If you want to know when the tea is a certain temp, then T, Ta, T0 are known, and t is unknown:
T = Ta + (T0-Ta)*e^(-kt)
(T-Ta)/(T0-Ta) = e^(-kt)
ln[(T-Ta)/(T0-Ta)] = -kt
ln[(T-Ta)/(T0-Ta)]/-k = t
Time for tea to reach temperature T:
t = 1/k*ln[(T0-Ta)/(T-Ta)]
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