Newton's law of cooling proposes that the rate of change of temperature is proportional to the temperature difference to the ambient (room) temperature.

Question

(continuation of title) This can be modelled using the equation dT/dt=-k(T-T_a). This can also be written as dT/T-T_a=-k dt.

Where

T=Temperature of material

T_a=Ambient (room) temperature

k =A cooling constant

a) Integrate both sides of the equation and show that the temperature difference is given by:

(T-T_a)=C_0e^(-kt)

b)Calculate C_0 if the initial temperature is 70 degrees Celsius and T_a= 20 degrees Celsius .

Answer 1

Hi Mason!

Let me try and help you out with this problem!

Starting Equation (Newton's Law of Cooling):

[dT / dt] = -k * [T - T_a]

...where T is the current temperature of our sample, T_a is the temperature of my ambient environment (oftentimes equal to room temperature), t is how much time has passed since we started "measuring" the temperature, and k is some constant value which is a kind of description of how quickly the environment can "suck away" heat from the sample (for example, we might expect different "k-values" depending on if the environment is "windy" or not!).

This equation is currently in the form of a derivative, and we are asked to integrate it so that we obtain an expression for "Temperature over Time".

Firstly, let's multiply both sides by "dt"...

[dT / dt] = -k * [T - T_a]

dT = -k * [T - T_a] * dt

Before we integrate each side, we need to make sure that all the "T-terms" are on the "dT side" (and, if there were any "t-terms", they'd need to be on the "dt side").

dT = -k * [T - T_a] * dt (Divide both sides by "[T - T_a]")

(1 / [T - T_a]) * dT = -k * dt

Now we can do indefinite integration on each side of this equation...

Note that integrating "1 / x" type expressions turns into natural logarithms, and that the only thing in the right integral is "k", which is a constant value with respect to time...

∫ (1 / [T - T_a]) * dT = ∫ -k * dt

ln( T - T_a ) = -kT + C₀

Remember that since we are doing indefinite integration, we get a +C₀ from each integral ... but since its some unknown constant we can just imagine combining them into one single +C₀ for simplicity...

Now, we want to solve for T in terms of t. To "break T out of the logarithm", we need to do e^ of each side of the equation...

ln( T - T_a ) = -kt + C₀ (exponentiate each side...)

T - T_a = e^{(-kt + C}0⁾ (use exponential rules to split the 1 exponential into two...)

T - T_a = e^-kt * e^C0 (C₀ is just some arbitrary unknown constant number, so e^C0 is ALSO just some arbitrary unknown constant number! Let's just "rename" this term C₀ for convenience!)

T - T_a = e^-kt * C₀

T - T_a = C₀e^-kT

T = C₀e^-kt + T_a

But, how can we evaluate this mystery number C₀? We need to use some information from the word problem to do this (we need to know at least ONE data point of Temperature and Time...)

In part (b) they tell us that the initial temperature (initial meaning t = 0) is 70C and the ambient temperature is 20C.

Plug these numbers into our equation...

T = C₀e^-kt + T_a

(70) = C₀e^-k(0) + (20)

(70) = C₀e⁰ + (20) (e⁰ = 1)

(70) = C₀ + (20)

C₀ = 50 C

Hope that helps!

--Zach