Question

Temperature Mixing:

In this problem we will build a model for mixing problems with dif-
ferent temperatures of water. Throughout we will assume that mixing
happens instantaneously and no heat is lost to the surroundings.

(a) Suppose we have v liters of water at temperature T in an urn and
we pour in u liters of water at temperature S. What is the tem-
perature of the mixture? (This is going to be a weighted average.)

b. Let v(t) denote the volume of the water at time t, T(t) denote the temperature at time t. Suppose after Gt seconds Gu gallons of water are added to the urn. Compute T(t+Gt)-T(t). Now assume the water is pouring in at a constant rate and temperature and use the limit definition of the derivative to compute dT/dt in term of du/dt,S,T and v(t)

Answer 1

Alright, let's take it step by step!

(a) When you mix water with different temperatures, the final temperature is like a weighted average. Imagine you have `v` liters of water at temperature `T` and `u` liters of water at temperature `S`. The amount of thermal energy in the first batch is `v*T` and in the second batch it's `u*S`. When you combine them, the total thermal energy is `v*T + u*S`. Since the total volume is now `v + u`, the average energy per liter (which is the final temperature) is `(v*T + u*S) / (v + u)`.

In equation form:

Final Temperature, F = (v*T + u*S) / (v + u).

(b) Now let's move to the changing volumes and temperatures. Let `v(t)` be the volume at time `t`, and `T(t)` the temperature at time `t`. Let's say that in `Gt` seconds, `Gu` gallons of water are added at temperature `S`. We’ll assume that 1 gallon is the same as 1 liter for simplicity, though in reality they are slightly different.

The new volume after `Gt` seconds is `v(t) + Gu`, and the total thermal energy is `v(t)*T(t) + Gu*S`. The new average temperature is:

T(t+Gt) = (v(t)*T(t) + Gu*S) / (v(t) + Gu).

Now, T(t+Gt) - T(t) = [(v(t)*T(t) + Gu*S) / (v(t) + Gu)] - T(t).

Now, let's think about water pouring at a constant rate. Let's use the limit definition of the derivative. Instead of `Gu` gallons in `Gt` seconds, let's say a tiny amount of water `du` is added in a tiny amount of time `dt`. So, `du/dt` is the rate at which water is poured into the urn.

Using the limit definition:

dT/dt = lim (dt -> 0) [(v(t)*T(t) + du*S) / (v(t) + du) - T(t)] / dt

= [(v(t)*T(t) + du*S) / (v(t) + du) - T(t)]' (derivative with respect to t)

= [v'(t)*T(t) + v(t)*T'(t) + du/dt*S - v'(t)*T(t) - v(t)*T'(t)] / (v(t) + du) (using product rule)

= (du/dt*S) / (v(t) + du).

As dt approaches 0, du becomes very small, and thus we can ignore it in comparison to v(t), so:

dT/dt ≈ (du/dt*S) / v(t).

This is the rate of change of temperature with respect to time, in terms of the rate at which water is poured, the temperature at which it is poured, and the volume of water already in the urn.