Before getting into the main topic, let me chat a bit about BoJone’s home...
As I have mentioned in some of my articles, I am a child from the countryside, and my home is currently in the countryside as well. Although our life cannot be described as "impoverished" and we have acquired many electrical appliances, we have never purchased a washing machine or a water heater. As for the washing machine, I feel that washing clothes by hand is quite good; at least it prevents me from being lazy. Regarding the water heater, since we live in the countryside, it is quite convenient to obtain firewood and straw. Furthermore, the stalks left after the rice harvest can also be used as fuel. We usually use firewood for cooking, so a water heater is really not very necessary. (Regrettably, biogas digesters have not become popular here, so please don’t blame me for emitting greenhouse gases... ^_^)
Since there is no water heater, we have to heat water manually. Usually, we boil a large pot of water, pour a basin full when taking a bath, and then add cold water to lower the temperature before washing up. The problem in this article originates precisely from this process of adjusting the water temperature. When the water is very hot, in order to speed up the cooling process, we often take a "two-pronged approach": simultaneously injecting cold water into the basin and letting hot water out of the basin. This leads to a question: what is the relationship between the water temperature and time?
To simplify this problem into a physical model, let us make the following assumptions:
A container holds water with an initial mass M_0 and an initial temperature T_0. Now, water at temperature T_1 is injected into the container at a rate of v_1 kg/s, and hot water is released from the container at a rate of v_2 kg/s. Assume that during this process, there is no heat loss to the air, and thermal equilibrium is achieved instantaneously upon injection. Furthermore, assume the thermal equilibrium only affects the water currently in the container (meaning the injection does not lower the temperature of the water already in the "drain pipe"). Find the relationship between the water temperature T and time t. (Temperature is in Kelvin).
Let the specific heat capacity of water be c. At time t, the mass of water in the container is M and the temperature is T. Thus, the thermal energy of the water in the container is: E = MTc After a time interval \Delta t, the change in water mass is \Delta M = (v_1 - v_2)\Delta t, and the change in thermal energy is: \Delta E = v_1 T_1 c \Delta t - v_2 T c \Delta t The change in water temperature is \Delta T = \frac{E + \Delta E}{(M + \Delta M)c} - T.
Expanding the above expression and neglecting second-order terms, we obtain: v_1 T \Delta t + M \Delta T = v_1 T_1 \Delta t Taking the limit, we have: \frac{dT}{dt} = \frac{v_1}{M}(T_1 - T)
It is not difficult to determine that the function for the water mass is M = M_0 + (v_1 - v_2)t. Substituting this in, we get: \frac{dT}{dt} = v_1 \frac{T_1 - T}{M_0 + (v_1 - v_2)t} \tag{1} Equation (1) is the differential equation we need to solve. This equation is quite simple; by separating the variables, we get: \frac{dT}{T_1 - T} = v_1 \frac{dt}{M_0 + (v_1 - v_2)t}
When v_1 = v_2, the right side becomes a constant, and we have: C - \ln(T_1 - T) = \frac{v_1}{M_0}t Here C is the integration constant. Using the initial condition t=0, T=T_0 and simplifying, we get: T = T_1 + (T_0 - T_1) \exp\left(-\frac{v_1 t}{M_0}\right) \tag{2}
When v_1 \neq v_2, the right side is a reciprocal of a linear function. Integrating gives: C - \ln(T_1 - T) = \frac{v_1}{v_1 - v_2} \ln[M_0 + (v_1 - v_2)t] Similarly, simplifying based on the initial conditions (would you allow me to leave this part of the work to you? ^_^), we get: T = T_1 + (T_0 - T_1) \left[1 + \left(\frac{v_1 - v_2}{M_0}\right)t\right]^{\left(\frac{v_1}{v_2 - v_1}\right)} \tag{3} One can observe that (2) and (3) are essentially the same, with (2) being a limit of (3).
Both solutions satisfy the following: 1) When T_0 = T_1, the temperature does not decrease; 2) When v_1 = 0, the temperature also does not decrease; 3) As t \to \infty, T \to T_1 (provided that v_1 > v_2).
The above is BoJone’s simple analysis of a problem from daily life. The model is quite crude, and corrections are welcome. Readers can and should apply the knowledge they have learned to understand life’s problems. In this regard, I believe one should "accumulate much and apply much, accumulate little and apply little," rather than just "accumulating much and applying little." Use the knowledge you have mastered to conduct research as much as possible.