Can someone explain this equation to me? Δm=μΔt

[PurpleRain]

I don't understand how this equation came about and what μ means.
And from there they said that μ=Δm/Δt=constant

I have to use it to solve a problem about finding the final temperature of two liquids.

I tried to google the equation, but I couldn't find anything.
Can someone please explain this to me?

Edit: μ is the mass of the coffee/milk, t is time and m is probably their mass combined.

 

[mfb]

Some context would help. What is m? Is t a time, a temperature, or something else?
Something (m) changes linearly in something else (t), and μ is defined as the rate of this change.

 

[PurpleRain]

Well...I don't know what is m, t, and μ. They only gave us this formula and didn't say anything else. That's why I'm confused. I think m is mass, though.

Edit: And t is probably time, since θ was used for temperature in the problem

 

[mfb]

Okay, forget the formula for now, undefined variables don’t help. What is the problem you are trying to solve?

 

[PurpleRain]

Ah...sorry. I just read the problem again, μ is the mass of milk/coffee. M is probably their mass combined (?)

I have a problem with a coffee machine that has two coaxial tubes. A tube has a length of 5 cm. It says that the milk flows through the space between the two tubes and the coffee through the central tube, in opposite directions. The milk enters with a temperature θ1= 10 °C, and the coffee with a temperature θ2= 90 °C.
It is known that if in the time unit through this device circulates in the same direction the same mass of liquid μ, then when it exists the machine, the milk warms to θ3 = 60 ° C. It is considered that, in a stationary state, the flow of the heat is the same everywhere.
At a) we had to find θ4, which is the final temperature of the coffee. θ4= 40 °C.
At b) we had to find s-the distance between the sections of the tubes where the temperature of the milk is equal to the temperature of the coffee. s=3m.
At c) we have to find θ3 and θ4 if the speed doubles.Edit: Another question. As an observation, it says that Δθ=(θ2-θ1-Δθ1). Why is that? Shouldn't it be only θ2-θ1?

 

[mfb]

It is known that if in the time unit through this device circulates in the same direction the same mass of liquid μ, then when it exists the machine, the milk warms to θ3 = 60 ° C.

I don't understand the grammar of this statement.

At b) we had to find s-the distance between the sections of the tubes where the temperature of the milk is equal to the temperature of the coffee. s=3m.

cm? And what does that mean?

As an observation, it says that Δθ=(θ2-θ1-Δθ1). Why is that? Shouldn't it be only θ2-θ1?

That depends on where the equation is used and what it calculates.
Formulas without context are useless.

 

[Chestermiller]

Ah...sorry. I just read the problem again, μ is the mass of milk/coffee. M is probably their mass combined (?)

I have a problem with a coffee machine that has two coaxial tubes. A tube has a length of 5 cm. It says that the milk flows through the space between the two tubes and the coffee through the central tube, in opposite directions. The milk enters with a temperature θ1= 10 °C, and the coffee with a temperature θ2= 90 °C.
It is known that if in the time unit through this device circulates in the same direction the same mass of liquid μ, then when it exists the machine, the milk warms to θ3 = 60 ° C. It is considered that, in a stationary state, the flow of the heat is the same everywhere.
At a) we had to find θ4, which is the final temperature of the coffee. θ4= 40 °C.
At b) we had to find s-the distance between the sections of the tubes where the temperature of the milk is equal to the temperature of the coffee. s=3m.
At c) we have to find θ3 and θ4 if the speed doubles.Edit: Another question. As an observation, it says that Δθ=(θ2-θ1-Δθ1). Why is that? Shouldn't it be only θ2-θ1?

This sounds like a counter-current heat heat exchanger formed by two concentric tubes. But your grammar is hard to understand. We need to know the flow rates of the two fluids.

 

[PurpleRain]

Oh, sorry. It's quite hard to translate from German to English, and the original wording is just as confusing. At one point I just used google translate, because I got a bit lazy.
Anyways, I'll reword the problem, without using Google Translate this time. I hope it will be a bit easier to understand.

A coffee machine that has two coaxial tubes. A tube has a length (L) of 5 cm (centimeters). The milk flows through the space between the two tubes, while the coffee flows through the central tube. They flow in opposite directions. The milk enters (It doesn't say where in the original text) at a temperature (θ1) of 10 °C (Celsius). The coffee enters the tube at a temperature (θ2) of 90 °C. If in the time unit the same mass of liquid μ circulates in both directions, then when the milk exists the machine it (the milk) warms to the temperature (θ3) of 60 ° C. In a stationary state, the flow of the heat is the same everywhere.

The machine doesn't exchange heat with the environment. The density and the specific heat of the coffee and the milk are considered to be the same.
Φ=Q/Δt, where Q=heat, t=time.That's all it says.

 

[mfb]

You can post the German problem statement in addition, maybe I can help translating.

The flow rates are not specified.

That is okay, as you know the relative flow rate ("the same mass of liquid μ circulates in both directions"). You can also assume that both liquids have the same specific heat capacity.

 

[Chestermiller]

This is a thermodynamics problem in which you are supposed to apply the open system version of the first law of thermodynamics to a control volume (the two pipes) operating at steady state. The parameter $\mu$ is the mass flow rate of the coffee and of the milk flowing through each of the pipes, and has units of kg/s. The form of the first law applicable to this situation is $$\sum{h_{in}}-\sum{h_{out}}=0$$ where $h_{in}$ is the enthalpy per unit mass of each inlet stream, and $h_{out}$ is the enthalpy per unit mass of each outlet stream. So, $$\sum{h_{in}}=\mu C (\theta_1-\theta_{ref})+\mu C(\theta_2-\theta_{ref})$$and $$\sum{h_{out}}=\mu C (\theta_3-\theta_{ref})+\mu C(\theta_4-\theta_{ref})$$ where $\theta_{ref}$ is an arbitrary reference temperature (which cancels out of the equation) and C is the heat capacity of water.

It isn't clear what part (b) is asking.