Help Finding Final Temp of Mixing Water Together

AI Thread Summary
To find the final temperature when mixing 1 liter of 20°C water with 4 liters of 40°C water, the conservation of energy principle is applied. The equation q = mc∆t is used, where the heat lost by the hot water equals the heat gained by the cold water. By setting up the equation 1(T-20) = 4(40-T), the specific heat 'c' cancels out, simplifying the calculation. Rearranging the equation leads to the final temperature, which is determined to be 32°C. This method effectively demonstrates the balance of heat transfer between the two water volumes.
absolutezero
Messages
2
Reaction score
0

Homework Statement



What would be the final temp if you mixed 1 liter of 20°C water with 4 liters of 40°C water?

This question is very straightforward, and is, of course, assuming total conservation of temp.
(given mass m1 of water at a temperature T1 and mass m2 of water at T2, mix them together)

Homework Equations



My book mentions q=mc∆t, but using that equation, I get calories as the units. And this equation I barely know how to use.

The Attempt at a Solution



My guess is the answer is 38.75 °C, but I figured that by reasoning only, as I have no clue what equation to use. I wondered if it might be a proportions prob, but haven't gotten a logical answer that way. I know the answer has to be over 20°C but under 40°C.

I am very very confused, so a clear answer with an good explanation would be the best! Thanks very much!
 
Physics news on Phys.org
absolutezero said:
My book mentions q=mc∆t, but using that equation, I get calories as the units. And this equation I barely know how to use.
The answer depends on the units of 'c'

My guess is the answer is 38.75 °C,
Not a very good guess

I know the answer has to be over 20°C but under 40°C.
A good start

Using the equation above we know that energy q=mc∆t,
Imagine the final temperature is T, we have a change in temperature of (T-20) for the cold water and (40-T) for the hot water - note the way around for these since one is going up and one is going down - we end up with 5litres at T.

We don't need to know 'c' and we don't need to worry about the mass of 1litre because that is all going to cancel, and we don't care about the actual value of q.

q = 1 c (T-20) + 4 c (40-T) = 5 c T

All you have to do is rearrange to find T.
 
Thank you very much for replying! :smile: Please forgive me, as I understand only a speck of physics.

This makes sence:
The answer depends on the units of 'c'

What do you mean by this?
All you have to do is rearrange to find T.

Is it a matter of using the distributive property in algebra?

q = 1 c (T-20) + 4 c (40-T) = 5 c T
q = 1cT -20c + 80c -4cT = 5cT
q = 60c - 3ct = 5ct
q = 60c = 8ct
q = 7.5 = t

In one of my attempts to figure out the answer, I came up with 8, but assuming 8 is correct, what do I do with it? It's 8 what? 8 bananas? 8 oranges? Is something multiplied by a factor of 8? Or 1/8?

Is this even remotely in the right direction? (Knowing my luck, probably not. :redface: )

I am very sorry for my ignorance! :rolleyes: Thank you for your help! :smile:
 
Last edited:
absolutezero said:
What do you mean by this?
C is the conversion between mass and temperature to energy, it's normally in J/kg/k but you could have it in calories/g/k if you like.

Is it a matter of using the distributive property in algebra?
Sorry I wasn't very clear.

The heat lost by the hot water in cooling from 40 -> T equals the heat gained by the cold water going from 20->T.

So: 1 c (T-20) = 4c(40-T)
Cancels the 'c' and rearrange
 
Last edited:
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top