what is the final temperature of coffee and mug once they come to thermal equilibrium?

benji

Here's the question:

You cascade 150 thousand of hot java at 86C into a 200 g glass cup at 22C. If they come to thermal equilibrium quickly, what is the last temperature?

I used Q=mc (delta)T.

So Q + Q = 0. I put the properties of the java in one and the properties of the glass in some other. Since I couldn't find a specific heat for coffee I just used that of water (I figure they're pretty similar?)...

Anyhow, for my terminal answer u git 39.42C when it should be something like 73C.

I don't understand this stuff besides well yet, then it'due south probably a logic mistake on my role, just if someone could explain that to me I'd really appreciate it. Cheers.

Also... (I figured I'd just add together on to this post instead of creating another thread...) How would I go almost doing this one:

Two hundred fifty grams of water at 80C is poured into a Styrofoam loving cup of negligible heat capacity containing 180 chiliad of water at 10C. After an additional 300 g of h2o is added to the loving cup, the mixture come up to an equilibrium temperature of 30C. What was the temperature of additional 300 g of water?

marlon

[tex]0.15c_{java}(86-10) = 0.2c_{glass}(x-22)[/tex]

Did you do this...ps what did you employ for c_glass ???

regards
marlon

marlon

The phylosophy is that the total amount of estrus is always conserved during the heath-send. So the coffee will give off rut and the drinking glass volition absorbe the heath until equilibrium has been reached...

Normally, T is in kelvin simply once y'all are talking near temperature differences, it doesn't thing what units you utilise as long as you use 1 unit all the fourth dimension...

regards
marlon

dextercioby

For the outset office you didn't give us the numbers,meaning the specific heat of drinking glass.The logics seems good,Hot coffee is hot h2o with a tiny amount of java (peradventure u like information technology without (also much)sugar,i hate coffe altogether),and then u migh have messed up either the computations,or the units for the concrete quantities.

And for the second problem,testify us what u did....

Daniel.

benji

For the get-go trouble this is what I did:

mc(deltaT) + mc(deltaT) = 0
.15(4187)(deltaT) + .ii(840)(deltaT) = 0

And then I merely solved for that.

For the 2nd problem I didn't do much, I but doing the exact same matter as I did in the first problem for the .25kg of water and the .18kg of water and and then I was planning on somehow interpreting that answer into some other equation with the .3kg of water and an equilibrium of 30C... But with the first part I ended up with -65719.152C and I knew I was way off...

dextercioby

For the get-go trouble this is what I did:
mc(deltaT) + mc(deltaT) = 0
.fifteen(4187)(deltaT) + .2(840)(deltaT) = 0

So I just solved for that.

This annotation is misleading.You lot might call back 'delta T' from the LHS is the same thing with the 'delta T' from the RHS.Too,information technology should read as an equality between heat absorbed and heat given.

Okay,the estrus rest is
[tex] 0.15Kg\cdot 4187J Kg^{-i}C^{-1}(86-t)C=0.2Kg\cdot 840J Kg^{-1}C^{-1} (t-22)C [/tex]
,which leads to the result
[tex] t\sim 72.5C [/tex]

For the second problem I didn't practise much, I simply doing the verbal aforementioned thing as I did in the first trouble for the .25kg of h2o and the .18kg of water and and so I was planning on somehow interpreting that respond into some other equation with the .3kg of water and an equilibrium of 30C... Only with the get-go role I ended upwardly with -65719.152C and I knew I was mode off...

This time u have two equations.The first one is
[tex] 0.25Kg\cdot 4187J Kg^{-1}C^{-one}(80-t_{1})C=0.180\cdot 4187J Kg^{-1}C^{-1} (t_{1}-10)C [/tex] (1)
The decond i is
[tex]0.430Kg \cdot 4187J Kg^{-1}C^{-i}(t_{i}*30)C=0.300Kg \cdot 4187J Kg^{-ane}C^{-one} (t_{two}*30) [/tex](2)

,where the sign "*" means "mathematically operated" and is a "-",the order of the 2 numbers in the subclass being chosen knowing that the result of the subtraction be >= to 0.

Daniel.

benji

Thank you a lot for the help guys!

benji

I didn't really understand what you were saying there in your last post dextercioby, most my 2d problem.

What I did was take mc(deltaT)+mc(deltaT)=0 for the get-go ii portions of water.

So I take .25(4187)(Tf-80)=-.one(4187)(Tf-10) which solves to 60C.

The for the 2d part of the trouble where the additional 300g of h2o is added and the entire portion of water comes to an equilibrium of 30C I did this:

mc(deltaT)+mc(deltaT)=0
.43(4187)(thirty-60)=-.3(4187)(30-Ti)

...which solves to -13C.

The reply is supposed to be 0.33C. Can you see where I went wrong?

apchemstudent

I didn't really understand what you were maxim there in your concluding post dextercioby, most my second problem.
What I did was take mc(deltaT)+mc(deltaT)=0 for the outset 2 portions of water.

And so I take .25(4187)(Tf-eighty)=-.1(4187)(Tf-ten) which solves to 60C.

The for the second part of the problem where the boosted 300g of water is added and the entire portion of h2o comes to an equilibrium of 30C I did this:

mc(deltaT)+mc(deltaT)=0
.43(4187)(30-lx)=-.3(4187)(xxx-Ti)

...which solves to -13C.

The answer is supposed to be 0.33C. Can you meet where I went wrong?

You lot have the wrong mass for your first equation on the right side of the equation... information technology's suppose to be .xviii kg

dextercioby

U missused the numbers.It's 0.180 Kg in the starting time case,not 0.100 Kg as u took information technology.

Daniel.

learningphysics

I didn't really understand what you were proverb at that place in your last mail dextercioby, almost my 2nd problem.
What I did was take mc(deltaT)+mc(deltaT)=0 for the first 2 portions of water.

And so I take .25(4187)(Tf-80)=-.i(4187)(Tf-10) which solves to 60C.

The for the 2d part of the problem where the additional 300g of h2o is added and the entire portion of water comes to an equilibrium of 30C I did this:

mc(deltaT)+mc(deltaT)=0
.43(4187)(30-sixty)=-.three(4187)(30-Ti)

...which solves to -13C.

The answer is supposed to exist 0.33C. Tin can you run across where I went wrong?

How you're doing it is fine. But you tin solve with with one equation. Simply remember that the total oestrus change of all 3 groups is 0.

Then
0.25(4187)(30-80) + 0.180(4187)(thirty-x) + 0.30(4187)(thirty-t)=0

and y'all get 0.33C.

dextercioby

2 hundred fifty grams of water at 80C is poured into a Styrofoam cup of negligible heat capacity containing 180 chiliad of water at 10C. After an additional 300 thousand of h2o is added to the cup, the mixture come to an equilibrium temperature of 30C. What was the temperature of additional 300 yard of water?

EDIT:I messed up calculations and wound up with a unlike event.Information technology's something bad happening with me taoday... :yuck:
EDIT:This method i proposed leads to the same outcome as the one proposed by learningphysics.

Daniel.

learningphysics

This text is cryptic.My equations hold in the assumption that the 300g of h2o are added to the original mixing after the original mixing reached its equilibrium point.That's why the results are unlike,coz the problem assumes that the all the iii water quantities are mixed from the beginning.
I would so take to advice you lot to follow the indications given by learningphysics.

Daniel.

Information technology shouldn't matter when they are mixed. I got the same answer with your equations.

benji

Thanks guys, stupid mistake on my role!

I actually had .18 written downwards but for some reason I erased and wrote .one.

what is the final temperature of coffee and mug once they come to thermal equilibrium?

Couldn't find a specific heat for coffee

Answers and Replies

Related Threads on Couldn't find a specific heat for coffee

0 Response to "what is the final temperature of coffee and mug once they come to thermal equilibrium?"

Post a Comment

Iklan Atas Artikel

Iklan Tengah Artikel 1

Iklan Tengah Artikel 2

Iklan Bawah Artikel