Couldn't find a specific heat for coffee

[benji]

Here's the question:

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

I used Q=mc (delta)T.

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

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

I don't understand this stuff too well yet, so it's probably a logic mistake on my part, but if someone could explain that to me I'd really appreciate it. Thanks.




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

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

 

[marlon]

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

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

regards
marlon

 

[marlon]

The phylosophy is that the total amount of heat is always conserved during the heath-transport. So the coffee will give off heat and the glass will absorbe the heath until equilibrium has been reached...

Normally, T is in kelvin but once you are talking about temperature differences, it doesn't matter what units you use as long as you use one unit all the time...


regards
marlon

 

[dextercioby]

For the first part you didn't give us the numbers,meaning the specific heat of glass.The logics seems good,Hot coffee is hot water with a tiny amount of coffee (maybe u like it without (too much)sugar,i hate coffe altogether),so u migh have messed up either the computations,or the units for the physical quantities.

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

Daniel.

 

[benji]

For the first problem this is what I did:

mc(deltaT) + mc(deltaT) = 0
.15(4187)(deltaT) + .2(840)(deltaT) = 0

Then I just solved for that.




For the second problem I didn't do much, I just doing the exact same thing as I did in the first problem for the .25kg of water and the .18kg of water and then I was planning on somehow interpreting that answer into another 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 first problem this is what I did:

mc(deltaT) + mc(deltaT) = 0
.15(4187)(deltaT) + .2(840)(deltaT) = 0

Then I just solved for that.

This notation is misleading.You might think 'delta T' from the LHS is the same thing with the 'delta T' from the RHS.Besides,it should read as an equality between heat absorbed and heat given.

Okay,the heat balance is
[tex] 0.15Kg\cdot 4187J Kg^{-1}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 do much, I just doing the exact same thing as I did in the first problem for the .25kg of water and the .18kg of water and then I was planning on somehow interpreting that answer into another 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...

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

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

Daniel.

 

[benji]

Thanks a lot for the help guys!

 

[benji]

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

What I did was take mc(deltaT)+mc(deltaT)=0 for the first 2 portions of water.

So I have .25(4187)(Tf-80)=-.1(4187)(Tf-10) which solves to 60C.

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

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

...which solves to -13C.

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

 

[apchemstudent]

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

What I did was take mc(deltaT)+mc(deltaT)=0 for the first 2 portions of water.

So I have .25(4187)(Tf-80)=-.1(4187)(Tf-10) which solves to 60C.

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

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

...which solves to -13C.

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

You have the wrong mass for your first equation on the right side of the equation... it's suppose to be .18 kg

 

[dextercioby]

U missused the numbers.It's 0.180 Kg in the first case,not 0.100 Kg as u took it.

Daniel.

 

[learningphysics]

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

What I did was take mc(deltaT)+mc(deltaT)=0 for the first 2 portions of water.

So I have .25(4187)(Tf-80)=-.1(4187)(Tf-10) which solves to 60C.

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

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

...which solves to -13C.

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

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

So
0.25(4187)(30-80) + 0.180(4187)(30-10) + 0.30(4187)(30-t)=0

and you get 0.33C.

 

[dextercioby]

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

EDIT:I messed up calculations and wound up with a different result.It's something bad happening with me taoday...
EDIT:This method i proposed leads to the same result as the one proposed by learningphysics.

Daniel.

 

[learningphysics]

This text is ambiguous.My equations hold in the assumption that the 300g of water are added to the original mixing after the original mixing reached its equilibrium point.That's why the results are different,coz the problem assumes that the all the three water quantities are mixed from the beginning.

I would then have to advice you to follow the indications given by learningphysics.

Daniel.

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

 

[benji]

Thanks guys, stupid mistake on my part!

I actually had .18 written down but for some reason I erased and wrote .1.