Specific heat capacity question....

[albertlee]

If an amount of water is added to another amount of water which is 3 times more in volume with temperature 10, what is the temperature of the water, which when added to another, the overall temperature is 20??

 

specific heat capacity for water = 4200J/kg c

 

below is my solution

 

Let's assume the amount of water is 1 and 3.

 

Say, the energy used to heat up the amount of water of 3 to 10c from 0c is:

3*4200*10 = 42000*3J

 

the energy used to heat up the total amount of water from 0c to 20c is:

4*4200*20 = 84000*4J = 42000*8J

 

Ok, subtract the nergy used to heat up the total amount by the energy used to heat up only the amount of water of 3, and the answer is: 5*42000J

 

and that's the energy used to heat up the water of 1...

so,

4200*1*t = 5*42000J

t = 50c = temperatue rises from 0 using the according amount of energy...

 

 

 

so, what I am asking is, can you people find a simpler method apart from the above??

 

thx

[Kedas]

Just write down the equation with at the left side the total sum of energy before the mix and at the right side total after the mix.

Then you will also notice that it's independant of the heat capacity since they are the same for both.

[albertlee]

the really tricky thing here is, how do you define the sum of energy??

 

since of course, water also has energy even when it's at 0c....

 

probably you people have to show me with equations to demonstrate your view..

[swansont]

the really tricky thing here is' date=' how do you define the sum of energy??

 

since of course, water also has energy even when it's at 0c....

 

probably you people have to show me with equations to demonstrate your view..[/quote']

 

 

Since the difference in energy is what's imortant here, you get to define zero. What you're interested in is changes in temperature.

 

Assume the base temperature is 10, so you are measuring how much energy is present above the base value present at 10 C; you have volumes V, 3V and 4V, but the 3V volume has no excess energy

 

VT' = 4V*10 (since the temperature went up by 10. T' is the temperature above 10)

 

T' = 40, meaning it started at 50 degrees

 

---

 

If you were measuring an absolute energy, you'd have to use an absolute temperature scale, like Kelvins

[albertlee]

thanks swansont, but a question on your wording..

 

what does T' mean? btw, why put a hypostrophe??

[J.C.MacSwell]

Just a note of caution: Beware of phase transitions.

 

Since you are assuming constant pressure of 1 atmosphere, there are none in this case.

[swansont]

thanks swansont' date=' but a question on your wording..

 

what does T' mean? btw, why put a hypostrophe??[/quote']

 

So that it would be less likely to be confused with the actual temperature.

[Kedas]

the really tricky thing here is' date=' how do you define the sum of energy??

 

since of course, water also has energy even when it's at 0c....

 

probably you people have to show me with equations to demonstrate your view..[/quote']

 

You don't need to know the energy at 0°C since that same energy is also present after you mix them.

If you assume that the energy in the water up to 0°c = x.V (x is an unknown number)

So the ernegy equation would be:

x.V1 + T1.c.V1 + x.V2 + T2.c.V2 = x.V1 + x.V2 + Tm.c.Vm

so you don't have to know a lot of math to see that you can just remove x.V1 and x.V2

T1.c.V1 + T2.c.V2 = Tm.c.Vm

now you will also see that you can remove the heat capacity c

T1.V1 + T2.V2 = Tm.Vm (Vm=V1+V2)

I'm sure you can handle it from here.

And like mentioned before this is only valid if there is no phase transitions. (so only water)

 

one more remark in case you want to use the joules, it's the mass of the water that has to be multiplied with the heat capacity not the volume but that is a constant multiplied with the volume so it won't change your result.