| Topic: |
Science > Physics |
| User: |
"Colin" |
| Date: |
15 Feb 2006 05:58:34 PM |
| Object: |
Easy thermochemistry question |
Hi,
If anyone can help with what (for those of you in this group) will be a
simple question I would most appreciate it. It regards mixing fluids at
different temperatures and the final temp - but it is the volume I need to
know. Here's the problem:
A hot water tank has 100 litres of water at 80 ºC. We want to run a bath
with 60 litres of water at 40ºC, for which we are going to add water at
15ºC. How much hot water do we need?
Please show how you arrived at the answer, because for sure the other guys
on the plumbing course will want to know!
Regards,
Colin.
.
|
|
| User: "RP" |
|
| Title: Re: Easy thermochemistry question! |
15 Feb 2006 11:50:48 PM |
|
|
Colin wrote:
Hi,
If anyone can help with what (for those of you in this group) will be a
simple question I would most appreciate it. It regards mixing fluids at
different temperatures and the final temp - but it is the volume I need to
know. Here's the problem:
A hot water tank has 100 litres of water at 80 ºC. We want to run a bath
with 60 litres of water at 40ºC, for which we are going to add water at
15ºC. How much hot water do we need?
Please show how you arrived at the answer, because for sure the other guys
on the plumbing course will want to know!
Regards,
Colin.
Ignoring thermal expansion and assuming a constant specific heat:
Temperature is a measure of energy per molecule, and for any given
substance this can be expressed alternately as an energy per unit volume
multiplied
by a constant of proportionality. We don't need to know the value of the
proportionality constant for this problem, it cancels out of the
solution, nor do we need to define the units of energy used, they also
cancel out.
Let k be the heat content per degree per liter gained by water for every
degree that it rises above 0C.
Thus
Water at 15C has heat content per liter, greater than preexisting 0C
heat content per liter, of 15k.
With this system, taking some liberties, we get
15C = 15k
40C=40k
80C=80k
60 liters of water at 40C thus has a heat content, above that at 0C, of
60 liters * 40k = 2400 k liter
Now we need to know how many liters of hot and cold water respectively
will provide this heat content when the total volume of the mixed water
is 60 liters.
Let x be the volume of hot water added, and let y be the volume of cold
water added, both measured in liters.
x + y = 60 liter
x = 60 liter - y
The total heat content will be the sum of the heat contents of the cold
and hot water used, or:
80xk + 15yk = 2400 k liter
Dividing through by k
80 x + 15 y = 2400 liter
By substitution:
[80 x + (60 liter - x)15] = 2400 liter
80 x + 900 liter - 15 x = 2400 liter
65 x = 1500 liter
x ~ 23.077 liter
In general, where V3 = mixed water volume and T3 = mixed water temp:
V1 = (V3T3 - V3T2) / (T1 - T2)
It doesn't matter whether volumes and temps 1 or 2 refer to the hot or
cold water. They are interchangeable in the equation. Thus if you want
the cold water volume instead, then just solve the equation using V1 as
the cold water volume and T1 as the cold water temp.
Richard Perry
.
|
|
|
| User: "Colin" |
|
| Title: Re: Easy thermochemistry question! |
16 Feb 2006 12:41:23 AM |
|
|
V1 = (V3T3 - V3T2) / (T1 - T2)
It doesn't matter whether volumes and temps 1 or 2 refer to the hot or
cold water. They are interchangeable in the equation. Thus if you want the
cold water volume instead, then just solve the equation using V1 as the
cold water volume and T1 as the cold water temp.
Richard Perry
Sincere thanks for your time.
Colin.
.
|
|
|
|
| User: "tadchem" |
|
| Title: Re: Easy thermochemistry question! |
16 Feb 2006 01:54:48 AM |
|
|
It would have been a little simpler if you had just left out the 'k'
stuff.
<Always keep your audience in mind - in this case plumbers.>
The volume equation:
V(final) = V(hot) + V(cold) [1]
The 'heat' equation (using Heat = Volume * Temperature)
V(final) * T(final) = V(hot) * T(hot) + V(cold) * T(cold) [2]
Given the values in the OP:
60 = V(hot) + V(cold) [3]
60 * 40 = V(hot) * 80 + V(cold) * 15 [4]
Multiply the top by 15 so we can subtract out the V(cold) terms
60 * 15 = V(hot) * 15 + V(cold) * 15 [5]
Subtract [5] from [4]
60 * (40 - 15) = V(hot) * (80 - 15) [6]
Do the arithmetic
1500 = V(hot) * 65 [7]
V(hot) = 1500 / 65 = 23.077
Plug this back into [3]
60 = 23.077 + V(cold) [8]
and do some more arithmetic
V(cold) = 36.923
Although V(hot) = 23 and V(cold) = 37 will get you as close as can be
measured with hand tools and a real bathtub.
Tom Davidson
Richmond, VA
.
|
|
|
| User: "RP" |
|
| Title: Re: Easy thermochemistry question! |
16 Feb 2006 08:55:39 AM |
|
|
tadchem wrote:
It would have been a little simpler if you had just left out the 'k'
stuff.
Sure, but if you didn't notice I provided the solution both ways :)
I went the technical route first time around to show how the equation
that you provided below, and that I provided also, was derived.
<Always keep your audience in mind - in this case plumbers.>
Believe it or not, at least in my state, plumbers have to attend college
in order to get a Master's License. In these days and times in order to
be proficient in any technical trade a technician must have engineering
skills at least to some extent. Don't kid yourself, plumbing isn't a
simpleton's trade. Though there may be plenty of simpletons in the
trade, they are serving only as manual laborers for the real brains
behind the operation, and are destined to failure if they ever attempt
to go it on their own. Any plumber should know how to calculate flow
rates, water mixing, etc., at the bare minimum.
Richard Perry
The volume equation:
V(final) = V(hot) + V(cold) [1]
The 'heat' equation (using Heat = Volume * Temperature)
V(final) * T(final) = V(hot) * T(hot) + V(cold) * T(cold) [2]
Given the values in the OP:
60 = V(hot) + V(cold) [3]
60 * 40 = V(hot) * 80 + V(cold) * 15 [4]
Multiply the top by 15 so we can subtract out the V(cold) terms
60 * 15 = V(hot) * 15 + V(cold) * 15 [5]
Subtract [5] from [4]
60 * (40 - 15) = V(hot) * (80 - 15) [6]
Do the arithmetic
1500 = V(hot) * 65 [7]
V(hot) = 1500 / 65 = 23.077
Plug this back into [3]
60 = 23.077 + V(cold) [8]
and do some more arithmetic
V(cold) = 36.923
Although V(hot) = 23 and V(cold) = 37 will get you as close as can be
measured with hand tools and a real bathtub.
Tom Davidson
Richmond, VA
.
|
|
|
| User: "Doune" |
|
| Title: Re: Easy thermochemistry question! |
18 Feb 2006 08:13:10 PM |
|
|
<Always keep your audience in mind - in this case plumbers.>
or Chemists...
Don't kid yourself, plumbing isn't a simpleton's trade. Though there
may be plenty of simpletons in the trade
See?
SCW
.
|
|
|
|
|
|
|
|
Related Articles |
|
|
