An unknown volume of water at 18.2°C is added to 24.3 mL water at 35.6°C. If the final temperature is 23.5°C, what was the unknown volume? (Assume that no heat is lost to the surroundings; d of water is 1.00 g/mL.)
2007-03-14 14:37:09 · 7 answers · asked by pinay804 2 in Science & Mathematics ➔ Chemistry
OK, since we know that no heat was lost to the surroundings, that means that all the heat from the hot water must have remained in the system. There is no net gain or loss of heat; it is maintained in the system. The hot water and the cold water will mix until they attain an equilibrium temperature. Any heat lost by the hot water must be gained by the cold water.
So, what formula do you use? (Bear with me here, I don't know how to do a greek Delta or a subscript on here!) The change in heat is represented by DeltaH = (mass)(specific heat)(DeltaTemp).
Since we know that the heat gained by the cold water MUST be equal to the heat lost by the hot water, we can use that fact to create an inequality:
-(DeltaH-hot) = (DeltaH-cold)
(Note that there's a minus-sign on the left: heat LOST by hot = heat gained by cold.)
Expand this out, and you get:
-(mass-hot)(specific heat-hot)(DeltaTemp-hot) = (mass-cold)(specific heat-cold)(DeltaTemp-cold)
For this problem, we don't really care about the specific heat of water, since it's going to be the same for both the hot and the cold. (If we were using two different substance, say, water and iron, then it would matter.) So let's simplify the equation:
-(mass-hot)(DeltaTemp-hot) = (mass-cold)(DeltaTemp-cold)
How do you find the mass of the water? Simple multiplication, since the density of water is given: multiply the number of mLs by the density, to give grams.
How do you find the change in temperature (DeltaTemp)? Take the final temperature minus the initial temperature.
With this equation above, and knowing how to find the mass and the DeltaTemp, you should be able to plug in the values provided in the problem, and solve for the one unknown.
Make sense?
2007-03-14 14:53:32 · answer #1 · answered by kittenpie 3 · 0⤊ 0⤋
You do it by looking at calories (units of heat energy). You know that 24.3ml of water at 35.6 deg has a certain number of calories. You know that the final volume has the temperature it does because of the number of calories contributed by the hotter water. The total amount of heat doesn't change in the problem.
So, you take the difference in temperature from the 35.6 deg and the final temperature, calculate the number of calories difference, and see how much water at 18.2 deg that number of calories will raise it to 23.5 deg. This will give you the volume of the first unknown, and you can add this to the 24.3ml to find your final volume.
If you still can't get it, post it again and we can work the numbers for you.
2007-03-14 14:47:15 · answer #2 · answered by xaviar_onasis 5 · 0⤊ 0⤋
This is a complex problem, because the amount of energy that it takes to heat or cool water is actually not the same at different temperatures. There is no analytical solution, you have to use tables of specific heat capacity data for water. Typically approximate equations are available for this purpose, but you will have to do some numerical integration.
One good thing is that we're talking about only liquid water here so you don't need to worry about latent heat of phase changes. A rough approximation would be to assume that water in this temperature range has a constant specific heat. Then you can just balance energy before and after you mix.
2007-03-14 15:19:04 · answer #3 · answered by Dorkus 2 · 0⤊ 0⤋
this is a straight math problem for finding the unknown volume..
first, the total volume of the final solution is 24.3ml+the unknown volume of water, therefore
x/(x+24.3)*18.2+24.3/(x+24.3)*35.6=23.5
18.2x+24.3*35.6=23.5(x+24.3)
18.2x+865.08=23.5x+571.05
5.3x=294.03
x=55.477ml
This does not take into account the possible expansion and contraction of water due to the change in temperature.
2007-03-14 14:54:08 · answer #4 · answered by j_con999 2 · 0⤊ 0⤋
when a cool mass (18 c water) is combined with a warm mass (35 c water) the temperature (molecule velocity or activity) blends. It is simply a matter of mass action, there is no reaction involved. Since both are water, it is a simple calculation. Had the liquids been not the same, then difference in the heat capacity would have to be considered, along with possible reactions for enthalpy changes.
2007-03-14 14:54:27 · answer #5 · answered by lare 7 · 0⤊ 0⤋
I think the explanation is not chemistry but mathmatics - algebra to be more specific.
For example, you are told to ignore heat loss, and the density is a bum steer - if you are adding one liquid to an identical one, then it doesn't matter WHAT the density is does it.
The same answer would apply to the same volumes and temps of salt water, condensed milk or cement.
View this as a problem for algebra and NOT chemistry I think you will find.
2007-03-14 14:50:23 · answer #6 · answered by Mark T 6 · 0⤊ 0⤋
i am going to assume equilibrium. ?H = T?S, so in the equation ?G = ?H - T?S, basically replace the H with that, clean up for S and get a numerical fee, then plug that fee in for S and clean up for H
2016-12-02 00:42:48 · answer #7 · answered by ? 4 · 0⤊ 0⤋