1. ## Physics

How can you heat $\displaystyle 1 \,\mathrm{kg}$ mineral water which is at a temperature of $\displaystyle 0 \,{}^\circ \mathrm{C}$ to at least $\displaystyle 60 \,{}^\circ \mathrm{C}$ with using $\displaystyle 1 \,\mathrm{kg}$ water at a temperature of $\displaystyle 100 \,{}^\circ \mathrm{C}$?

2. Thanks but it isn't said how to do it, it's the question. Obviously there is a trick in it otherwise it cannot be done as you said correctly.

3. Originally Posted by james_bond
Thanks but it isn't said how to do it, it's the question. Obviously there is a trick in it otherwise it cannot be done as you said correctly.
Originally Posted by topsquark
Heat gained = Heat lost.

I am assuming you need to heat the water in a bath or something, so what we need to do is find out how much of the 100 C water to put in the bath. The concept is that once equilibrium has been reached the bath and the 1 kg of water will both be at the same temperature.

$\displaystyle \begin{matrix} \underbrace{(1000 g)(4.18~J/g~C)(60~C)} \\ \text{heat gained} \end{matrix} = \begin{matrix} \underbrace{m(4.18~J/g~C)(100~C - 40~C)} \\ \text{heat lost} \end{matrix}$
This will represent the amount of water at 100 C to increase the termperature of the 1 kg of water at 0 to 60 degrees.
Oops! I made mistake. (Tsk, tsk! You should have caught it too.) The final temperature of the 100 C water is 60 C, not 40 C like I had put into the equation. (I thought my logic there was a bit suspect, but I was trusting my equation.) So
$\displaystyle \underbrace{(1000 g)(4.18~J/g~C)(60~C)}_{\text{heat gained}} = \underbrace{m(4.18~J/g~C)(100~C - 60~C)}_{\text{heat lost}}$

Solve for m.

In the end we are getting as much heat as possible out of the mass m. We can always presume we are doing this with less than perfect efficiency and require more than this mass, but certainly no less than it.

-Dan

4. Pour away the water which is at 0 degrees.
Replace it with the water that is at 100 degrees.
The new water in the container is at temperature 100 degrees which is greater than 60 degrees.

5. Sorry Oli, but I made a significant mistake in the question.

Originally Posted by james_bond
How can you heat $\displaystyle 1 \,\mathrm{kg}$ mineral water which is at a temperature of $\displaystyle 0 \,{}^\circ \mathrm{C}$ to at least $\displaystyle 60 \,{}^\circ \mathrm{C}$ with using $\displaystyle 1 \,\mathrm{kg}$ "normal" water at a temperature of $\displaystyle 100 \,{}^\circ \mathrm{C}$?

6. Originally Posted by james_bond
Sorry Oli, but I made a significant mistake in the question.
Actually that's going to make a change in the solution and I'm not sure where to tell you to get the number. Mineral water will have a different specific heat capacity than plain water. You will still be able to warm it to 60 C because odds are the specific heat of mineral water is less than that of pure water. (This is true for most materials, actually. Water has a large specific heat.) But we can't calculate how much water must be used with the information available.

-Dan

7. Originally Posted by topsquark
Actually that's going to make a change in the solution and I'm not sure where to tell you to get the number. Mineral water will have a different specific heat capacity than plain water. You will still be able to warm it to 60 C because odds are the specific heat of mineral water is less than that of pure water. (This is true for most materials, actually. Water has a large specific heat.) But we can't calculate how much water must be used with the information available.

-Dan
I don't think that this is the question... the specific heat capacity of pure water won't be that different from that of mineral water. And 1kg of water at 100 degrees is no where near enough to heat 1kg of water at 0 degrees to 60. This is why I thought it must be some sort of lateral thinking problem.

Where did you find the question?

Another possibility is that the hot water is gas, but the cold water is still liquid: it takes energy in order to change from state to state, so when you put heat energy into water it sort of gets stuck at 100 degrees until it actually boils.

It is a bit of a silly assumption to make though... but the only possibility that springs to mind, unless that specific heat capacity of pure water really is much greater than the specific heat capacity of mineral water. I really doubt this though: after all, mineral water has more stuff in it than pure water, so I would of thought mineral water had a higher specific heat capacity.

8. I just looked up: the specific heat capacity of a sugar solution is greater
than that of pure water.

So I think if anything, the mineral water would have the greater heat capacity.

9. This problem is an optional homework (by Friday) but I'm really interested in solving it.

10. Originally Posted by james_bond
This problem is an optional homework (by Friday) but I'm really interested in solving it.
Well, the solution is to stick your mineral water in a metal container, pour in the 100 C water and stick the whole thing in an insulated container. The bath will warm the mineral water. All the stuff I did earlier was to show that it could be done. But again, we don't know how much hot water to put in to make the final temperature 60 C, all we know is that it can be done.

-Dan

11. Originally Posted by topsquark
Well, the solution is to stick your mineral water in a metal container, pour in the 100 C water and stick the whole thing in an insulated container. The bath will warm the mineral water. All the stuff I did earlier was to show that it could be done. But again, we don't know how much hot water to put in to make the final temperature 60 C, all we know is that it can be done.

-Dan
The problem is that I don't think 1kg of 100 degree water is enough to heat 1kg of 0 degree water to 60 degrees: if all the water ends up at the same temperature, won't it end up at 50 degrees?

12. Originally Posted by Oli
The problem is that I don't think 1kg of 100 degree water is enough to heat 1kg of 0 degree water to 60 degrees: if all the water ends up at the same temperature, won't it end up at 50 degrees?
Originally Posted by topsquark
$\displaystyle (1000 g)(4.18~J/g~C)(60~C) = m(4.18~J/g~C)(100~C - 60~C)$
Well, this can't be done unless the specific heat of mineral water has a specific heat of 2.79 J/g C or smaller. It would depend on the properties of the mineral water.

-Dan