if i get 1 litre of 10 degree water and another at 90 degrees and mixed them will the heat balance?

Question

basically i get 1 litre of boiling hot water and another cold bottle and mixed them in a big tub will the heat be exactly at the middle tempreture of these 2 bottles. e.g 1 at 10 degrees and 1 at 90 degrees after being mixed will it be exactly 50 degrees?

Answer 1

On paper that has to be the answer but there is a concept of heat loss when boiling water comes in contact with cold water.
The loss is in the nature of steam.
Steam is at a the temperature of boiling water.
So there is a loss of heat. The cold water ends up being less in quantity due to steam loss.
So the cold water cannot offset the heat of the boiling water, so the temperture will be higher than that of 50 Degrees.

Answer 2

When mixed, the hot water will give up enough heat to cool down 40 degrees while the cold water absorbs enough heat to raise its temperature 40 degrees so the mixture will be at 50 degrees.

If your hot water is at 90 C, then, since it isn't boiling, there will be no steam. Air that is cooled by its proximity to the cold water may cause water vapor in the hot air near the hot water to condense and become visible.

As far as "exactly", we are talking theoretical numbers. In the real world as soon as you remove the heat from the hot water and/or remove the cold water from refrigeration, the temperatures will start to move to the temperature of the room where you are conducting this experiment.

Answer 3

If we do not bother about minor difference in the resulting temperature, then they will reach the equilibrium temperature of 50 degree.

For accurate measurements the above formula is valid only if the temperature difference is small say, 2 or 3 degree.

The specific heat capacity of water at various temperatures varies to a great extent.

Answer 4

In theory this would be very nearly so. However the specific heat capacity of water is nearly a constant but not nearly so. In other words heated to raise the temperature of water from 21 oC to 22 oC is slightly different from that needed to raise the temperature from 85 to 86 oC.

So you would need a not so accurate thermocouple or thermometer to take your readings. i.e. not reading say not higher than the 5th decimal.

Secondly you would need to use a vessel that would not lose any heat certainly not a big tub. the vessel surface itself must initially be at 50 degrees.

Answer 5

If you make the assumption that the 2l of water remain in the liquid state, then the temperature will be the average of the two volumes (given that the specific heat capacity is the same).

at the point of homogenous mixing, the temperature is exactly 40 deg C.

The poster who believes that vapourisation increases heat in the system is wrong. Vapourisation requires energy, it doesn't create it, otherwise it would be an explosive process

Answer 6

yes the heat will be balanced. It is the principle of mixture.
"Heat Loss by hot body is equal to heat gained by cold body"

Net temperature for this case is 50 degree.

Answer 7

Assuming you do not gain or lose energy to the "big tub" that you pour them into...yes!
Heat(calories) =Specific heat(cal/g C)*mass(grams)*change in Temp(C).
The specific heat of water is 1.00cal/g C and your mass/volume is the same for both=1000g, they would both change temp by 40 degrees(one positively, the other negatively), so that the amount of heat gained will equal heat lost in a closed, isolated system.
Hope this helps!

Answer 8

Assuming no secondary losses/ evaporation, YES.

Answer 9

yes