Question 1: How many liters of 1 liter of 100 degrees of water will become 70 degrees Celsius when y

Formula: q cmδt, q, heat absorbed and released;

c, specific heat capacity of water, 4200j kg·

t, temperature difference.

Solution: 1 liter 1dm3

The mass of these waters is: m v 1000 water indicates the density of water and is 1000kg m3

Q suck q put.

After mixing, if up to 70:

cm cold (70 25) cm hot (100 70) m cold 30 45

That is: Q suck Q release.

After mixing if up to 80:

cm cold (80 25) cm hot (100 80) m cold 20 55

That is, if it is 90 hot water, you only need to change 100 in the above two calculations to 90 and repeat the calculation.

The specific heat capacity c (as if 4200 kg.) is known. c), the mass of the water m, the initial temperature t.;

c*m*t。It's energy, and it's enough to list the equation according to the total energy before and after.

Without considering heat loss: the heat absorbed or released can be found according to the formula cm t (where c is a constant 4200).

For example, if 1l100 water is exothermic to 70 degrees Celsius, the heat released is cm1 (100-70).

XL25 water endothermic temperature rises to 70, and the absorbed heat is cm2 (70-50) heat released = heat absorbed, then m1 (100-70) = m2 (70-25) m1 m2 = (70-25) (100-70) = 45 30=, m1 = 1L

m2=2/3l

In the same way, find the other data as.

Hope, thank you.

This will change according to different outside temperatures and different containers. If you want to get him cool quickly, just add some cold water.

After turning off the heat of 100°C boiling water, it immediately becomes 90°C. Depending on the season, it can change from 10 minutes to 70-80°C in winter and about 20 minutes in summer.

There is also a condition to see how much water, 500 ml of boiling water becomes 70-80 ° for a maximum of 3 minutes, the water temperature changes depending on the amount of water, room temperature container size and other factors.

Set the water that is added to the jujube to prepare for disturbance is x liters, according to the meaning of the stool:

1×100+20x=60(1+x)

>100+20x=60+60x

>40x=40

>x=1

So, the water that needs to be added is 1l = 1000 milliliters.

If you add 100 degrees of water to 0 degrees of water, will 50 degrees of water come out of 2 liters? It is important to note that water remains liquid at both 0 degrees and 100 degrees. 0 degrees is the freezing point of water, but at this temperature the liquid and solid phases of water can coexist.

If the liquid water at 0 degrees turns into ice, a large amount of heat needs to be released to keep the liquid from cooling. Similarly, at 100 degrees, the vapor and liquid phases of water coexist, and if the heat is not absorbed further, and if it is not heated continuously, the water will remain the liquid phase as superheated water.

<> other hand, the purity of water is not easy to freeze even at 0 degrees Celsius. The freezing of water is dependent on out-of-phase nuclei, which lack nucleation points such as bubbles, impurities in pure water. As a result, pure water does not freeze easily and remains liquid even below 0 degrees, allowing scientists to cool ultrapure water to tens of degrees below zero.

Supercooled water is in a sub-stable state, but if you shake the supercooled water, it enters the bubbles and forms crystal nuclei, so it can freeze immediately in a cold place. Similarly, the vaporization of water requires a core, and water with high purity does not boil easily and can become superheated water.

Let's get back to that. However, this is not the case. Because the density and specific heat capacity of water are not fixed constants, but vary according to temperature, such changes need to be taken into account in practical calculations.

At atmospheric pressure, water has the highest density at 4 degrees Celsius, which is 1000 kilograms of cubic meters, but the density of water at other temperatures is lower than this, for example, water 0 degrees Celsius is kilogram cubic meters, and water 100 degrees Celsius is kilogram cubic meters. Therefore, 1 liter of water at 0 degrees Celsius is heavier than 1 liter of water at 100 degrees Celsius, and the difference becomes 41.47 grams, too.

On the other hand, in the range of 0 degrees to 100 degrees, the specific heat capacity of water decreases with increasing temperature, with the lowest around 50 degrees Celsius, and after that, it rises as the temperature increases. The change from 0 degrees to 50 degrees and from 50 degrees to 100 degrees is not linear, but there is no big difference overall, and the magnitude of the change is small. In addition, as you can see from the above formula, the volume of water can be eliminated on both sides of the equation, so even if 100 degrees of water is mixed with a few liters and 0 degrees of water, the temperature of the water will reach about degrees if they are equal in volume.

No, because the temperature is relatively high, pouring it into a relatively low temperature water will only neutralize its temperature, and will not synthesize to 50 degrees because it is zero.

No, because the density and specific heat capacity of water change with temperature, and it is not possible to mix the two to get 50 degrees, so you need to take this change into account.

You won't get two liters of water at 50 degrees; Because the temperature may affect the specific heat capacity of the water and the density of the water, the temperature of the water cannot reach exactly 50 degrees.

will not get two liters of water at 50 degrees; Because according to the experiment, the temperature of the obtained water is about 48 degrees.

One liter, the heated water vapor will still return to the pot, and the amount of water will not change 1 kW * 60 minutes = 1 kW·h = 3,600,000 joules (remember).

According to the specific heat capacity, the formula is carefully shouted: c= e(q) m jujube t, e is the heat absorbed, and it is q in middle school textbooks; m is the mass of the object (kg), and t is the value of the temperature rising (falling) after endothermic (exothermy).

Specific heat capacity of water c

4200 joules per kilogram Celsius.

One liter of water = 1kg

Therefore 4200 = 3600000 1 * t, so t = Celsius, obviously more than 100 degrees Celsius, because the stool filial piety sedan car is in the standard atmospheric pressure.

At this time, the water can only be 100 degrees Celsius.

1 liter of water at 100 degrees Celsius is lowered to 50 degrees Celsius and released: 4200 * 1 * (100-50) = 210000J of heat.

1 liter of water at 0 degrees to 50 degrees absorbs: 4200 * 1 * (50-0) = 210000J of heat.

So, without counting heat loss, it's 50 degrees.

If the heat loss is calculated, it will not be 50 degrees.

Summary. The average temperature of the water after mixing is about Celsius. First of all, you need to consider the total heat of the water after mixing.

The heat of 100 liters of water at 20 degrees Celsius is: 100 liters 1 kilogram joule (kilogram Celsius) 20 degrees Celsius = 8360 joules 10 liters, 100 degrees water with 100 degrees Celsius 1 kilogram joules (kilograms Celsius) 100 degrees Celsius = 41800 joules The total heat of water after mixing is 8360 joules + 41800 joules = 50160 joules.

Secondly, the total quality of the water after mixing needs to be considered. The mass of 100 liters of water at 20 degrees is 100 liters 1 kg liter = 100 kg, the mass of water at 10 liters at 100 degrees is 10 liters 1 kilogram liter = 10 kg, and the total mass of water after mixing is 100 kg + 10 kg = 110 kg. Therefore, the average temperature of the water after mixing is:

8360 joules 20 degrees Celsius + 41800 joules 100 degrees Celsius) 50160 joules) Celsius Therefore, the average temperature of the water after mixing is about 100 degrees Celsius.

100 liters of 20 degrees + 10 liters of 100 degrees of water mixing What is the final degree.

The average temperature of the water after mixing is about Celsius. First of all, you need to consider the total heat of the water after mixing. The calories of 100 liters of water at 20 degrees are:

100 liters 1 kilogram joules (kilogram Celsius) 20 degrees Celsius = 8360 joules 10 liters, the heat of 100 degrees of water is: 10 liters 1 kilogram joules (kilograms Celsius) 100 degrees Celsius = 41800 joules The total heat of water after mixing is 8360 joules + 41800 joules = 50160 joules. Secondly, the total quality of the mixed water needs to be considered.

The mass of 100 liters of water at 20 degrees is 100 liters of water with 1 kilogram of energy = 100 kilograms of weight, the mass of 10 liters of water at 100 degrees Celsius is 10 liters of 1 kilogram liter = 10 kilograms, and the total mass of the mixed water is 100 kilograms + 10 kilograms = 110 kilograms. Therefore, the average temperature of the water after mixing is: (8360 joules 20 degrees Celsius + 41800 joules 100 degrees Celsius) 50160 joules Therefore, the average temperature of the mixed water is about Celsius.

Such an exaggeration can reach more than 70 degrees...

Through the calculations, the answer is yes.

Add 100 degrees of water to 20 degrees of water, and what will be the final degree.

This depends on the ratio of the two, if the ratio is m:n (100 degrees:20 degrees:), then the temperature after mixing is.

100*m+20*n)/(m+n).

The formula is already simple.