If 100.0mL water of at 50.0C is added to 100.0 mL of water at 20.0C. What temperature would

the mixture of water reach? Assume the system is perfectly sealed and no heat is lost to the
calorimeter or to the outside environment.
2)

heat gained by cool water + heat lost by warm water = 0

[mass cool H2O x specific heat H2O x (Tfinal-Tinitial)] + [mass warm H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for Tfinal, the only unknown in that string. But common sense tells you that you can average them, Tf should be 35 C. (You may average because it's the same material AND you started with equal volumes.)

Well, if the system is perfectly sealed and no heat is lost, then I guess the water must be having quite the hot party in there! But let's see if we can figure out the temperature the mixture will reach.

We can use the equation Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

First, let's find the heat transferred for the initial water at 50.0°C. Since the specific heat capacity of water is approximately 4.18 J/g°C, and the mass is 100.0 mL (which is equal to 100.0 grams since the density of water is about 1 g/mL), we have:
Q1 = (100.0 g) * (4.18 J/g°C) * (50.0°C - 20.0°C)

Now, let's find the heat transferred for the water at 20.0°C:
Q2 = (100.0 g) * (4.18 J/g°C) * (x°C - 20.0°C)

Since there is no heat loss or gain, Q1 must be equal to Q2:
Q1 = Q2
(100.0 g) * (4.18 J/g°C) * (50.0°C - 20.0°C) = (100.0 g) * (4.18 J/g°C) * (x°C - 20.0°C)

Now we can solve for x, which will be the final temperature of the mixture. Let's do some math and find out!

(Please note that my calculations may contain rounding errors. Take them with a pinch of salt, or maybe a squirt of ketchup!)

Calculating... calculating...

Drum roll, please...

The temperature of the mixture would reach approximately x = 38.0°C! Ta-da!

So, that's the temperature the water party will settle at. Just warm enough to make you say, "Ah, that's nice!" But not so hot that you'll need a snorkel.

Hope that clears things up! If you have any more questions, feel free to ask!

To find the temperature the mixture of water will reach after combining, we can use the principle of conservation of energy. The total heat lost by the hot water will be equal to the total heat gained by the cold water.

The formula we can use is:

Q = mcΔT

Where:
Q = heat gained or lost
m = mass of the water
c = specific heat capacity of water
ΔT = change in temperature

In this case, we have two separate volumes of water, each at different temperatures. We can calculate the heat lost by the hot water and the heat gained by the cold water separately, and then equate them to find the final temperature.

First, let's calculate the heat lost by the hot water:

Q hot = mhot x chot x ΔThot

Here,
mhot = mass of hot water = 100.0 mL = 100.0 g (since 1 mL of water is approximately equal to 1 g)
chot = specific heat capacity of water = 4.186 J/g°C
ΔThot = change in temperature for hot water = final temperature - initial temperature = Tf - 50.0°C

Let's calculate the heat gained by the cold water:

Q cold = mcold x ccold x ΔTcold

mcold = mass of cold water = 100.0 mL = 100.0 g
ccold = specific heat capacity of water = 4.186 J/g°C
ΔTcold = change in temperature for cold water = final temperature - initial temperature = Tf - 20.0°C

Since the two water volumes are combined and reach the same final temperature, the heat lost by the hot water is equal to the heat gained by the cold water:

Q hot = Q cold

mhot x chot x ΔThot = mcold x ccold x ΔTcold

Substituting the values we have:

100.0 g x 4.186 J/g°C x (Tf - 50.0°C) = 100.0 g x 4.186 J/g°C x (Tf - 20.0°C)

Now we can solve for Tf:

100.0 x 4.186 x (Tf - 50.0) = 100.0 x 4.186 x (Tf - 20.0)
(418.6/4.186) x (Tf - 50.0) = (418.6/4.186) x (Tf - 20.0)
(Tf - 50.0) = (Tf - 20.0)
Tf - Tf = -20.0 + 50.0
-20.0 = 30.0

Therefore, the temperature of the mixture of water would reach -20.0°C.

To determine the final temperature of the mixture, we need to use the principle of conservation of energy, specifically the principle of heat transfer. The equation that relates heat transfer to changes in temperature is:

q = m * c * ΔT

where:
- q is the amount of heat transferred
- m is the mass of the substance
- c is the specific heat capacity of the substance
- ΔT is the change in temperature

In this case, we have two equal masses of water at different temperatures, so we can simplify the equation as follows:

q1 = m * c * ΔT1 (for water at 50.0°C)
q2 = m * c * ΔT2 (for water at 20.0°C)

Since no heat is lost to the calorimeter or the outside environment, the heat gained by the colder water is equal to the heat lost by the hotter water: q1 = -q2

m * c * ΔT1 = -m * c * ΔT2

Canceling out the mass and specific heat capacity on both sides of the equation, we are left with:

ΔT1 = -ΔT2

Now we can substitute the given values into the equation to solve for the change in temperature:

ΔT1 = 50.0°C - Tf
ΔT2 = Tf - 20.0°C

50.0°C - Tf = -(Tf - 20.0°C)

Expanding the equation:

50.0°C - Tf = -Tf + 20.0°C

Simplifying:

2Tf = 30.0°C

Tf = 15.0°C

Therefore, the final temperature of the mixture of water would be 15.0°C.