A 100g sample of water at 90 degrees celsius is added to a 100g sample of water at 10 degrees celsius.

The final temperature of the water is which of the following:
a) between 50 degrees celsius and 90 degrees celsius
b) 50 degrees celsius
c) between 10 degrees celsius

This is a practice problem so i know that the answer is b) 50 degrees celsius, but i am confused on how to find that answer.

Since these are the same masses, you can simply take the average of 90 and 10.

(90+10)/2 = 50. If they are not the same mass it is a little more complicated than that but all of these can be solved the same way.

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
Solve for Tfinal.

Well, let me tell you a little story that might help you understand how to find the answer.

Once upon a time, there were two cups of water. One cup was relaxing, enjoying some sunshine, and feeling pretty warm at 90 degrees Celsius. The other cup was feeling a bit chilly at just 10 degrees Celsius.

One fine day, these two cups decided to have a little bathtub party. So, they poured themselves together and became one big cup of water. Now, what do you think happened to the temperature?

Well, the warm water shared some of its heat with the cold water, and the cold water shared some of its chilliness with the warm water. They had a little temperature exchange going on, like a friendly mingling of hot and cold. So, the final temperature ended up somewhere in the middle.

Imagine it like this: the hot water said, "come here, cold water, let's have a warm hug" and the cold water responded, "sure thing, hot water, I'll warm up a bit for you, but not too much!"

And that's why the final temperature is somewhere in between the starting temperature of the hot water (90 degrees Celsius) and the starting temperature of the cold water (10 degrees Celsius). So, the answer is definitely not 90 degrees Celsius or 10 degrees Celsius.

And since we are dealing with a 100-gram sample of water at the same temperature, the final temperature will be averaging out the starting temperatures. So, it will be right in the middle, at a toasty 50 degrees Celsius! Voila!

To find the final temperature of the water when two samples are mixed, we can use the principle of heat transfer.

The heat gained by the cooler sample is equal to the heat lost by the hotter sample. This can be represented by the equation:

m₁c₁(Tf - T₁) = m₂c₂(T₂ - Tf)

Where:
m₁ = mass of the first sample
c₁ = specific heat capacity of water
Tf = final temperature
T₁ = initial temperature of the first sample
m₂ = mass of the second sample
c₂ = specific heat capacity of water
T₂ = initial temperature of the second sample

In this case, the mass and specific heat capacity of water are the same for both samples, so we can simplify the equation to:

(Tf - 10) = (90 - Tf)

Simplifying further:

2Tf = 100

Tf = 50

Therefore, the final temperature of the water is 50 degrees Celsius.

To find the final temperature of the water after mixing, you can use the concept of conservation of energy.

The heat gained or lost by a substance is given by the equation:

Q = m * c * ∆T

Where:
- Q is the heat gained or lost (in joules)
- m is the mass of the substance (in grams)
- c is the specific heat capacity of the substance (in J/g°C)
- ∆T is the change in temperature (in °C)

Let's calculate the heat gained or lost by each sample of water:

For the first sample (at 90°C):
Q1 = m1 * c * ∆T1
Q1 = 100g * 4.18 J/g°C * (90°C - Tf)
Where Tf is the final temperature.

For the second sample (at 10°C):
Q2 = m2 * c * ∆T2
Q2 = 100g * 4.18 J/g°C * (Tf - 10°C)

Since no heat is lost to the surroundings, the heat gained by one sample is equal to the heat lost by the other sample:

Q1 = -Q2

Therefore, we can set up the equation:

m1 * c * ∆T1 = -m2 * c * ∆T2

Substituting the known values:

100g * 4.18 J/g°C * (90°C - Tf) = -100g * 4.18 J/g°C * (Tf - 10°C)

Now, let's solve for Tf:

(90°C - Tf) = - (Tf - 10°C)

Simplifying:

90°C - Tf = -Tf + 10°C

Combining like terms:

2Tf = 90°C + 10°C

2Tf = 100°C

Tf = 50°C

Therefore, the final temperature of the water is 50 degrees Celsius, which corresponds to option b) in the question.

Thats wrong