I need help in Chemistry!

[ashitaria]

Does anyone know the 3 tenets of KMT?

What is standard pressure?

And how do you get the temperature of when 30g of 0 degrees ice is dropped into 30g of 100 degrees celcius water?

 

[Subotai]

Does anyone know the 3 tenets of KMT?

What is standard pressure?

And how do you get the temperature of when 30g of 0 degrees ice is dropped into 30g of 100 degrees celcius water?

1. Idk

2. 1013 hPa

3. Will post tomorrow ( it's 00:37 am )

 

[y4r5xeym5]

Wouldn't the 3rd one just be 50 degrees? Because, you know...water and ice are the same thing.....

Also, I prefer using mmHg (or torrs) for my measurements. 760 is a much easier number to remember.

As for KMT:
1. A gas consists of small particles (atoms or molecules) that move randomly with rapid velocities.
2. The attractive forces between the particles of a gas can be neglected.
3. The actual volume occupied by gas molecules is extremely small compared to the volume that the gas occupies.
4. The average kinetic energy of gas molecules is proportional to the Kelvin temperature.
5. Gas particles are in constant motion, moving rapidly in straight paths.

 

[ashitaria]

Wouldn't the 3rd one just be 50 degrees? Because, you know...water and ice are the same thing.....

Also, I prefer using mmHg (or torrs) for my measurements. 760 is a much easier number to remember.

As for KMT:
1. A gas consists of small particles (atoms or molecules) that move randomly with rapid velocities.
2. The attractive forces between the particles of a gas can be neglected.
3. The actual volume occupied by gas molecules is extremely small compared to the volume that the gas occupies.
4. The average kinetic energy of gas molecules is proportional to the Kelvin temperature.
5. Gas particles are in constant motion, moving rapidly in straight paths.

Um...You forget that Ice requires heat to melt before turning into liquid. Ice and liquid is not the same thing.

As for KMT, I wish to know the three tenets...not five, sorry.

 

[ProxyAmenRa]

q=energy
m=mass
dt=change in temp
ql= energy of change of state
lf=latent heat of fusion (freezing)

q=mc(dt)
ql=m*lf

Energy is going to be lost from the water, exothermic. Energy is going to be gained by the ice, endothermic.

The working out is set into three parts. There is heat loss by the hot water, reducing the temp to the equivalence point. Heat gain from melting the ice and heat gain from heating the cold to equivalence point. At the equivalence point everything equates to 0.

heat gain = heat loss

0 = m1*c1*(t2-100) + m1*c1*(t2-0) + 334*30
-334*30 = m1*c1*(t2-100) + m1c1*(t2)
-334/4.18 = 2*t2 - 100
100 - 334/4.18 = t2
t2 = 10.0475 celcius or 283.1975 K

I think it is correct. I have not done this sort of thing in 4 years.

 

[Jill BioSkop]

http://www.vigoschools.net/~mmc3/c1...MT & Gases/L1 - Properties of Gases & KMT.pdf

The above might help. It has standard pressure in a variety of units (page 4, second row). As for the tenets, I don't know what level you're studying this at, but could it be the three states a chemical can be found in (liquid, solid or gaseous)?

 

[ashitaria]

I'm sorry proxy, but try as I might, I cannot make sense of your equations. Would you mine going more into depth?

And though I know the tenets of KMT, I do not know the three tenets of KMT...

 

[ProxyAmenRa]

change in heat energy of an object = q

q is proportional to the change in temperature

q= mass * heat capacity * delta T

mass = m
heat capacity = c
delta T= dt = t2 - t1

heat capacity of water = 4.184 J/K/g
q=m*c*dt
q=m*c*(t2-t1)

t2 = final temp
t1 = initial temp

The energy required for change in state is proportional to the mass of the object.

ql = energy required to change state
lf = latent heat of fusion kJ/kg or J/g
latent heat of fusion of water = 334 J/g

ql=lf*m

Solving in multiple Stages

The first stage is the exothermic release of energy from the hot water and the endothermic intake of the ice to change state.

heat loss = heat gain
mass of hot water * specific heat of water * temperature change of water = latent heat of fusion of water * mass of water

Here is a slight point of confusion exothermic reacts always result in a negative (q) which means energy is leaving the object. Therefore to remedy this we swamp the t2 and t1 or we could solve the problem as a systems process.

Here is an example; how much heat is required to be transfered to alter the temperature of 30g of water from 100 degrees celcius to 10 degrees celcius.

t2 = 10 degrees celcius
t1 = 100 degrees celcius

q=m*c*dt
q=m*c*(t2-t1)
q=30*4.18*(10-100)
q=30*4.18*(-90)
q= -11286 Joules

The negative sign means that the transference of energy is exothermic. Hence, from the object to the surroundings.

Swapping t2 and t1. Hence dt = t1 - t2 (only for heat loss)

m*c*(t1-t2) = m*lf
30*4.18*(100 - t2)= 30 * 334

We are finding t2.

100-t2=(30*334)/(30*4.18)
-t2= ((30*334)/(4.18*30)) - 100
-t2 = -20.09
t = 20.09 degrees Celsius <-- heat loss due to melting ice

Systems Approach

Q = is heat entering of leaving system which equals processes in the system.

Since the system is isolated Q=0.

Q= m*c*(t2-100) + lf*m
0=30*4.18*(t2-100) + 334*30

Divide both sides by (30*4.18)

0=t2-100 + 334/4.18
-t2=-100+334/4.18
-t2=-20.095
t2=20.095

I prefer the systems approach and hence I will use it in the next part.

Finding Final Temperature

Here we are going to find out what will be the final temperature of the water.

Temperature of water body 1 = 0 degrees Celsius
Mass of water body 1 = 30 grams
Temperature of water body 2 = 20.095 degrees Celsius
Mass water body 2 = 30 grams

Q=0 because it is an isolated system

0=m1*c1*(t2-t1)+m2*c1*(t2-t1)
0=30*4.18*(t2-20.095)+30*4.18*(t2-0)

Divide both sides by (30*4.18). Remember 0 divided by anything is 0.

0=t2-20.095 + t2
-2*t2=-20.095
t2 = 10.0475 degrees Celsius

Previous Solving

I previously solved everything in one step.

Q=0
0=m1*c*(t2-t1)+m2*c*(t2-t1)+lf*m2
0 = m1*c1*(t2-100) + m1*c1*(t2-0) + 334*30
-334*30 = m1*c1*(t2-100) + m1c1*(t2)
-334/4.18 = 2*t2 - 100
100 - 334/4.18 = t2
t2 = 10.0475 celcius or 283.1975 K

 

[ashitaria]

change in heat energy of an object = q

q is proportional to the change in temperature

q= mass * heat capacity * delta T

mass = m
heat capacity = c
delta T= dt = t2 - t1

heat capacity of water = 4.184 J/K/g
q=m*c*dt
q=m*c*(t2-t1)

t2 = final temp
t1 = initial temp

The energy required for change in state is proportional to the mass of the object.

ql = energy required to change state
lf = latent heat of fusion kJ/kg or J/g
latent heat of fusion of water = 334 J/g

ql=lf*m

Solving in multiple Stages

The first stage is the exothermic release of energy from the hot water and the endothermic intake of the ice to change state.

heat loss = heat gain
mass of hot water * specific heat of water * temperature change of water = latent heat of fusion of water * mass of water

Here is a slight point of confusion exothermic reacts always result in a negative (q) which means energy is leaving the object. Therefore to remedy this we swamp the t2 and t1 or we could solve the problem as a systems process.

Here is an example; how much heat is required to be transfered to alter the temperature of 30g of water from 100 degrees celcius to 10 degrees celcius.

t2 = 10 degrees celcius
t1 = 100 degrees celcius

q=m*c*dt
q=m*c*(t2-t1)
q=30*4.18*(10-100)
q=30*4.18*(-90)
q= -11286 Joules

The negative sign means that the transference of energy is exothermic. Hence, from the object to the surroundings.

Swapping t2 and t1. Hence dt = t1 - t2 (only for heat loss)

m*c*(t1-t2) = m*lf
30*4.18*(100 - t2)= 30 * 334

We are finding t2.

100-t2=(30*334)/(30*4.18)
-t2= ((30*334)/(4.18*30)) - 100
-t2 = -20.09
t = 20.09 degrees Celsius <-- heat loss due to melting ice

Systems Approach

Q = is heat entering of leaving system which equals processes in the system.

Since the system is isolated Q=0.

Q= m*c*(t2-100) + lf*m
0=30*4.18*(t2-100) + 334*30

Divide both sides by (30*4.18)

0=t2-100 + 334/4.18
-t2=-100+334/4.18
-t2=-20.095
t2=20.095

I prefer the systems approach and hence I will use it in the next part.

Finding Final Temperature

Here we are going to find out what will be the final temperature of the water.

Temperature of water body 1 = 0 degrees Celsius
Mass of water body 1 = 30 grams
Temperature of water body 2 = 20.095 degrees Celsius
Mass water body 2 = 30 grams

Q=0 because it is an isolated system

0=m1*c1*(t2-t1)+m2*c1*(t2-t1)
0=30*4.18*(t2-20.095)+30*4.18*(t2-0)

Divide both sides by (30*4.18). Remember 0 divided by anything is 0.

0=t2-20.095 + t2
-2*t2=-20.095
t2 = 10.0475 degrees Celsius

Previous Solving

I previously solved everything in one step.

Q=0
0=m1*c*(t2-t1)+m2*c*(t2-t1)+lf*m2
0 = m1*c1*(t2-100) + m1*c1*(t2-0) + 334*30
-334*30 = m1*c1*(t2-100) + m1c1*(t2)
-334/4.18 = 2*t2 - 100
100 - 334/4.18 = t2
t2 = 10.0475 celcius or 283.1975 K

Yes, that makes a whole lot more sense now. I completely forgot that c= specific heat XP.

Now that I know it does, I get it, but wow, I think that the level of chemistry that you presented is way too different than the one my teacher taught. Oh well, I'll figure out how to modify it somehow...

 

[Melkor]

When I was a childer' I did my own homework.

Afore' we had this internet thing.

Oh wait...we had internet... I was just clever. (Deprived)

 

[ProxyAmenRa]

I recently designed an industrial distiller. I will check out what methods of calculation I used.

 

[Sparrow]

Umm. There's google...there's wikipedia...there's yahoo answers...wikianswers...

It would be much more efficient and time-saving if you'd use the resources available to you :P

 

[Agent Intellect]

Umm. There's google...there's wikipedia...there's yahoo answers...wikianswers...

It would be much more efficient and time-saving if you'd use the resources available to you :P

Yeah, it's an enormous burden on the people of this forum to have a thread like this existing - they might accidentally click it, read the first line of the OP and decide it's nothing that interests them. I guess it's a better use of time to argue about whether Ti is subjective or objective, though, than to actually help people with real questions about science.

Proxy pretty much said everything I would have said (I just took a chem course studying this same kind of stuff) so there isn't much I can say here. Good luck with chemistry, though, I found it a very interesting and challenging course.

 

[ashitaria]

The only reason why I decided to post here was because I couldn't exactly find the stuff I need on yahoo, wikipedia or google.
They probably did mention something like this, but it wasn't specified clearly enough and/or I wasn't sure if it was factually accurate.