q1 = heat to move T ice at -19 C to zero C. The heat comes from the water at 85 C.
q1 = mass x specific heat ice x [Tfinal-Tinitial) =
24.0 x 2.08 x [0-(-19)]
q2 = heat to melt the ice. Heat comes from the water at 85 C.
q2 = mass ice x heat fusion ice =
q2 = 24.0 x 6.02 kJ/mol x 18.015 g/mol = (the 18.015 is the molar mass of water and that conversion is done because heat of fusion is listed in kJ/MOL but q1 specific heat was listed in J/GRAM.
q3. So now we have water at zero mixing with water at something less than 85 (some of the heat has been removed to move Ice from -19 and to melt it).
q3 = mass ice x specific heat water (the ice is now a liquid) x (Tfinal-Tinitial) = 24.0 x 4.18 x (Tfinal - 0) =
q4 = heat from water at 85 (it will be losing heat) to the final T.
mass water x specific heat water x (Tfinal-Tinitial) =
118.0 x 4.18 x (Tfinal-85).
Now just sum all of these and the sum must equal to zero; that is, heat gained by ice = heat lost by water.
q1 + q2 + q3 + q4 = 0.
Solve for Tf = final T. Check my work but I obtained between 60 and 70 C for the final T. I shall be happy to check your work if you post it.
If you are having a test Monday, there are two or three things to remember that will work all of these problems.
1. heat lost + heat gained = 0
2. mass x specific heat x (Tfinal-Tinitial) = q (everything is in the same phase).
3. phase change (solid to liquid or liquid to a gas).
mass x heat fusion for solid to liquid.
mass x heat of vaporization for liquid to gas.
Those three concepts will work 99.9% of the problems you are likely to have.
A 24.0 g sample of ice at -19.0°C is mixed with 118.0 g of water at 85.0°C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of H2O(s) and H2O(l) are 2.08 and 4.18 J/g°C, respectively, and the enthalpy of fusion for ice is 6.02 kJ/mol.
i really need help doing this. what formulas do i use? we didn't even learn this and we are getting tested on monday =(
6 answers
wow, thank you.
when do you use the equation -qhot=qcold?
I don't really understand fully q3 and q4
for q3 why do u have to use 24g ice but u use the Cp of water and not ice.
and for q4 and we used another water equation... why do we have to do this twice?
thanks so much!
when do you use the equation -qhot=qcold?
I don't really understand fully q3 and q4
for q3 why do u have to use 24g ice but u use the Cp of water and not ice.
and for q4 and we used another water equation... why do we have to do this twice?
thanks so much!
when do you use the equation -qhot=qcold?
-qhot=qcold is a rearrangement of
heat gained (the cold in your equation) + heat lost (the hot in your equation) = 0 and that one, the one I listed as #1 above, is the one I prefer. It takes care of the signs for me and I don't need to figure out which one is losing heat and which is gaining heat.
I don't really understand fully q3 and q4
for q3 why do u have to use 24g ice but u use the Cp of water and not ice.
You use 24.0 because that's the amount of ice in the problem and you use Cp for WATER because the ice has melted and it is now a liquid so the q3 part of the problem is to find the heat absorbed by the zero degree melted ice on its way to the final T, whatever that may be.
and for q4 and we used another water equation... why do we have to do this twice?
You must understand that we are dealing with two different sources of water. The first one, from q3, is the 24.0 g ice that has melted at zero degrees C and is absorbing heat to some final temperature, designated as Tfinal. The other source of water (you know we don't have two kinds of water, just water from two different places) is from the 118.0 grams of water that was initially at 85 degrees. It has been the "heat pump" if you will, that has supplied heat to move the ice from -19 to zero, to melt the ice at zero, and to move the zero degree water formed to Tfinal.
-qhot=qcold is a rearrangement of
heat gained (the cold in your equation) + heat lost (the hot in your equation) = 0 and that one, the one I listed as #1 above, is the one I prefer. It takes care of the signs for me and I don't need to figure out which one is losing heat and which is gaining heat.
I don't really understand fully q3 and q4
for q3 why do u have to use 24g ice but u use the Cp of water and not ice.
You use 24.0 because that's the amount of ice in the problem and you use Cp for WATER because the ice has melted and it is now a liquid so the q3 part of the problem is to find the heat absorbed by the zero degree melted ice on its way to the final T, whatever that may be.
and for q4 and we used another water equation... why do we have to do this twice?
You must understand that we are dealing with two different sources of water. The first one, from q3, is the 24.0 g ice that has melted at zero degrees C and is absorbing heat to some final temperature, designated as Tfinal. The other source of water (you know we don't have two kinds of water, just water from two different places) is from the 118.0 grams of water that was initially at 85 degrees. It has been the "heat pump" if you will, that has supplied heat to move the ice from -19 to zero, to melt the ice at zero, and to move the zero degree water formed to Tfinal.
Please note that I made an error in the above. The heat of fusion is 6.02 kJ/mol. To change to J/g we must do this.
6.02 kJ/mol x (1 mol/18.015 g) x (1000 J/kJ) = 334 J/g. Sorry about that.
6.02 kJ/mol x (1 mol/18.015 g) x (1000 J/kJ) = 334 J/g. Sorry about that.
So the answer may NOT be between 60 and 70.
mct+DfM+mct=0
1 answer