I will give a lot of points for this since I know it is a lot of work. But pleas
ID: 842317 • Letter: I
Question
I will give a lot of points for this since I know it is a lot of work. But please show me how to do all the problems I listed on here so I can learn. This is practice for my final and I need help learning. Thanks!
So I did a lab called the enthalpy of ice.
Here is my data:
Mass of empty stacked cups for insulation=5.27 grams
Mass of water in the stacked cups=50 grams
total mass of cups, water and ice=64.47 grams
mass of ice and water=59.2 grams
mass of ice grams
Okay that was in my first table.
The second table is about temperatures.
So i do not know what to put in the other two parts of the table...I think it would be the same data. IDK.
3. Next question says use the equation- Q=mcw(triangle)T and the data in your tables to calculate the gain or loss of heat in the water. Repeat this for the melted ice. Remember that Cw is the specific heat of water (4.17 Joules/g degrees celcius) Use this table below for data.
4. Use equation Qwater+Qf+Qmelted ice=zero and the data in the table below in rows 1 and 2 to calculate teh heat energy that melted the ice. Enter this value into the table.
3rd table
Melted ice warming(Q melted ice)
5. Use equation Qf=mLf and the data in your table to calculate the heat of fusion of ice.
6. The accepted value for enthalpy of fusion of ice is 334 J/g. Calculate your error.
7. (triangle)H is normally given in units of kJ/mole. Use your answer to question 5 and grams/mole of water to find the experimental value (triangle)H for the fusion of ice.
Thank you sooooo much in advance for helping me! Ill give lots of points!!!! Please help. Thanks!
Temp. Initial, degrees celcius Final degrees celcius Change in temperature Water 45 degrees celcius 23 degrees celcius 22 degrees celcius Melted ice 0 degrees celciusExplanation / Answer
Mass of empty stacked cups for insulation=5.27 grams
Mass of water in the stacked cups=50 grams
total mass of cups, water and ice=64.47 grams
mass of ice and water=59.2 grams
mass of ice grams
Okay that was in my first table.
The second table is about temperatures.
23 degree celcius
Since mass of ice water is very less than that of normal water, thus it will have same final temperature as that of water initially at 45 degree celcius
So i do not know what to put in the other two parts of the table...I think it would be the same data. IDK.
3. Next question says use the equation- Q=mcw(triangle)T and the data in your tables to calculate the gain or loss of heat in the water. Repeat this for the melted ice. Remember that Cw is the specific heat of water (4.17 Joules/g degrees celcius) Use this table below for data.
Loss of heat of water = mcw*change in temperature = 50*4.17*22 = 4587 Joules
Gain of heat for ice water = 9.2*23*4.17 = 882.37 Joules
4. Use equation Qwater+Qf+Qmelted ice=zero and the data in the table below in rows 1 and 2 to calculate teh heat energy that melted the ice. Enter this value into the table.
3rd table
Melted ice warming(Q melted ice)
5. Use equation Qf=mLf and the data in your table to calculate the heat of fusion of ice.
Lf = Qf/m = 3704.63 /9.2 = 402.67 Joules/g
6. The accepted value for enthalpy of fusion of ice is 334 J/g. Calculate your error.
Error = 100*(402.67-334)/334 =20.55 percent
7. (triangle)H is normally given in units of kJ/mole. Use your answer to question 5 and grams/mole of water to find the experimental value (triangle)H for the fusion of ice.
(triangle)H = 402.67 Joules/g *(1000/1000) = 402.65 kJ/(1000*g) = 402.65 kJ/(1000/18)mol = 402.65*18
= 7.2477 kJ/mol
Temp. Initial, degrees celcius Final degrees celcius Change in temperature Water 45 degrees celcius 23 degrees celcius 22 degrees celcius Melted ice 0 degrees celcius23 degree celcius
Since mass of ice water is very less than that of normal water, thus it will have same final temperature as that of water initially at 45 degree celcius
23 degree celcius